Solution

The anti derivative of $\sin 2 x$ is a function of $x$ whose derivative is $\sin 2 x$. It is known that,

$\frac{d}{d x}(\cos 2 x)=-2 \sin 2 x$

$\Rightarrow \sin 2 x=-\frac{1}{2} \frac{d}{d x}(\cos 2 x)$

$\therefore \sin 2 x=\frac{d}{d x}(-\frac{1}{2} \cos 2 x)$

Therefore, the anti derivative of $\sin 2 x$ is $-\frac{1}{2} \cos 2 x$

Solution

The anti derivative of $\cos 3 x$ is a function of $x$ whose derivative is $\cos 3 x$.

It is known that,

$\frac{d}{d x}(\sin 3 x)=3 \cos 3 x$

$\Rightarrow \cos 3 x=\frac{1}{3} \frac{d}{d x}(\sin 3 x)$

$\therefore \cos 3 x=\frac{d}{d x}(\frac{1}{3} \sin 3 x)$

Therefore, the anti derivative of $\cos 3 x$ is $\frac{1}{3} \sin 3 x$.

Solution

The anti derivative of $e^{2 x}$ is the function of $x$ whose derivative is $e^{2 x}$.

It is known that, $\frac{d}{d x}(e^{2 x})=2 e^{2 x}$

$\Rightarrow e^{2 x}=\frac{1}{2} \frac{d}{d x}(e^{2 x})$

$\therefore e^{2 x}=\frac{d}{d x}(\frac{1}{2} e^{2 x})$

Therefore, the anti derivative of $e^{2 x}$ is $\frac{1}{2} e^{2 x}$.

Solution

The anti derivative of $(a x+b)^{2}$ is the function of $x$ whose derivative is $(a x+b)^{2}$.

It is known that,

$\frac{d}{d x}(a x+b)^{3}=3 a(a x+b)^{2}$

$\Rightarrow(a x+b)^{2}=\frac{1}{3 a} \frac{d}{d x}(a x+b)^{3}$

$\therefore(a x+b)^{2}=\frac{d}{d x}(\frac{1}{3 a}(a x+b)^{3})$

Therefore, the anti derivative of $(a x+b)^{2}$ is $\frac{1}{3 a}(a x+b)^{3}$.

Solution

The anti derivative of $(\sin 2 x-4 e^{3 x})$ is the function of $x$ whose derivative is $(\sin 2 x-4 e^{3 x})$.

It is known that, $\frac{d}{d x}(-\frac{1}{2} \cos 2 x-\frac{4}{3} e^{3 x})=\sin 2 x-4 e^{3 x}$

Therefore, the anti derivative of $(\sin 2 x-4 e^{3 x})$ is $(-\frac{1}{2} \cos 2 x-\frac{4}{3} e^{3 x})$.

Solution

$ \begin{aligned} & \int(4 e^{3 x}+1) d x \\ & =4 \int e^{3 x} d x+\int 1 d x \\ & =4(\frac{e^{3 x}}{3})+x+C \\ & =\frac{4}{3} e^{3 x}+x+C \end{aligned} $

Solution

$\int x^{2}(1-\frac{1}{x^{2}}) d x$

$=\int(x^{2}-1) d x$

$=\int x^{2} d x-\int 1 d x$

$=\frac{x^{3}}{3}-x+C$

Solution

$\int(a x^{2}+b x+c) d x$

$=a \int x^{2} d x+b \int x d x+c \int 1 \cdot d x$

$=a(\frac{x^{3}}{3})+b(\frac{x^{2}}{2})+c x+C$

$=\frac{a x^{3}}{3}+\frac{b x^{2}}{2}+c x+C$

Solution

$\int(2 x^{2}+e^{x}) d x$

$=2 \int x^{2} d x+\int e^{x} d x$

$=2(\frac{x^{3}}{3})+e^{x}+C$

$=\frac{2}{3} x^{3}+e^{x}+C$

Solution

$ \begin{aligned} & \int(\sqrt{x}-\frac{1}{\sqrt{x}})^{2} d x \\ & =\int(x+\frac{1}{x}-2) d x \\ & =\int x d x+\int \frac{1}{x} d x-2 \int 1 \cdot d x \\ & =\frac{x^{2}}{2}+\log |x|-2 x+C \end{aligned} $

Solution

$ \begin{aligned} & \int \frac{x^{3}+5 x^{2}-4}{x^{2}} d x \\ & =\int(x+5-4 x^{-2}) d x \\ & =\int x d x+5 \int 1 \cdot d x-4 \int x^{-2} d x \\ & =\frac{x^{2}}{2}+5 x-4(\frac{x^{-1}}{-1})+C \\ & =\frac{x^{2}}{2}+5 x+\frac{4}{x}+C \end{aligned} $

Solution

$ \begin{aligned} & \int \frac{x^{3}+3 x+4}{\sqrt{x}} d x \\ & =\int(x^{\frac{5}{2}}+3 x^{\frac{1}{2}}+4 x^{-\frac{1}{2}}) d x \\ & =\frac{x^{\frac{7}{2}}}{\frac{7}{2}}+\frac{3(x^{\frac{3}{2}})}{\frac{3}{2}}+\frac{4(x^{\frac{1}{2}})}{\frac{1}{2}}+C \\ & =\frac{2}{7} x^{\frac{7}{2}}+2 x^{\frac{3}{2}}+8 x^{\frac{1}{2}}+C \\ & =\frac{2}{7} x^{\frac{7}{2}}+2 x^{\frac{3}{2}}+8 \sqrt{x}+C \end{aligned} $

Solution

$\int \frac{x^{3}-x^{2}+x-1}{x-1} d x$

On dividing, we obtain

$=\int(x^{2}+1) d x$

$=\int x^{2} d x+\int 1 d x$

$=\frac{x^{3}}{3}+x+C$

Solution

$\int(1-x) \sqrt{x} d x$

$=\int(\sqrt{x}-x^{\frac{3}{2}}) d x$

$=\int x^{\frac{1}{2}} d x-\int x^{\frac{3}{2}} d x$

$=\frac{x^{\frac{3}{2}}}{3}-\frac{x^{\frac{5}{2}}}{5}+C$

$=\frac{2}{3} x^{\frac{3}{2}}-\frac{2}{5} x^{\frac{5}{2}}+C$

Solution

$ \int \sqrt{x}(3 x^{2}+2 x+3) d x $

$=\int(3 x^{\frac{5}{2}}+2 x^{\frac{3}{2}}+3 x^{\frac{1}{2}}) d x$

$=3 \int x^{\frac{5}{2}} d x+2 \int x^{\frac{3}{2}} d x+3 \int x^{\frac{1}{2}} d x$

$=3(\frac{x^{\frac{7}{2}}}{\frac{7}{2}})+2(\frac{x^{\frac{5}{2}}}{\frac{5}{2}})+3 \frac{(x^{\frac{3}{2}})}{\frac{3}{2}}+C$

$=\frac{6}{7} x^{\frac{7}{2}}+\frac{4}{5} x^{\frac{5}{2}}+2 x^{\frac{3}{2}}+C$

Solution

$\int(2 x-3 \cos x+e^{x}) d x$

$=2 \int x d x-3 \int \cos x d x+\int e^{x} d x$

$=\frac{2 x^{2}}{2}-3(\sin x)+e^{x}+C$

$=x^{2}-3 \sin x+e^{x}+C$

Solution

$\int(2 x^{2}-3 \sin x+5 \sqrt{x}) d x$ $=2 \int x^{2} d x-3 \int \sin x d x+5 \int x^{\frac{1}{2}} d x$

$=\frac{2 x^{3}}{3}-3(-\cos x)+5(\frac{x^{\frac{3}{2}}}{\frac{3}{2}})+C$

$=\frac{2}{3} x^{3}+3 \cos x+\frac{10}{3} x^{\frac{3}{2}}+C$

Solution

$\int \sec x(\sec x+\tan x) d x$

$=\int(\sec ^{2} x+\sec x \tan x) d x$

$=\int \sec ^{2} x d x+\int \sec x \tan x d x$

$=\tan x+\sec x+C$

Solution

$\int \frac{\sec ^{2} x}{cosec 2} d x$

$ \begin{aligned} & =\int \frac{\frac{1}{\cos ^{2} x}}{\frac{1}{\sin ^{2} x}} d x \\ & =\int \frac{\sin ^{2} x}{\cos ^{2} x} d x \\ & =\int \tan ^{2} x d x \\ & =\int(\sec ^{2} x-1) d x \\ & =\int \sec ^{2} x d x-\int 1 d x \\ & =\tan x-x+C \end{aligned} $

Solution

$\int \frac{2-3 \sin x}{\cos ^{2} x} d x$

$=\int(\frac{2}{\cos ^{2} x}-\frac{3 \sin x}{\cos ^{2} x}) d x$

$=\int 2 \sec ^{2} x d x-3 \int \tan x \sec x d x$

$=2 \tan x-3 \sec x+C$

Solution

$ (\sqrt{x}+\frac{1}{\sqrt{x}}) d x $

$=\int x^{\frac{1}{2}} d x+\int x^{-\frac{1}{2}} d x$

$=\frac{x^{\frac{3}{2}}}{3}+\frac{x^{\frac{1}{2}}}{1}+C$

2

$=\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+C$

Hence, the correct Answer is C.

Solution

It is given that,

$\frac{d}{d x} f(x)=4 x^{3}-\frac{3}{x^{4}}$

$\therefore$ Anti derivative of $4 x^{3}-\frac{3}{x^{4}}=f(x)$ $\therefore f(x)=\int 4 x^{3}-\frac{3}{x^{4}} d x$

$f(x)=4 \int x^{3} d x-3 \int(x^{-4}) d x$

$f(x)=4(\frac{x^{4}}{4})-3(\frac{x^{-3}}{-3})+C$

$\therefore(x)=x^{4}+\frac{1}{x^{3}}+C$

Also,

$f(2)=0$

$\therefore f(2)=(2)^{4}+\frac{1}{(2)^{3}}+C=0$

$\Rightarrow 16+\frac{1}{8}+C=0$

$\Rightarrow C=-(16+\frac{1}{8})$

$\Rightarrow C=\frac{-129}{8}$

$\therefore f(x)=x^{4}+\frac{1}{x^{3}}-\frac{129}{8}$

Hence, the correct Answer is A.

Solution

Let $1+x^{2}=t$

$\therefore 2 x d x=d t$

$\Rightarrow \int \frac{2 x}{1+x^{2}} d x=\int \frac{1}{t} d t$

$=\log |t|+C$

$=\log |1+x^{2}|+C$

$=\log (1+x^{2})+C$

Solution

Let $\log |x|=t$

$\therefore \frac{1}{x} d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{(\log |x|)^{2}}{x} d x & =\int t^{2} d t \\ & =\frac{t^{3}}{3}+C \\ & =\frac{(\log |x|)^{3}}{3}+C \end{aligned} $

Solution

$\frac{1}{x+x \log x}=\frac{1}{x(1+\log x)}$

Let $1+\log x=t$

$\therefore \frac{1}{x} d x=d t$

$\Rightarrow \int \frac{1}{x(1+\log x)} d x=\int \frac{1}{t} d t$

$=\log |t|+C$

$=\log |1+\log x|+C$

Solution

$\sin x \cdot \sin (\cos x)$

Let $\cos x=t$

$\therefore-\sin x d x=d t$

$ \begin{aligned} \Rightarrow \int \sin x \cdot \sin (\cos x) d x & =-\int \sin t d t \\ & =-[-\cos t]+C \\ & =\cos t+C \\ & =\cos (\cos x)+C \end{aligned} $

Solution

$\sin (a x+b) \cos (a x+b)=\frac{2 \sin (a x+b) \cos (a x+b)}{2}=\frac{\sin 2(a x+b)}{2}$

Let $2(a x+b)=t$

$\therefore 2 a d x=d t$ $\Rightarrow \int \frac{\sin 2(a x+b)}{2} d x=\frac{1}{2} \int \frac{\sin t d t}{2 a}$

$ \begin{aligned} & =\frac{1}{4 a}[-\cos t]+C \\ & =\frac{-1}{4 a} \cos 2(a x+b)+C \end{aligned} $

Solution

Let $a x+b=t$

$\Rightarrow a d x=d t$

$\therefore d x=\frac{1}{a} d t$

$\Rightarrow \int(a x+b)^{\frac{1}{2}} d x=\frac{1}{a} \int t^{\frac{1}{2}} d t$

$=\frac{1}{a}(\frac{t^{\frac{3}{2}}}{\frac{3}{2}})+C$

$=\frac{2}{3 a}(a x+b)^{\frac{3}{2}}+C$

Solution

Let $(x+2)=t$ $\therefore d x=d t$

$ \begin{aligned} \Rightarrow \int x \sqrt{x+2} d x & =\int(t-2) \sqrt{t} d t \\ & =\int(t^{\frac{3}{2}}-2 t^{\frac{1}{2}}) d t \\ & =\int t^{\frac{3}{2}} d t-2 \int t^{\frac{1}{2}} d t \\ & =\frac{t^{\frac{5}{2}}}{\frac{5}{2}}-2(\frac{t^{\frac{3}{2}}}{\frac{3}{2}})+C \\ & =\frac{2}{5} t^{\frac{5}{2}}-\frac{4}{3} t^{\frac{3}{2}}+C \\ & =\frac{2}{5}(x+2)^{\frac{5}{2}}-\frac{4}{3}(x+2)^{\frac{3}{2}}+C \end{aligned} $

Solution

Let $1+2 x^{2}=t$

$\therefore 4 x d x=d t$ $\Rightarrow \int x \sqrt{1+2 x^{2}} d x=\int \frac{\sqrt{t} d t}{4}$

$=\frac{1}{4} \int t^{\frac{1}{2}} d t$

$=\frac{1}{4}(\frac{t^{\frac{3}{2}}}{\frac{3}{2}})+C$

$=\frac{1}{6}(1+2 x^{2})^{\frac{3}{2}}+C$

Solution

Let $x^{2}+x+1=t$

$\therefore(2 x+1) d x=d t$

$\int(4 x+2) \sqrt{x^{2}+x+1} d x$

$=\int 2 \sqrt{t} d t$

$=2 \int \sqrt{t} d t$

$=2(\frac{t^{\frac{3}{2}}}{\frac{3}{2}})+C$

$=\frac{4}{3}(x^{2}+x+1)^{\frac{3}{2}}+C$

Solution

$\frac{1}{x-\sqrt{x}}=\frac{1}{\sqrt{x}(\sqrt{x}-1)}$

Let $(\sqrt{x}-1)=t$

$\therefore \frac{1}{2 \sqrt{x}} d x=d t$

$\Rightarrow \int \frac{1}{\sqrt{x}(\sqrt{x}-1)} d x=\int \frac{2}{t} d t$

$=2 \log |t|+C$

$=2 \log |\sqrt{x}-1|+C$

Solution

Let $x+4=t$

$\therefore d x=d t$

$ \begin{aligned} \int \frac{x}{\sqrt{x+4}} d x & =\int \frac{(t-4)}{\sqrt{t}} d t \\ & =\int(\sqrt{t}-\frac{4}{\sqrt{t}}) d t \\ & =\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-4(\frac{t^{\frac{1}{2}}}{\frac{1}{2}})+C \\ & =\frac{2}{3}(t)^{\frac{3}{2}}-8(t)^{\frac{1}{2}}+C \\ & =\frac{2}{3} t \cdot t^{\frac{1}{2}}-8 t^{\frac{1}{2}}+C \\ & =\frac{2}{3} t^{\frac{1}{2}}(t-12)+C \\ & =\frac{2}{3}(x+4)^{\frac{1}{2}}(x+4-12)+C \\ & =\frac{2}{3} \sqrt{x+4}(x-8)+C \end{aligned} $

Solution

Let $x^{3}-1=t$

$\therefore 3 x^{2} d x=d t$ $\Rightarrow \int(x^{3}-1)^{\frac{1}{3}} x^{5} d x=\int(x^{3}-1)^{\frac{1}{3}} x^{3} \cdot x^{2} d x$

$=\int t^{\frac{1}{3}}(t+1) \frac{d t}{3}$

$=\frac{1}{3} \int(t^{\frac{4}{3}}+t^{\frac{1}{3}}) d t$

$=\frac{1}{3}[\frac{t^{\frac{7}{3}}}{\frac{7}{3}}+\frac{t^{\frac{4}{3}}}{\frac{4}{3}}]+C$

$=\frac{1}{3}[\frac{3}{7} t^{\frac{7}{3}}+\frac{3}{4} t^{\frac{4}{3}}]+C$

$=\frac{1}{7}(x^{3}-1)^{\frac{7}{3}}+\frac{1}{4}(x^{3}-1)^{\frac{4}{3}}+C$

Solution

Let $2+3 x^{3}=t$

$\therefore 9 x^{2} d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{x^{2}}{(2+3 x^{3})^{3}} d x & =\frac{1}{9} \int \frac{d t}{(t)^{3}} \\ & =\frac{1}{9}[\frac{t^{-2}}{-2}]+C \\ & =\frac{-1}{18}(\frac{1}{t^{2}})+C \\ & =\frac{-1}{18(2+3 x^{3})^{2}}+C \end{aligned} $

Solution

Let $\log x=t$

$\therefore \frac{1}{x} d x=d t$

$\Rightarrow \int \frac{1}{x(\log x)^{m}} d x=\int \frac{d t}{(t)^{m}}$

$=(\frac{t^{-m+1}}{1-m})+C$

$=\frac{(\log x)^{1-m}}{(1-m)}+C$

Solution

Let $9-4 x^{2}=t$

$\therefore-8 x d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{x}{9-4 x^{2}} d x & =\frac{-1}{8} \int_t^{1} d t \\ & =\frac{-1}{8} \log |t|+C \\ & =\frac{-1}{8} \log |9-4 x^{2}|+C \end{aligned} $

Solution

Let $2 x+3=t$

$\therefore 2 d x=d t$

$ \begin{aligned} \Rightarrow \int e^{2 x+3} d x & =\frac{1}{2} \int e^{t} d t \\ & =\frac{1}{2}(e^{t})+C \\ & =\frac{1}{2} e^{(2 x+3)}+C \end{aligned} $

Solution

Let $x^{2}=t$

$\therefore 2 x d x=d t$

$\Rightarrow \int \frac{x}{e^{x^{2}}} d x=\frac{1}{2} \int \frac{1}{e^{t}} d t$

$=\frac{1}{2} \int e^{-t} d t$

$=\frac{1}{2}(\frac{e^{-t}}{-1})+C$

$=-\frac{1}{2} e^{-x^{2}}+C$

$=\frac{-1}{2 e^{x^{2}}}+C$

Solution

Let $\tan ^{-1} x=t$

$\therefore \frac{1}{1+x^{2}} d x=d t$ $\Rightarrow \int \frac{e^{\tan ^{-1} x}}{1+x^{2}} d x=\int e^{t} d t$

$=e^{t}+C$

$=e^{\tan ^{-1} x}+C$

Solution

$\frac{e^{2 x}-1}{e^{2 x}+1}$

Dividing numerator and denominator by $e^{x}$, we obtain

$ \frac{\frac{(e^{2 x}-1)}{e^{x}}}{(e^{2 x}+1)}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} $

Let $e^{x}+e^{-x}=t$

$\therefore(e^{x}-e^{-x}) d x=d t$

$\Rightarrow \int \frac{e^{2 x}-1}{e^{2 x}+1} d x=\int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} d x$

$=\int \frac{d t}{t}$

$=\log |t|+C$

$=\log |e^{x}+e^{-x}|+C$

Solution

Let $e^{2 x}+e^{-2 x}=t$

$\therefore(2 e^{2 x}-2 e^{-2 x}) d x=d t$

$\Rightarrow 2(e^{2 x}-e^{-2 x}) d x=d t$

$\Rightarrow \int(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}) d x=\int \frac{d t}{2 t}$

$=\frac{1}{2} \int_t^{1} d t$

$=\frac{1}{2} \log |t|+C$

$=\frac{1}{2} \log |e^{2 x}+e^{-2 x}|+C$

Solution

$\tan ^{2}(2 x-3)=\sec ^{2}(2 x-3)-1$

Let $2 x-3=t$

$\therefore 2 d x=d t$

$ \begin{aligned} \Rightarrow \int \tan ^{2}(2 x-3) d x & =\int[(\sec ^{2}(2 x-3))-1] d x \\ & =\frac{1}{2} \int(\sec ^{2} t) d t-\int 1 d x \\ & =\frac{1}{2} \int \sec ^{2} t d t-\int 1 d x \\ & =\frac{1}{2} \tan t-x+C \\ & =\frac{1}{2} \tan (2 x-3)-x+C \end{aligned} $

Solution

Let $7-4 x=t$

$\therefore-4 d x=d t$

$ \begin{aligned} \therefore \int \sec ^{2}(7-4 x) d x & =\frac{-1}{4} \int \sec ^{2} t d t \\ & =\frac{-1}{4}(\tan t)+C \\ & =\frac{-1}{4} \tan (7-4 x)+C \end{aligned} $

Solution

Let $\sin ^{-1} x=t$ $\therefore \frac{1}{\sqrt{1-x^{2}}} d x=d t$

$\Rightarrow \int \frac{\sin ^{-1} x}{\sqrt{1-x^{2}}} d x=\int t d t$

$=\frac{t^{2}}{2}+C$

$=\frac{(\sin ^{-1} x)^{2}}{2}+C$

Solution

$\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}=\frac{2 \cos x-3 \sin x}{2(3 \cos x+2 \sin x)}$

Let $3 \cos x+2 \sin x=t$

$\therefore(-3 \sin x+2 \cos x) d x=d t$

$ \begin{aligned} \int \frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x} d x & =\int \frac{d t}{2 t} \\ & =\frac{1}{2} \int \frac{1}{t} d t \\ & =\frac{1}{2} \log |t|+C \\ & =\frac{1}{2} \log |2 \sin x+3 \cos x|+C \end{aligned} $

Solution

$\frac{1}{\cos ^{2} x(1-\tan x)^{2}}=\frac{\sec ^{2} x}{(1-\tan x)^{2}}$

Let $(1-\tan x)=t$

$\therefore-\sec ^{2} x d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{\sec ^{2} x}{(1-\tan x)^{2}} d x & =\int \frac{-d t}{t^{2}} \\ & =-\int t^{-2} d t \\ & =+\frac{1}{t}+C \\ & =\frac{1}{(1-\tan x)}+C \end{aligned} $

Solution

Let $\sqrt{x}=t$

$\therefore \frac{1}{2 \sqrt{x}} d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{\cos \sqrt{x}}{\sqrt{x}} d x & =2 \int \cos t d t \\ & =2 \sin t+C \\ & =2 \sin \sqrt{x}+C \end{aligned} $

Solution

Let $\sin 2 x=t$

$\therefore 2 \cos 2 x d x=d t$

$ \begin{aligned} \Rightarrow \int \sqrt{\sin 2 x} \cos 2 x d x & =\frac{1}{2} \int \sqrt{t} d t \\ & =\frac{1}{2}(\frac{t^{\frac{3}{2}}}{\frac{3}{2}})+C \\ & =\frac{1}{3} t^{\frac{3}{2}}+C \\ & =\frac{1}{3}(\sin 2 x)^{\frac{3}{2}}+C \end{aligned} $

Solution

Let $1+\sin x=t$ $\therefore \cos x d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{\cos x}{\sqrt{1+\sin x}} d x & =\int \frac{d t}{\sqrt{t}} \\ & =\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C \\ & =2 \sqrt{t}+C \\ & =2 \sqrt{1+\sin x}+C \end{aligned} $

Solution

Let $\log \sin x=t$

$\Rightarrow \frac{1}{\sin x} \cdot \cos x d x=d t$

$\therefore \cot x d x=d t$

$\Rightarrow \int \cot x \log \sin x d x=\int t d t$

$ \begin{aligned} & =\frac{t^{2}}{2}+C \\ & =\frac{1}{2}(\log \sin x)^{2}+C \end{aligned} $

Solution

Let $1+\cos x=t$ $\therefore-\sin x d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{\sin x}{1+\cos x} d x & =\int-\frac{d t}{t} \\ & =-\log |t|+C \\ & =-\log |1+\cos x|+C \end{aligned} $

Solution

Let $1+\cos x=t$

$\therefore-\sin x d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{\sin x}{(1+\cos x)^{2}} d x & =\int-\frac{d t}{t^{2}} \\ & =-\int t^{-2} d t \\ & =\frac{1}{t}+C \\ & =\frac{1}{1+\cos x}+\mathbf{C} \end{aligned} $

Solution

$ \text{ Let } \begin{aligned} I & =\int \frac{1}{1+\cot x} d x \\ & =\int \frac{1}{1+\frac{\cos x}{\sin x}} d x \\ & =\int \frac{\sin x}{\sin x+\cos x} d x \\ & =\frac{1}{2} \int \frac{2 \sin x}{\sin x+\cos x} d x \\ & =\frac{1}{2} \int \frac{(\sin x+\cos x)+(\sin x-\cos x)}{(\sin x+\cos x)} d x \\ & =\frac{1}{2} \int 1 d x+\frac{1}{2} \int \frac{\sin x-\cos x}{\sin x+\cos x} d x \\ & =\frac{1}{2}(x)+\frac{1}{2} \int \frac{\sin x-\cos x}{\sin x+\cos x} d x \end{aligned} $

Let $\sin x+\cos x=t \Rightarrow(\cos x-\sin x) d x=d t$

$ \begin{aligned} \therefore I & =\frac{x}{2}+\frac{1}{2} \int \frac{-(d t)}{t} \\ & =\frac{x}{2}-\frac{1}{2} \log |t|+C \\ & =\frac{x}{2}-\frac{1}{2} \log |\sin x+\cos x|+C \end{aligned} $

Solution

$ \text{ Let } \begin{aligned} I & =\int \frac{1}{1-\tan x} d x \\ & =\int \frac{1}{1-\frac{\sin x}{\cos x}} d x \\ & =\int \frac{\cos x}{\cos x-\sin x} d x \\ & =\frac{1}{2} \int \frac{2 \cos x}{\cos x-\sin x} d x \\ & =\frac{1}{2} \int \frac{(\cos x-\sin x)+(\cos x+\sin x)}{(\cos x-\sin x)} d x \\ & =\frac{1}{2} \int 1 d x+\frac{1}{2} \int \frac{\cos x+\sin x}{\cos x-\sin x} d x \\ & =\frac{x}{2}+\frac{1}{2} \int \frac{\cos x+\sin x}{\cos x-\sin x} d x \end{aligned} $

Put $\cos x-\sin x=t \Rightarrow(-\sin x-\cos x) d x=d t$

$ \begin{aligned} \therefore I & =\frac{x}{2}+\frac{1}{2} \int \frac{-(d t)}{t} \\ & =\frac{x}{2}-\frac{1}{2} \log |t|+C \\ & =\frac{x}{2}-\frac{1}{2} \log |\cos x-\sin x|+C \end{aligned} $

Solution

$ \text{ Let } \begin{aligned} I & =\int \frac{\sqrt{\tan x}}{\sin x \cos x} d x \\ & =\int \frac{\sqrt{\tan x} \times \cos x}{\sin x \cos x \times \cos x} d x \\ & =\int \frac{\sqrt{\tan x}}{\tan x \cos ^{2} x} d x \\ & =\int \frac{\sec ^{2} x d x}{\sqrt{\tan x}} \end{aligned} $

Let $\tan x=t \Rightarrow \sec ^{2} x d x=d t$

$ \begin{aligned} \therefore I & =\int \frac{d t}{\sqrt{t}} \\ & =2 \sqrt{t}+C \\ & =2 \sqrt{\tan x}+C \end{aligned} $

Solution

$\frac{(x+1)(x+\log x)^{2}}{x}=(\frac{x+1}{x})(x+\log x)^{2}=(1+\frac{1}{x})(x+\log x)^{2}$

Let $(x+\log x)=t$

$\therefore(1+\frac{1}{x}) d x=d t$

$\Rightarrow \int(1+\frac{1}{x})(x+\log x)^{2} d x=\int t^{2} d t$

$ \begin{aligned} & =\frac{t^{3}}{3}+C \\ & =\frac{1}{3}(x+\log x)^{3}+C \end{aligned} $

Solution

$\frac{(x+1)(x+\log x)^{2}}{x}=(\frac{x+1}{x})(x+\log x)^{2}=(1+\frac{1}{x})(x+\log x)^{2}$

Let $(x+\log x)=t$

$\therefore(1+\frac{1}{x}) d x=d t$

$\Rightarrow \int(1+\frac{1}{x})(x+\log x)^{2} d x=\int t^{2} d t$

$ \begin{aligned} & =\frac{t^{3}}{3}+C \\ & =\frac{1}{3}(x+\log x)^{3}+C \end{aligned} $

Solution

Let $x^{4}=t$

$\therefore 4 x^{3} d x=d t$ $\Rightarrow \int \frac{x^{3} \sin (\tan ^{-1} x^{4})}{1+x^{8}} d x=\frac{1}{4} \int \frac{\sin (\tan ^{-1} t)}{1+t^{2}} d t$

Let $\tan ^{-1} t=u$

$\therefore \frac{1}{1+t^{2}} d t=d u$

From (1), we obtain

$ \begin{aligned} & \begin{aligned} & \int \frac{x^{3} \sin (\tan ^{-1} x^{4}) d x}{1+x^{8}}=\frac{1}{4} \int \sin u d u \\ &=\frac{1}{4}(-\cos u)+C \\ &=\frac{-1}{4} \cos (\tan ^{-1} t)+C \\ &= \frac{-1}{4} \cos (\tan ^{-1} x^{4})+C \end{aligned} \end{aligned} $

Solution

Let $x^{10}+10^{x}=t$

$\therefore(10 x^{9}+10^{x} \log _{e} 10) d x=d t$ $\Rightarrow \int \frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}} d x=\int \frac{d t}{t}$

$=\log t+C$

$=\log (10^{x}+x^{10})+C$

Hence, the correct Answer is D.

Solution

Let $I=\int \frac{d x}{\sin ^{2} x \cos ^{2} x}$

$=\int \frac{1}{\sin ^{2} x \cos ^{2} x} d x$

$=\int \frac{\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x$

$=\int \frac{\sin ^{2} x}{\sin ^{2} x \cos ^{2} x} d x+\int \frac{\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x$

$=\int \sec ^{2} x d x+\int cosec^{2} x d x$

$=\tan x-\cot x+C$

Hence, the correct Answer is B.

Solution

$ \begin{aligned} & \sin ^{2}(2 x+5)=\frac{1-\cos 2(2 x+5)}{2}=\frac{1-\cos (4 x+10)}{2} \\ & \Rightarrow \int \sin ^{2}(2 x+5) d x=\int \frac{1-\cos (4 x+10)}{2} d x \\ & =\frac{1}{2} \int 1 d x-\frac{1}{2} \int \cos (4 x+10) d x \\ & =\frac{1}{2} x-\frac{1}{2}(\frac{\sin (4 x+10)}{4})+C \\ & =\frac{1}{2} x-\frac{1}{8} \sin (4 x+10)+C \end{aligned} $

Solution

It is known that, $\sin A \cos B=\frac{1}{2}{\sin (A+B)+\sin (A-B)}$

$\therefore \int \sin 3 x \cos 4 x d x=\frac{1}{2} \int{\sin (3 x+4 x)+\sin (3 x-4 x)} d x$

$ =\frac{1}{2} \int{\sin 7 x+\sin (-x)} d x $

$ \begin{aligned} & =\frac{1}{2} \int{\sin 7 x-\sin x} d x \\ & =\frac{1}{2} \int \sin 7 x d x-\frac{1}{2} \int \sin x d x \\ & =\frac{1}{2}(\frac{-\cos 7 x}{7})-\frac{1}{2}(-\cos x)+C \\ & =\frac{-\cos 7 x}{14}+\frac{\cos x}{2}+C \end{aligned} $

Solution

It is known that, $\cos A \cos B=\frac{1}{2}{\cos (A+B)+\cos (A-B)}$

$ \begin{aligned} \therefore \int \cos 2 x(\cos 4 x \cos 6 x) d x & =\int \cos 2 x[\frac{1}{2}{\cos (4 x+6 x)+\cos (4 x-6 x)}] d x \\ & =\frac{1}{2} \int{\cos 2 x \cos 10 x+\cos 2 x \cos (-2 x)} d x \\ & =\frac{1}{2} \int{\cos 2 x \cos 10 x+\cos ^{2} 2 x} d x \\ & =\frac{1}{2} \int[{\frac{1}{2} \cos (2 x+10 x)+\cos (2 x-10 x)}+(\frac{1+\cos 4 x}{2})] d x \\ & =\frac{1}{4} \int(\cos 12 x+\cos 8 x+1+\cos 4 x) d x \\ & =\frac{1}{4}[\frac{\sin 12 x}{12}+\frac{\sin 8 x}{8}+x+\frac{\sin 4 x}{4}]+C \end{aligned} $

Solution

$ \begin{aligned} & \text{ Let } I=\int \sin ^{3}(2 x+1) \\ & \Rightarrow \int \sin ^{3}(2 x+1) d x=\int \sin ^{2}(2 x+1) \cdot \sin (2 x+1) d x \\ & =\int(1-\cos ^{2}(2 x+1)) \sin (2 x+1) d x \end{aligned} $

Let $\cos (2 x+1)=t$

$\Rightarrow-2 \sin (2 x+1) d x=d t$

$\Rightarrow \sin (2 x+1) d x=\frac{-d t}{2}$

$ \begin{aligned} \Rightarrow I & =\frac{-1}{2} \int(1-t^{2}) d t \\ & =\frac{-1}{2}{t-\frac{t^{3}}{3}} \\ & =\frac{-1}{2}{\cos (2 x+1)-\frac{\cos ^{3}(2 x+1)}{3}} \\ & =\frac{-\cos (2 x+1)}{2}+\frac{\cos ^{3}(2 x+1)}{6}+C \end{aligned} $

Solution

$ \text{ Let } \begin{aligned} I & =\int \sin ^{3} x \cos ^{3} x \cdot d x \\ & =\int \cos ^{3} x \cdot \sin ^{2} x \cdot \sin x \cdot d x \\ & =\int \cos ^{3} x(1-\cos ^{2} x) \sin x \cdot d x \end{aligned} $

Let $\cos x=t$

$ \begin{aligned} \Rightarrow & -\sin x \cdot d x=d t \\ \Rightarrow I & =-\int t^{3}(1-t^{2}) d t \\ & =-\int(t^{3}-t^{5}) d t \\ & =-{\frac{t^{4}}{4}-\frac{t^{6}}{6}}+C \\ & =-{\frac{\cos ^{4} x}{4}-\frac{\cos ^{6} x}{6}}+C \\ & =\frac{\cos ^{6} x}{6}-\frac{\cos ^{4} x}{4}+C \end{aligned} $

Solution

It is known that, $\sin A \sin B=\frac{1}{2}{\cos (A-B)-\cos (A+B)}$

$\therefore \int \sin x \sin 2 x \sin 3 x d x=\int[\sin x \cdot \frac{1}{2}{\cos (2 x-3 x)-\cos (2 x+3 x)}] d x$

$=\frac{1}{2} \int(\sin x \cos (-x)-\sin x \cos 5 x) d x$

$=\frac{1}{2} \int(\sin x \cos x-\sin x \cos 5 x) d x$

$=\frac{1}{2} \int \frac{\sin 2 x}{2} d x-\frac{1}{2} \int \sin x \cos 5 x d x$

$=\frac{1}{4}[\frac{-\cos 2 x}{2}]-\frac{1}{2} \int{\frac{1}{2} \sin (x+5 x)+\sin (x-5 x)} d x$

$=\frac{-\cos 2 x}{8}-\frac{1}{4} \int(\sin 6 x+\sin (-4 x)) d x$

$=\frac{-\cos 2 x}{8}-\frac{1}{4}[\frac{-\cos 6 x}{3}+\frac{\cos 4 x}{4}]+C$

$=\frac{-\cos 2 x}{8}-\frac{1}{8}[\frac{-\cos 6 x}{3}+\frac{\cos 4 x}{2}]+C$

$=\frac{1}{8}[\frac{\cos 6 x}{3}-\frac{\cos 4 x}{2}-\cos 2 x]+C$

Solution

It is known that, $\sin A \sin B=\frac{1}{2} \cos (A-B)-\cos (A+B)$ $\therefore \int \sin 4 x \sin 8 x d x=\int{\frac{1}{2} \cos (4 x-8 x)-\cos (4 x+8 x)} d x$

$ \begin{aligned} & =\frac{1}{2} \int(\cos (-4 x)-\cos 12 x) d x \\ & =\frac{1}{2} \int(\cos 4 x-\cos 12 x) d x \\ & =\frac{1}{2}[\frac{\sin 4 x}{4}-\frac{\sin 12 x}{12}] \end{aligned} $

Solution

$ \begin{aligned} & \frac{1-\cos x}{1+\cos x}=\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} x} \\ & {[2 \sin ^{2} \frac{x}{2}=1-\cos x \text{ and } 2 \cos ^{2} \frac{x}{2}=1+\cos x]} \\ & =\tan ^{2} \frac{x}{2} \\ & =(\sec ^{2} \frac{x}{2}-1) \\ & \therefore \int \frac{1-\cos x}{1+\cos x} d x=\int(\sec ^{2} \frac{x}{2}-1) d x \\ & =[\frac{\tan \frac{x}{2}}{1}-x]+C \\ & =2 \tan \frac{x}{2}-x+C \end{aligned} $

Solution

$ \begin{aligned} & \frac{\cos x}{1+\cos x}=\frac{\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}} \quad[\cos x=\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2} \text{ and } \cos x=2 \cos ^{2} \frac{x}{2}-1] \\ & =\frac{1}{2}[1-\tan ^{2} \frac{x}{2}] \\ & \begin{aligned} \therefore \int \frac{\cos x}{1+\cos x} d x & =\frac{1}{2} \int(1-\tan ^{2} \frac{x}{2}) d x \\ & =\frac{1}{2} \int(1-\sec ^{2} \frac{x}{2}+1) d x \\ & =\frac{1}{2} \int(2-\sec ^{2} \frac{x}{2}) d x \\ & =\frac{1}{2}[2 x-\frac{\tan ^{2}}{2}]+C \\ & =x-\tan \frac{x}{2}+C \end{aligned} \end{aligned} $

Solution

$ \begin{aligned} & \sin ^{4} x=\sin ^{2} x \sin ^{2} x \\ & =(\frac{1-\cos 2 x}{2})(\frac{1-\cos 2 x}{2}) \\ & =\frac{1}{4}(1-\cos 2 x)^{2} \\ & =\frac{1}{4}[1+\cos ^{2} 2 x-2 \cos 2 x] \\ & =\frac{1}{4}[1+(\frac{1+\cos 4 x}{2})-2 \cos 2 x] \\ & =\frac{1}{4}[1+\frac{1}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x] \\ & =\frac{1}{4}[\frac{3}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x] \\ & \therefore \int \sin ^{4} x d x=\frac{1}{4} \int[\frac{3}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x] d x \\ & =\frac{1}{4}[\frac{3}{2} x+\frac{1}{2}(\frac{\sin 4 x}{4})-\frac{2 \sin 2 x}{2}]+C \\ & =\frac{1}{8}[3 x+\frac{\sin 4 x}{4}-2 \sin 2 x]+C \\ & =\frac{3 x}{8}-\frac{1}{4} \sin 2 x+\frac{1}{32} \sin 4 x+C \end{aligned} $

Solution

$ \begin{aligned} \cos ^{4} 2 x & =(\cos ^{2} 2 x)^{2} \\ & =(\frac{1+\cos 4 x}{2})^{2} \\ & =\frac{1}{4}[1+\cos ^{2} 4 x+2 \cos 4 x] \\ & =\frac{1}{4}[1+(\frac{1+\cos 8 x}{2})+2 \cos 4 x] \\ & =\frac{1}{4}[1+\frac{1}{2}+\frac{\cos 8 x}{2}+2 \cos 4 x] \\ & =\frac{1}{4}[\frac{3}{2}+\frac{\cos 8 x}{2}+2 \cos 4 x] \end{aligned} $

$ \begin{aligned} \therefore \int \cos ^{4} 2 x d x & =\int(\frac{3}{8}+\frac{\cos 8 x}{8}+\frac{\cos 4 x}{2}) d x \\ & =\frac{3}{8} x+\frac{\sin 8 x}{64}+\frac{\sin 4 x}{8}+C \end{aligned} $

Solution

$ \begin{aligned} \frac{\sin ^{2} x}{1+\cos x} & =\frac{(2 \sin \frac{x}{2} \cos \frac{x}{2})^{2}}{2 \cos ^{2} \frac{x}{2}}[\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2} ; \cos x=2 \cos ^{2} \frac{x}{2}-1] \\ & =\frac{4 \sin ^{2} \frac{x}{2} \cos ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}} \\ & =2 \sin ^{2} \frac{x}{2} \\ & =1-\cos ^{2} x \\ \therefore \int \frac{\sin ^{2} x}{1+\cos x} d x & =\int(1-\cos x) d x \\ & =x-\sin x+C \end{aligned} $

Solution

$ \begin{aligned} & \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}=\frac{-2 \sin \frac{2 x+2 \alpha}{2} \sin \frac{2 x-2 \alpha}{2}}{-2 \sin \frac{x+\alpha}{2} \sin \frac{x-\alpha}{2}} \quad[\cos C-\cos D=-2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}] \\ &=\frac{\sin (x+\alpha) \sin (x-\alpha)}{\sin (\frac{x+\alpha}{2}) \sin (\frac{x-\alpha}{2})} \\ &=\frac{[2 \sin (\frac{x+\alpha}{2}) \cos (\frac{x+\alpha}{2})][2 \sin (\frac{x-\alpha}{2}) \cos (\frac{x-\alpha}{2})]}{\sin (\frac{x+\alpha}{2}) \sin (\frac{x-\alpha}{2})} \\ &=4 \cos (\frac{x+\alpha}{2}) \cos (\frac{x-\alpha}{2}) \\ &=2[\cos (\frac{x+\alpha}{2}+\frac{x-\alpha}{2})+\cos \frac{x+\alpha}{2}-\frac{x-\alpha}{2}] \\ &=2[\cos (x)+\cos \alpha] \\ &=2 \cos x+2 \cos \alpha \\ & \therefore \int \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} d x=\int 2 \cos x+2 \cos \alpha \\ &=2[\sin x+x \cos \alpha]+C \end{aligned} $

Solution

$ \begin{aligned} \frac{\cos x-\sin x}{1+\sin 2 x} & =\frac{\cos x-\sin x}{(\sin ^{2} x+\cos ^{2} x)+2 \sin x \cos x} \\ & =\frac{\cos x-\sin x}{.(\sin x+\cos x)^{2} x+\cos ^{2} x=1 ; \sin 2 x=2 \sin x \cos x]} \end{aligned} $

Let $\sin x+\cos x=t$

$\therefore(\cos x-\sin x) d x=d t$

$\Rightarrow \int \frac{\cos x-\sin x}{1+\sin 2 x} d x=\int \frac{\cos x-\sin x}{(\sin x+\cos x)^{2}} d x$

$ =\int \frac{d t}{t^{2}} $

$ \begin{aligned} & =\int t^{-2} d t \\ & =-t^{-1}+C \\ & =-\frac{1}{t}+C \\ & =\frac{-1}{\sin x+\cos x}+C \end{aligned} $

Solution

$ \begin{aligned} & \tan ^{3} 2 x \sec 2 x=\tan ^{2} 2 x \tan 2 x \sec 2 x \\ & =(\sec ^{2} 2 x-1) \tan 2 x \sec 2 x \\ & =\sec ^{2} 2 x \cdot \tan 2 x \sec 2 x-\tan 2 x \sec 2 x \\ & \therefore \int \tan ^{3} 2 x \sec 2 x d x=\int \sec ^{2} 2 x \tan 2 x \sec 2 x d x-\int \tan 2 x \sec 2 x d x \\ & =\int \sec ^{2} 2 x \tan 2 x \sec 2 x d x-\frac{\sec 2 x}{2}+C \end{aligned} $

Let $\sec 2 x=t$

$\therefore 2 \sec 2 x \tan 2 x d x=d t$

$ \begin{aligned} \therefore \int \tan ^{3} 2 x \sec 2 x d x & =\frac{1}{2} \int t^{2} d t-\frac{\sec 2 x}{2}+C \\ & =\frac{t^{3}}{6}-\frac{\sec 2 x}{2}+C \\ & =\frac{(\sec 2 x)^{3}}{6}-\frac{\sec 2 x}{2}+C \end{aligned} $

Solution

$\tan ^{4} x$

$=\tan ^{2} x \cdot \tan ^{2} x$

$=(\sec ^{2} x-1) \tan ^{2} x$

$=\sec ^{2} x \tan ^{2} x-\tan ^{2} x$

$=\sec ^{2} x \tan ^{2} x-(\sec ^{2} x-1)$

$=\sec ^{2} x \tan ^{2} x-\sec ^{2} x+1$

$\therefore \int \tan ^{4} x d x=\int \sec ^{2} x \tan ^{2} x d x-\int \sec ^{2} x d x+\int 1 \cdot d x$

$=\int \sec ^{2} x \tan ^{2} x d x-\tan x+x+C$

Consider $\int \sec ^{2} x \tan ^{2} x d x$

Let $\tan x=t \Rightarrow \sec ^{2} x d x=d t$

$\Rightarrow \int \sec ^{2} x \tan ^{2} x d x=\int t^{2} d t=\frac{t^{3}}{3}=\frac{\tan ^{3} x}{3}$

From equation (1), we obtain

$\int \tan ^{4} x d x=\frac{1}{3} \tan ^{3} x-\tan x+x+C$

Solution

$ \begin{aligned} & \frac{\sin ^{3} x+\cos ^{3} x}{\sin ^{2} x \cos ^{2} x}=\frac{\sin ^{3} x}{\sin ^{2} x \cos ^{2} x}+\frac{\cos ^{3} x}{\sin ^{2} x \cos ^{2} x} \\ & =\frac{\sin x}{\cos ^{2} x}+\frac{\cos x}{\sin ^{2} x} \\ & =\tan x \sec x+\cot x cosec x \\ & \therefore \quad \int \frac{\sin ^{3} x+\cos ^{3} x}{\sin ^{2} x \cos ^{2} x} d x=\int(\tan x \sec x+\cot x cosec x) d x \\ & =\sec x-cosec x+C \end{aligned} $

Solution

$ \begin{aligned} & \frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x} \\ & =\frac{\cos 2 x+(1-\cos 2 x)}{\cos ^{2} x} \quad[\cos 2 x=1-2 \sin ^{2} x] \\ & =\frac{1}{\cos ^{2} x} \\ & =\sec ^{2} x \\ & \therefore \int \frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x} d x=\int \sec ^{2} x d x=\tan x+C \end{aligned} $

Solution

$ \begin{aligned} \frac{1}{\sin x \cos ^{3} x} & =\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos ^{3} x} \\ & =\frac{\sin x}{\cos ^{3} x}+\frac{1}{\sin x \cos x} \\ & =\tan x \sec ^{2} x+\frac{1 \cos ^{2} x}{\frac{\sin x \cos x}{\cos ^{2} x}} \\ & =\tan x \sec ^{2} x+\frac{\sec ^{2} x}{\tan x} \end{aligned} $

$\therefore \int \frac{1}{\sin x \cos ^{3} x} d x=\int \tan x \sec ^{2} x d x+\int \frac{\sec ^{2} x}{\tan x} d x$

Let $\tan x=t \Rightarrow \sec ^{2} x d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{1}{\sin x \cos ^{3} x} d x & =\int t d t+\int_t^{1} d t \\ & =\frac{t^{2}}{2}+\log |t|+C \\ & =\frac{1}{2} \tan ^{2} x+\log |\tan x|+C \end{aligned} $

Solution

$\begin{aligned} & \frac{\cos 2 \mathrm{x}}{(\cos \mathrm{x}+\sin \mathrm{x})^2}=\frac{\cos 2 \mathrm{x}}{\cos ^2 \mathrm{x}+\sin ^2 \mathrm{x}+2 \sin \mathrm{x} \cos \mathrm{x}}=\frac{\cos 2 \mathrm{x}}{1+\sin 2 \mathrm{x}} \\ & \therefore \int \frac{\cos 2 \mathrm{x}}{(\cos \mathrm{x}+\sin \mathrm{x})^2} \mathrm{dx}=\int \frac{\cos 2 \mathrm{x}}{(1+\sin 2 \mathrm{x})} \mathrm{dx} \end{aligned}$

Let $1+\sin 2 x=t \Rightarrow 2 \cos 2 x d x=d t$

$\begin{aligned} & \therefore \int \frac{\cos 2 x}{(\cos x+\sin x)^2} d x=\frac{1}{2} \int \frac{1}{t} d t \\ & =\frac{1}{2} \log |t|+C \\ & =\frac{1}{2} \log |1+\sin 2 x|+C \\ & =\frac{1}{2} \log \left|(\sin x+\cos x)^2\right|+C \\ & =\log |\sin x+\cos x|+C \\ \end{aligned}$

Solution

$\sin ^{-1}(\cos x)$

Let $\cos x=t$

Then, $\sin x=\sqrt{1-t^{2}}$ $\Rightarrow(-\sin x) d x=d t$

$d x=\frac{-d t}{\sin x}$

$d x=\frac{-d t}{\sqrt{1-t^{2}}}$

$\therefore \int \sin ^{-1}(\cos x) d x=\int \sin ^{-1} t(\frac{-d t}{\sqrt{1-t^{2}}})$

$=-\int \frac{\sin ^{-1} t}{\sqrt{1-t^{2}}} d t$

Let $\sin ^{-1} t=u$

$\Rightarrow \frac{1}{\sqrt{1-t^{2}}} d t=d u$

$\therefore \int \sin ^{-1}(\cos x) d x=\int 4 d u$

$$ \begin{align*} & =-\frac{u^{2}}{2}+C \\ & =\frac{-(\sin ^{1} t)^{2}}{2}+C \\ & =\frac{-[\sin ^{-1}(\cos x)]^{2}}{2}+C \tag{1} \end{align*} $$

It is known that,

$ \begin{aligned} & \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} \\ & \therefore \sin ^{-1}(\cos x)=\frac{\pi}{2}-\cos ^{-1}(\cos x)=(\frac{\pi}{2}-x) \end{aligned} $

Substituting in equation (1), we obtain

$ \begin{aligned} \int \sin ^{-1}(\cos x) d x & =\frac{-[\frac{\pi}{2}-x]^{2}}{2}+C \\ & =-\frac{1}{2}(\frac{\pi^{2}}{2}+x^{2}-\pi x)+C \\ & =-\frac{\pi^{2}}{8}-\frac{x^{2}}{2}+\frac{1}{2} \pi x+C \\ & =\frac{\pi x}{2}-\frac{x^{2}}{2}+(C-\frac{\pi^{2}}{8}) \\ & =\frac{\pi x}{2}-\frac{x^{2}}{2}+C_1 \end{aligned} $

Solution

$ \begin{aligned} \frac{1}{\cos (x-a) \cos (x-b)} & =\frac{1}{\sin (a-b)}[\frac{\sin (a-b)}{\cos (x-a) \cos (x-b)}] \\ & =\frac{1}{\sin (a-b)}[\frac{\sin [(x-b)-(x-a)]}{\cos (x-a) \cos (x-b)}] \\ & =\frac{1}{\sin (a-b)} \frac{[\sin (x-b) \cos (x-a)-\cos (x-b) \sin (x-a)]}{\cos (x-a) \cos (x-b)} \\ & =\frac{1}{\sin (a-b)}[\tan (x-b)-\tan (x-a)] \\ \Rightarrow \int \frac{1}{\cos (x-a) \cos (x-b)} d x & =\frac{1}{\sin (a-b)} \int[\tan (x-b)-\tan (x-a)] d x \\ & =\frac{1}{\sin (a-b)}[-\log |\cos (x-b)|+\log |\cos (x-a)|] \\ & =\frac{1}{\sin (a-b)}[\log |\frac{\cos (x-a)}{\cos (x-b)}|]+C \end{aligned} $

Solution

$ \begin{aligned} \int \frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x & =\int(\frac{\sin ^{2} x}{\sin ^{2} x \cos ^{2} x}-\frac{\cos ^{2} x}{\sin ^{2} x \cos ^{2} x}) d x \\ & =\int(\sec ^{2} x-cosec^{2} x) d x \\ & =\tan x+\cot x+C \end{aligned} $

Hence, the correct Answer is A.

Solution

$\int \frac{e^{x}(1+x)}{\cos ^{2}(e^{x} x)} d x$

Let $e x^{x}=t$

$ \begin{aligned} & \Rightarrow(e^{x} \cdot x+e^{x} \cdot 1) d x=d t \\ & e^{x}(x+1) d x=d t \\ & \begin{aligned} \therefore \int \frac{e^{x}(1+x)}{\cos ^{2}(e^{x} x)} d x & =\int \frac{d t}{\cos ^{2} t} \\ & =\int \sec ^{2} t d t \\ & =\tan t+C \\ & =\tan (e^{x} \cdot x)+C \end{aligned} \end{aligned} $

Hence, the correct Answer is B.

Solution

Let $x^{3}=t$

$\therefore 3 x^{2} d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{3 x^{2}}{x^{6}+1} d x & =\int \frac{d t}{t^{2}+1} \\ & =\tan ^{1} t+C \\ & =\tan ^{-1}(x^{3})+C \end{aligned} $

Solution

Let $2 x=t$

$\therefore 2 d x=d t$ $\Rightarrow \int \frac{1}{\sqrt{1+4 x^{2}}} d x=\frac{1}{2} \int \frac{d t}{\sqrt{1+t^{2}}}$

$ \begin{matrix} =\frac{1}{2}[\log |t+\sqrt{t^{2}+1}|]+C & {[\int \frac{1}{\sqrt{x^{2}+a^{2}}} d t=\log |x+\sqrt{x^{2}+a^{2}}|]} \\ =\frac{1}{2} \log |2 x+\sqrt{4 x^{2}+1}|+C & \end{matrix} $

Solution

Let $2-x=t$

$\Rightarrow-d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{1}{\sqrt{(2-x)^{2}+1}} d x & =-\int \frac{1}{\sqrt{t^{2}+1}} d t \\ & =-\log |t+\sqrt{t^{2}+1}|+C \quad[\int \frac{1}{\sqrt{x^{2}+a^{2}}} d t=\log |x+\sqrt{x^{2}+a^{2}}|] \\ & =-\log |2-x+\sqrt{(2-x)^{2}+1}|+C \\ & =\log |\frac{1}{(2-x)+\sqrt{x^{2}-4 x+5}}|+C \end{aligned} $

Solution

Let $5 x=t$ $\therefore 5 d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{1}{\sqrt{9-25 x^{2}}} d x & =\frac{1}{5} \int \frac{1}{9-t^{2}} d t \\ & =\frac{1}{5} \int \frac{1}{\sqrt{3^{2}-t^{2}}} d t \\ & =\frac{1}{5} \sin ^{-1}(\frac{t}{3})+C \\ & =\frac{1}{5} \sin ^{-1}(\frac{5 x}{3})+C \end{aligned} $

Solution

Let $\sqrt{2} x^{2}=t$

$\therefore 2 \sqrt{2} x d x=d t$

$\Rightarrow \int \frac{3 x}{1+2 x^{4}} d x=\frac{3}{2 \sqrt{2}} \int \frac{d t}{1+t^{2}}$

$ =\frac{3}{2 \sqrt{2}}[\tan ^{-1} t]+C $

$=\frac{3}{2 \sqrt{2}} \tan ^{-1}(\sqrt{2} x^{2})+C$

Solution

Let $x^{3}=t$ $\therefore 3 x^{2} d x=d t$

$$ \begin{aligned} \Rightarrow \int \frac{x^{2}}{1-x^{6}} d x & =\frac{1}{3} \int \frac{d t}{1-t^{2}} \\ & =\frac{1}{3}[\frac{1}{2} \log |\frac{1+t}{1-t}|]+C \\ & =\frac{1}{6} \log |\frac{1+x^{3}}{1-x^{3}}|+C \end{aligned} $$

Solution

$$ \begin{equation*} \int \frac{x-1}{\sqrt{x^{2}-1}} d x=\int \frac{x}{\sqrt{x^{2}-1}} d x-\int \frac{1}{\sqrt{x^{2}-1}} d x \tag{1} \end{equation*} $$

For $\int \frac{x}{\sqrt{x^{2}-1}} d x$, let $x^{2}-1=t \Rightarrow 2 x d x=d t$

$\therefore \int \frac{x}{\sqrt{x^{2}-1}} d x=\frac{1}{2} \int \frac{d t}{\sqrt{t}}$

$=\frac{1}{2} \int t^{-\frac{1}{2}} d t$

$=\frac{1}{2}[2 t^{\frac{1}{2}}]$

$=\sqrt{t}$

$=\sqrt{x^{2}-1}$

From (1), we obtain

$ \begin{aligned} \int \frac{x-1}{\sqrt{x^{2}-1}} d x & =\int \frac{x}{\sqrt{x^{2}-1}} d x-\int \frac{1}{\sqrt{x^{2}-1}} d x \quad \quad[\int \frac{1}{\sqrt{x^{2}-a^{2}}} d t=\log |x+\sqrt{x^{2}-a^{2}}|] \\ & =\sqrt{x^{2}-1}-\log |x+\sqrt{x^{2}-1}|+C \end{aligned} $

Solution

Let $x^{3}=t$

$\Rightarrow 3 x^{2} d x=d t$

$ \begin{aligned} \therefore \int \frac{x^{2}}{\sqrt{x^{6}+a^{6}}} d x & =\frac{1}{3} \int \frac{d t}{\sqrt{t^{2}+(a^{3})^{2}}} \\ & =\frac{1}{3} \log |t+\sqrt{t^{2}+a^{6}}|+C \\ & =\frac{1}{3} \log |x^{3}+\sqrt{x^{6}+a^{6}}|+C \end{aligned} $

Solution

Let $\tan x=t$

$\therefore \sec ^{2} x d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{\sec ^{2} x}{\sqrt{\tan ^{2} x+4}} d x & =\int \frac{d t}{\sqrt{t^{2}+2^{2}}} \\ & =\log |t+\sqrt{t^{2}+4}|+C \\ & =\log |\tan x+\sqrt{\tan ^{2} x+4}|+C \end{aligned} $

Solution

$ \int \frac{1}{\sqrt{x^{2}+2 x+2}} d x=\int \frac{1}{\sqrt{(x+1)^{2}+(1)^{2}}} d x $

Let $x+1=t$

$\therefore d x=d t$

$\Rightarrow \int \frac{1}{\sqrt{x^{2}+2 x+2}} d x=\int \frac{1}{\sqrt{t^{2}+1}} d t$

$ \begin{aligned} & =\log |t+\sqrt{t^{2}+1}|+C \\ & =\log |(x+1)+\sqrt{(x+1)^{2}+1}|+C \\ & =\log |(x+1)+\sqrt{x^{2}+2 x+2}|+C \end{aligned} $

Solution

$ \int \frac{1}{9 x^{2}+6 x+5} d x=\int \frac{1}{(3 x+1)^{2}+(2)^{2}} d x $

Let $(3 x+1)=t$

$\therefore 3 d x=d t$

$\Rightarrow \int \frac{1}{(3 x+1)^{2}+(2)^{2}} d x=\frac{1}{3} \int \frac{1}{t^{2}+2^{2}} d t$

$ \begin{aligned} & =\frac{1}{3}[\frac{1}{2} \tan ^{-1}(\frac{t}{2})]+C \\ & =\frac{1}{6} \tan ^{-1}(\frac{3 x+1}{2})+C \end{aligned} $

Solution

$$ \begin{aligned} & =\int \frac{1}{\sqrt{7-(x+3)^2-(3)^2}} d x \\ & =\int \frac{1}{\sqrt{7+9-(x+3)^2}} d x \\ & =\int \frac{1}{\sqrt{16-(x+3)^2}} d x \\ & =\int \frac{1}{\sqrt{(4)^2-(x+3)^2}} d x \end{aligned} $$

It is of form $$ \int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1} \frac{x}{a}+C $$ $\therefore$ Replacing $x$ by $a$ by 4 , we get $$ =\sin ^{-1}\left(\frac{x+3}{4}\right)+C $$

Solution

$(x-1)(x-2)$ can be written as $x^{2}-3 x+2$.

Therefore,

$x^{2}-3 x+2$

$=x^{2}-3 x+\frac{9}{4}-\frac{9}{4}+2$

$=(x-\frac{3}{2})^{2}-\frac{1}{4}$

$=(x-\frac{3}{2})^{2}-(\frac{1}{2})^{2}$

$\therefore \int \frac{1}{\sqrt{(x-1)(x-2)}} d x=\int \frac{1}{\sqrt{(x-\frac{3}{2})^{2}-(\frac{1}{2})^{2}}} d x$

Let $x-\frac{3}{2}=t$

$\therefore d x=d t$

$\Rightarrow \int \frac{1}{\sqrt{(x-\frac{3}{2})^{2}-(\frac{1}{2})^{2}}} d x=\int \frac{1}{\sqrt{t^{2}-(\frac{1}{2})^{2}}} d t$

$=\log |t+\sqrt{t^{2}-(\frac{1}{2})^{2}}|+C$

$=\log |(x-\frac{3}{2})+\sqrt{x^{2}-3 x+2}|+C$

Solution

$8+3 x-x^{2}$ can be written as $8-(x^{2}-3 x+\frac{9}{4}-\frac{9}{4})$.

Therefore,

$8-(x^{2}-3 x+\frac{9}{4}-\frac{9}{4})$

$=\frac{41}{4}-(x-\frac{3}{2})^{2}$

$\Rightarrow \int \frac{1}{\sqrt{8+3 x-x^{2}}} d x=\int \frac{1}{\sqrt{\frac{41}{4}-(x-\frac{3}{2})^{2}}} d x$

Let $x-\frac{3}{2}=t$

$\therefore d x=d t$

$\Rightarrow \int \frac{1}{\sqrt{\frac{41}{4}-(x-\frac{3}{2})^{2}}} d x=\int \frac{1}{\sqrt{(\frac{\sqrt{41}}{2})^{2}-t^{2}}} d t$

$=\sin ^{-1}(\frac{t}{\frac{\sqrt{41}}{2}})+C$

$=\sin ^{-1}(\frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}})+C$

$=\sin ^{-1}(\frac{2 x-3}{\sqrt{41}})+C$

Solution

$$ \begin{aligned} & =\int \frac{1}{\sqrt{x^2-2(x)\left(\frac{a+b}{2}\right)+\left(\frac{a+b}{2}\right)^2-\left(\frac{a+b}{2}\right)^2+a b}} d x \\ & =\int \frac{1}{\sqrt{\left(x-\frac{a+b}{2}\right)^2-\left(\frac{a+b}{2}\right)^2+a b}} d x \\ & =\int \frac{1}{\sqrt{\left(x-\frac{a+b}{2}\right)^2-\left(\frac{a^2+b^2+2 a b}{4}\right)+a b}} d x \\ & =\int \frac{1}{\sqrt{\left(x-\frac{a+b}{2}\right)^2+\frac{-a^2-b^2-2 a b+4 a b}{4}}} d x \\ & =\int \frac{1}{\sqrt{\left(x-\frac{a+b}{2}\right)^2+\frac{-a^2-b^2+2 a b}{4}}} d x \end{aligned} $$

Solution

Let $4 x+1=A \frac{d}{d x}(2 x^{2}+x-3)+B$

$\Rightarrow 4 x+1=A(4 x+1)+B$

$\Rightarrow 4 x+1=4 A x+A+B$

Equating the coefficients of $x$ and constant term on both sides, we obtain $4 A=4 \Rightarrow A=1$

$A+B=1 \Rightarrow B=0$

Let $2 x^{2}+x-3=t$

$\therefore(4 x+1) d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{4 x+1}{\sqrt{2 x^{2}+x-3}} d x & =\int \frac{1}{\sqrt{t}} d t \\ & =2 \sqrt{t}+C \\ & =2 \sqrt{2 x^{2}+x-3}+C \end{aligned} $

Solution

Let $x+2=A \frac{d}{d x}(x^{2}-1)+B$

$\Rightarrow x+2=A(2 x)+B$

Equating the coefficients of $x$ and constant term on both sides, we obtain

$2 A=1 \Rightarrow A=\frac{1}{2}$

$B=2$

From (1), we obtain $(x+2)=\frac{1}{2}(2 x)+2$

Then, $\int \frac{x+2}{\sqrt{x^{2}-1}} d x=\int \frac{\frac{1}{2}(2 x)+2}{\sqrt{x^{2}-1}} d x$

$$ \begin{equation*} =\frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}-1}} d x+\int \frac{2}{\sqrt{x^{2}-1}} d x \tag{2} \end{equation*} $$

In $\frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}-1}} d x$, let $x^{2}-1=t \Rightarrow 2 x d x=d t$

$ \begin{aligned} \frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}-1}} d x & =\frac{1}{2} \int \frac{d t}{\sqrt{t}} \\ & =\frac{1}{2}[2 \sqrt{t}] \\ & =\sqrt{t} \\ & =\sqrt{x^{2}-1} \end{aligned} $

Then, $\int \frac{2}{\sqrt{x^{2}-1}} d x=2 \int \frac{1}{\sqrt{x^{2}-1}} d x=2 \log |x+\sqrt{x^{2}-1}|$

From equation (2), we obtain

$ \int \frac{x+2}{\sqrt{x^{2}-1}} d x=\sqrt{x^{2}-1}+2 \log |x+\sqrt{x^{2}-1}|+C $

$$ \begin{aligned} & \int \frac{5 x-2}{1+2 x+3 x^2} d x \\ & =5 \int \frac{x-\frac{2}{5}}{1+2 x+3 x^2} d x \\ & =\frac{5}{6} \int \frac{6 x-\frac{12}{5}}{1+2 x+3 x^2} d x \\ & =\frac{5}{6} \int \frac{6 x+2-\frac{12}{5}-2}{1+2 x+3 x^2} d x \\ & =\frac{5}{6} \int \frac{6 x+2-\frac{22}{5}}{1+2 x+3 x^2} d x \end{aligned} $$

$$ \begin{gathered} =\frac{5}{6} \int \frac{6 x+2}{1+2 x+3 x^2} d x-\frac{5}{6} \times \frac{22}{5} \int \frac{d x}{1+2 x+3 x^2} \\ =\frac{5}{6} \int \frac{6 x+2}{1+2 x+3 x^2} d x-\frac{11}{3} \int \frac{d x}{1+2 x+3 x^2} \\ \downarrow \end{gathered} $$

Solving $I_1$ $$ \mathrm{I}_1=\frac{5}{6} \int \frac{2+6 x}{3 x^2+2 x+1} d x $$

Let $$ 3 x^2+2 x+1=t $$

Diff both sides w.r.t.x $$ 6 x+2+0=\frac{d t}{d x} $$

$$ d x=\frac{d t}{6 x+2} $$

Thus, our equation becomes $$ \mathrm{I}_1=\frac{5}{6} \int \frac{2+6 x}{3 x^2+2 x+1} d x $$

Put the values of $\left(3 x^2+2 x+1\right)$ and $d x$, we get

$$ \begin{aligned} & \mathrm{I}_1=\frac{5}{6} \int \frac{6 x+2}{t} d x \\ & \mathrm{I}_1=\frac{5}{6} \int \frac{6 x+2}{t} \times \frac{d t}{6 x+2} \\ & \mathrm{I}_1=\frac{5}{6} \int \frac{1}{t} \cdot d t \\ & \mathrm{I}_1=\frac{5}{6} \log |t|+C_1 \\ & \mathrm{I}_1=\frac{5}{6} \log \left|3 x^2+2 x+1\right|+C_1 \quad\left(\text { Using } t=3 x^2+2 x+1\right) \end{aligned} $$

$\left(U\right.$ sing $\left.t=3 x^2+2 x+1\right)$

$$ \begin{aligned} & =\frac{11}{9} \cdot \frac{3}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right)+\frac{11}{9} C_2 \\ & =\frac{11}{3 \sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right)+C_3 \end{aligned} $$ (Where $C_3=\frac{11}{9} C_2$ )

Now, putting the values of $\mathrm{I}_1$ and $\mathrm{I}_2$ in (1) $$ \begin{aligned} & \int \frac{5 x-2}{1+2 x+3 x^2} \cdot d x \\ & \quad=\frac{5}{6} \int \frac{2+6 x}{1+2 x+3 x^2} \cdot d x-\frac{11}{3} \int \frac{1}{1+2 x+3 x^2} \cdot d x \\ & =\frac{5}{6} \log \left|3 x^2+2 x+1\right|-\frac{11}{3 \sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right)+C \end{aligned} $$

Solution

$\frac{6 x+7}{\sqrt{(x-5)(x-4)}}=\frac{6 x+7}{\sqrt{x^{2}-9 x+20}}$

Let $6 x+7=A \frac{d}{d x}(x^{2}-9 x+20)+B$

$\Rightarrow 6 x+7=A(2 x-9)+B$

Equating the coefficients of $x$ and constant term, we obtain

$2 A=6 \Rightarrow A=3$

$-9 A+B=7 \Rightarrow B=34$

$\therefore 6 x+7=3(2 x-9)+34$

$ \begin{aligned} \int \frac{6 x+7}{\sqrt{x^{2}-9 x+20}} & =\int \frac{3(2 x-9)+34}{\sqrt{x^{2}-9 x+20}} d x \\ & =3 \int \frac{2 x-9}{\sqrt{x^{2}-9 x+20}} d x+34 \int \frac{1}{\sqrt{x^{2}-9 x+20}} d x \end{aligned} $

Let $I_1=\int \frac{2 x-9}{\sqrt{x^{2}-9 x+20}} d x$ and $I_2=\int \frac{1}{\sqrt{x^{2}-9 x+20}} d x$

$\therefore \int \frac{6 x+7}{\sqrt{x^{2}-9 x+20}}=3 I_1+34 I_2$

Then,

$I_1=\int \frac{2 x-9}{\sqrt{x^{2}-9 x+20}} d x$

Let $x^{2}-9 x+20=t$

$\Rightarrow(2 x-9) d x=d t$

$\Rightarrow I_1=\frac{d t}{\sqrt{t}}$

$I_1=2 \sqrt{t}$

$I_1=2 \sqrt{x^{2}-9 x+20}$

and $I_2=\int \frac{1}{\sqrt{x^{2}-9 x+20}} d x$ $x^{2}-9 x+20$ can be written as $x^{2}-9 x+20+\frac{81}{4}-\frac{81}{4}$.

Therefore,

$x^{2}-9 x+20+\frac{81}{4}-\frac{81}{4}$

$=(x-\frac{9}{2})^{2}-\frac{1}{4}$

$=(x-\frac{9}{2})^{2}-(\frac{1}{2})^{2}$

$\Rightarrow I_2=\int \frac{1}{\sqrt{(x-\frac{9}{2})^{2}-(\frac{1}{2})^{2}}} d x$

$I_2=\log |(x-\frac{9}{2})+\sqrt{x^{2}-9 x+20}|$

Substituting equations (2) and (3) in (1), we obtain

$ \begin{aligned} \int \frac{6 x+7}{\sqrt{x^{2}-9 x+20}} d x & =3[2 \sqrt{x^{2}-9 x+20}]+34 \log [(x-\frac{9}{2})+\sqrt{x^{2}-9 x+20}]+C \\ & =6 \sqrt{x^{2}-9 x+20}+34 \log [(x-\frac{9}{2})+\sqrt{x^{2}-9 x+20}]+C \end{aligned} $

Solution

Let $x+2=A \frac{d}{d x}(4 x-x^{2})+B$

$\Rightarrow x+2=A(4-2 x)+B$

Equating the coefficients of $x$ and constant term on both sides, we obtain $-2 A=1 \Rightarrow A=-\frac{1}{2}$

$4 A+B=2 \Rightarrow B=4$

$\Rightarrow(x+2)=-\frac{1}{2}(4-2 x)+4$

$\therefore \int \frac{x+2}{\sqrt{4 x-x^{2}}} d x=\int \frac{-\frac{1}{2}(4-2 x)+4}{\sqrt{4 x-x^{2}}} d x$ $=-\frac{1}{2} \int \frac{4-2 x}{\sqrt{4 x-x^{2}}} d x+4 \int \frac{1}{\sqrt{4 x-x^{2}}} d x$

Let $I_1=\int \frac{4-2 x}{\sqrt{4 x-x^{2}}} d x$ and $I_2 \int \frac{1}{\sqrt{4 x-x^{2}}} d x$

$\therefore \int \frac{x+2}{\sqrt{4 x-x^{2}}} d x=-\frac{1}{2} I_1+4 I_2$

Then, $I_1=\int \frac{4-2 x}{\sqrt{4 x-x^{2}}} d x$

Let $4 x-x^{2}=1$

$\Rightarrow(4-2 x) d x=d t$

$\Rightarrow l_1=\int _{\sqrt{t}}^{d t}=2 \sqrt{t}=2 \sqrt{4 x-x^{2}}$

$I_2=\int \frac{1}{\sqrt{4 x-x^{2}}} d x$

$\Rightarrow 4 x-x^{2}=-(-4 x+x^{2})$

$=(-4 x+x^{2}+4-4)$

$=4-(x-2)^{2}$

$=(2)^{2}-(x-2)^{2}$

$\therefore I_2=\int \frac{1}{\sqrt{(2)^{2}-(x-2)^{2}}} d x=\sin ^{-1}(\frac{x-2}{2})$

Using equations (2) and (3) in (1), we obtain

$ \begin{aligned} \int \frac{x+2}{\sqrt{4 x-x^{2}}} d x & =-\frac{1}{2}(2 \sqrt{4 x-x^{2}})+4 \sin ^{-1}(\frac{x-2}{2})+C \\ & =-\sqrt{4 x-x^{2}}+4 \sin ^{-1}(\frac{x-2}{2})+C \end{aligned} $

Solution

$$ I=\int \frac{x+2}{\sqrt{x^2+2 x+3}} d x $$

Let, $x+2=\lambda \frac{d}{d x}\left(x^2+2 x+3\right)+\mu$, then $$ x+2=(2 x+2) \lambda+\mu $$

Comparing coefficients of like powers of $x$, we get and $2 \lambda=1$ or $\lambda=1 / 2$ or $$ \begin{aligned} 2 \lambda+\mu =2 \ 1+\mu =2 \end{aligned} $$

or $\quad \mu=1$ Hence, $\quad I=\int \frac{x+2}{\sqrt{x^2+2 x+3}} d x$

$$ \begin{aligned} & \qquad=\int \frac{\frac{1}{2}(2 x+2)+1}{\sqrt{x^2+2 x+3}} d x \\ & =\frac{1}{2} \int \frac{(2 x+2)}{\sqrt{x^2+2 x+3}} d x+\int \frac{1}{\sqrt{x^2+2 x+3}} d x \\ & =\int \frac{d t}{2 \sqrt{t}}+\int \frac{d x}{\sqrt{\left(x^2\right)+2 x+1+2}} \quad \text { where } \quad t=x^2+2 x+3 \end{aligned} $$

$$ \begin{aligned} & \text { or } d t=(2 x+2) d x \\ & =\sqrt{t}+\int \frac{d x}{\sqrt{(x+1)^2+(\sqrt{2})^2}} \\ & =\sqrt{x^2+2 x+3}+\log \left\lvert, \frac{(x+1)}{2}+\sqrt{(x+1)^2+(\sqrt{2})^2}+C\right. \\ & \left.=\sqrt{x^2+2 x+3}+\log \frac{(x+1)}{2}+\sqrt{x^2+2 x+3} \right\rvert,+C \end{aligned} $$

Solution

Let $(x+3)=A \frac{d}{d x}(x^{2}-2 x-5)+B$

$(x+3)=A(2 x-2)+B$

Equating the coefficients of $x$ and constant term on both sides, we obtain

$ \begin{aligned} & 2 A=1 \Rightarrow A=\frac{1}{2} \\ & -2 A+B=3 \Rightarrow B=4 \\ & \therefore(x+3)=\frac{1}{2}(2 x-2)+4 \\ & \Rightarrow \int \frac{x+3}{x^{2}-2 x-5} d x=\int \frac{\frac{1}{2}(2 x-2)+4}{x^{2}-2 x-5} d x \\ & \quad=\frac{1}{2} \int \frac{2 x-2}{x^{2}-2 x-5} d x+4 \int \frac{1}{x^{2}-2 x-5} d x \end{aligned} $

Let $I_1=\int \frac{2 x-2}{x^{2}-2 x-5} d x$ and $I_2=\int \frac{1}{x^{2}-2 x-5} d x$

$\therefore \int \frac{x+3}{(x^{2}-2 x-5)} d x=\frac{1}{2} I_1+4 I_2$

Then, $I_1=\int \frac{2 x-2}{x^{2}-2 x-5} d x$

Let $x^{2}-2 x-5=t$

$\Rightarrow(2 x-2) d x=d t$

$\Rightarrow I_1=\int \frac{d t}{t}=\log |t|=\log |x^{2}-2 x-5|$

$I_2=\int \frac{1}{x^{2}-2 x-5} d x$

$=\int \frac{1}{(x^{2}-2 x+1)-6} d x$

$=\int \frac{1}{(x-1)^{2}+(\sqrt{6})^{2}} d x$

$=\frac{1}{2 \sqrt{6}} \log (\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}})$

Substituting (2) and (3) in (1), we obtain

$ \begin{aligned} \int \frac{x+3}{x^{2}-2 x-5} d x & =\frac{1}{2} \log |x^{2}-2 x-5|+\frac{4}{2 \sqrt{6}} \log |\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}|+C \\ & =\frac{1}{2} \log |x^{2}-2 x-5|+\frac{2}{\sqrt{6}} \log |\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}|+C \end{aligned} $

Solution

Let $5 x+3=A \frac{d}{d x}(x^{2}+4 x+10)+B$

$\Rightarrow 5 x+3=A(2 x+4)+B$

Equating the coefficients of $x$ and constant term, we obtain

$ \begin{aligned} & 2 A=5 \Rightarrow A=\frac{5}{2} \\ & 4 A+B=3 \Rightarrow B=-7 \\ & \therefore 5 x+3=\frac{5}{2}(2 x+4)-7 \\ & \Rightarrow \int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=\int \frac{\frac{5}{2}(2 x+4)-7}{\sqrt{x^{2}+4 x+10}} d x \\ & \quad=\frac{5}{2} \int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x-7 \int \frac{1}{\sqrt{x^{2}+4 x+10}} d x \end{aligned} $

Let $I_1=\int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x$ and $I_2=\int \frac{1}{\sqrt{x^{2}+4 x+10}} d x$

$\therefore \int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=\frac{5}{2} I_1-7 I_2$

Then, $I_1=\int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x$

Let $x^{2}+4 x+10=t$

$\therefore(2 x+4) d x=d t$

$\Rightarrow I_1=\int \frac{d t}{t}=2 \sqrt{t}=2 \sqrt{x^{2}+4 x+10}$

$I_2=\int \frac{1}{\sqrt{x^{2}+4 x+10}} d x$

$=\int \frac{1}{\sqrt{(x^{2}+4 x+4)+6}} d x$

$=\int \frac{1}{(x+2)^{2}+(\sqrt{6})^{2}} d x$

$=\log |(x+2) \sqrt{x^{2}+4 x+10}|$

Using equations (2) and (3) in (1), we obtain

$ \begin{aligned} \int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x & =\frac{5}{2}[2 \sqrt{x^{2}+4 x+10}]-7 \log |(x+2)+\sqrt{x^{2}+4 x+10}|+C \\ & =5 \sqrt{x^{2}+4 x+10}-7 \log (x+2)+\sqrt{x^{2}+4 x+10} \mid+C \end{aligned} $

Solution

$ \begin{aligned} \int \frac{d x}{x^{2}+2 x+2} & =\int \frac{d x}{(x^{2}+2 x+1)+1} \\ & =\int \frac{1}{(x+1)^{2}+(1)^{2}} d x \\ & =[\tan ^{-1}(x+1)]+C \end{aligned} $

Hence, the correct Answer is B.

Solution

$\int \frac{d x}{\sqrt{9 x-4 x^{2}}}$

$=\int \frac{1}{\sqrt{-4(x^{2}-\frac{9}{4} x)}} d x$

$=\int \frac{1}{-4(x^{2}-\frac{9}{4} x+\frac{81}{64}-\frac{81}{64})} d x$

$=\int \frac{1}{\sqrt{-4[(x-\frac{9}{8})^{2}-(\frac{9}{8})^{2}]}} d x$

$=\frac{1}{2} \int \frac{1}{\sqrt{(\frac{9}{8})^{2}-(x-\frac{9}{8})^{2}}} d x$

$=\frac{1}{2}[\sin ^{-1}(\frac{x-\frac{9}{8}}{\frac{9}{8}})]+C$

$(\int \frac{d y}{\sqrt{a^{2}-y^{2}}}=\sin ^{-1} \frac{y}{a}+C)$

$=\frac{1}{2} \sin ^{-1}(\frac{8 x-9}{9})+C$

Hence, the correct Answer is B.

Solution

Let $\frac{x}{(x+1)(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}$

$\Rightarrow x=A(x+2)+B(x+1)$

Equating the coefficients of $x$ and constant term, we obtain

$A+B=1$

$2 A+B=0$

On solving, we obtain

$A=-1$ and $B=2$

$\therefore \frac{x}{(x+1)(x+2)}=\frac{-1}{(x+1)}+\frac{2}{(x+2)}$

$\Rightarrow \int \frac{x}{(x+1)(x+2)} d x=\int \frac{-1}{(x+1)}+\frac{2}{(x+2)} d x$

$=-\log |x+1|+2 \log |x+2|+C$

$=\log (x+2)^{2}-\log |x+1|+C$

$=\log \frac{(x+2)^{2}}{(x+1)}+C$

Solution

Let $\frac{1}{(x+3)(x-3)}=\frac{A}{(x+3)}+\frac{B}{(x-3)}$

$1=A(x-3)+B(x+3)$

Equating the coefficients of $x$ and constant term, we obtain

$A+B=0$

$-3 A+3 B=1$

On solving, we obtain

$A=-\frac{1}{6}$ and $B=\frac{1}{6}$

$\therefore \frac{1}{(x+3)(x-3)}=\frac{-1}{6(x+3)}+\frac{1}{6(x-3)}$

$\Rightarrow \int \frac{1}{(x^{2}-9)} d x=\int(\frac{-1}{6(x+3)}+\frac{1}{6(x-3)}) d x$

$ =-\frac{1}{6} \log |x+3|+\frac{1}{6} \log |x-3|+C $

$ =\frac{1}{6} \log |\frac{(x-3)}{(x+3)}|+C $

Solution

Let $\frac{3 x-1}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}$

$3 x-1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)$

Substituting $x=1,2$, and 3 respectively in equation (1), we obtain

$A=1, B=-5$, and $C=4$

$\therefore \frac{3 x-1}{(x-1)(x-2)(x-3)}=\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)}$

$\Rightarrow \int \frac{3 x-1}{(x-1)(x-2)(x-3)} d x=\int{\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)}} d x$

$=\log |x-1|-5 \log |x-2|+4 \log |x-3|+C$

Solution

Let $\frac{x}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}$

$x=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)$

Substituting $x=1,2$, and 3 respectively in equation (1), we obtain

$A=\frac{1}{2}, B=-2$, and $C=\frac{3}{2}$

$\therefore \frac{x}{(x-1)(x-2)(x-3)}=\frac{1}{2(x-1)}-\frac{2}{(x-2)}+\frac{3}{2(x-3)}$

$\Rightarrow \int \frac{x}{(x-1)(x-2)(x-3)} d x=\int{\frac{1}{2(x-1)}-\frac{2}{(x-2)}+\frac{3}{2(x-3)}} d x$

$ =\frac{1}{2} \log |x-1|-2 \log |x-2|+\frac{3}{2} \log |x-3|+C $

Solution

Let $\frac{2 x}{x^{2}+3 x+2}=\frac{A}{(x+1)}+\frac{B}{(x+2)}$

$2 x=A(x+2)+B(x+1)$

Substituting $x=-1$ and -2 in equation (1), we obtain

$A=-2$ and $B=4$ $\therefore \frac{2 x}{(x+1)(x+2)}=\frac{-2}{(x+1)}+\frac{4}{(x+2)}$

$\Rightarrow \int \frac{2 x}{(x+1)(x+2)} d x=\int{\frac{4}{(x+2)}-\frac{2}{(x+1)}} d x$

$ =4 \log |x+2|-2 \log |x+1|+C $

Solution

It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing $(1-x^{2})$ by $x(1-2 x)$, we obtain

$\frac{1-x^{2}}{x(1-2 x)}=\frac{1}{2}+\frac{1}{2}(\frac{2-x}{x(1-2 x)})$

Let $\frac{2-x}{x(1-2 x)}=\frac{A}{x}+\frac{B}{(1-2 x)}$

$\Rightarrow(2-x)=A(1-2 x)+B x$

Substituting $x=0$ and $\frac{1}{2}$ in equation (1), we obtain

$A=2$ and $B=3$

$\therefore \frac{2-x}{x(1-2 x)}=\frac{2}{x}+\frac{3}{1-2 x}$

Substituting in equation (1), we obtain

$ \begin{aligned} & \frac{1-x^{2}}{x(1-2 x)}=\frac{1}{2}+\frac{1}{2}{\frac{2}{x}+\frac{3}{(1-2 x)}} \\ & \Rightarrow \int \frac{1-x^{2}}{x(1-2 x)} d x=\int{\frac{1}{2}+\frac{1}{2}(\frac{2}{x}+\frac{3}{1-2 x})} d x \\ & =\frac{x}{2}+\log |x|+\frac{3}{2(-2)} \log |1-2 x|+C \\ & =\frac{x}{2}+\log |x|-\frac{3}{4} \log |1-2 x|+C \end{aligned} $

Solution

Let $\frac{x}{(x^{2}+1)(x-1)}=\frac{A x+B}{(x^{2}+1)}+\frac{C}{(x-1)}$

$x=(A x+B)(x-1)+C(x^{2}+1)$

$x=A x^{2}-A x+B x-B+C x^{2}+C$

Equating the coefficients of $x^{2}, x$, and constant term, we obtain

$A+C=0$

$-A+B=1$

$-B+C=0$

On solving these equations, we obtain

$A=-\frac{1}{2}, B=\frac{1}{2}$, and $C=\frac{1}{2}$

From equation (1), we obtain

$ \begin{aligned} & \therefore \frac{x}{(x^{2}+1)(x-1)}=\frac{(-\frac{1}{2} x+\frac{1}{2})}{x^{2}+1}+\frac{\frac{1}{2}}{(x-1)} \\ & \Rightarrow \int \frac{x}{(x^{2}+1)(x-1)}=-\frac{1}{2} \int \frac{x}{x^{2}+1} d x+\frac{1}{2} \int \frac{1}{x^{2}+1} d x+\frac{1}{2} \int \frac{1}{x-1} d x \\ & \quad=-\frac{1}{4} \int \frac{2 x}{x^{2}+1} d x+\frac{1}{2} \tan ^{-1} x+\frac{1}{2} \log |x-1|+C \end{aligned} $

Consider $\int \frac{2 x}{x^{2}+1} d x$, let $(x^{2}+1)=t \Rightarrow 2 x d x=d t$

$\Rightarrow \int \frac{2 x}{x^{2}+1} d x=\int \frac{d t}{t}=\log |t|=\log |x^{2}+1|$

$ \begin{aligned} \therefore \int \frac{x}{(x^{2}+1)(x-1)} & =-\frac{1}{4} \log |x^{2}+1|+\frac{1}{2} \tan ^{-1} x+\frac{1}{2} \log |x-1|+C \\ & =\frac{1}{2} \log |x-1|-\frac{1}{4} \log |x^{2}+1|+\frac{1}{2} \tan ^{-1} x+C \end{aligned} $

Solution

Let $\frac{x}{(x-1)^{2}(x+2)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^{2}}+\frac{C}{(x+2)}$

$x=A(x-1)(x+2)+B(x+2)+C(x-1)^{2}$

Substituting $x=1$, we obtain

$B=\frac{1}{3}$

Equating the coefficients of $x^{2}$ and constant term, we obtain

$A+C=0$

$-2 A+2 B+C=0$

On solving, we obtain

$ \begin{aligned} & A=\frac{2}{9} \text{ and } C=\frac{-2}{9} \\ & \begin{aligned} & \therefore \frac{x}{(x-1)^{2}(x+2)}=\frac{2}{9(x-1)}+\frac{1}{3(x-1)^{2}}-\frac{2}{9(x+2)} \\ & \Rightarrow \int \frac{x}{(x-1)^{2}(x+2)} d x=\frac{2}{9} \int \frac{1}{(x-1)} d x+\frac{1}{3} \int \frac{1}{(x-1)^{2}} d x-\frac{2}{9} \int \frac{1}{(x+2)} d x \\ &=\frac{2}{9} \log |x-1|+\frac{1}{3}(\frac{-1}{x-1})-\frac{2}{9} \log |x+2|+C \\ &=\frac{2}{9} \log |\frac{x-1}{x+2}|-\frac{1}{3(x-1)}+C \end{aligned} \end{aligned} $

Solution

$\frac{3 x+5}{x^{3}-x^{2}-x+1}=\frac{3 x+5}{(x-1)^{2}(x+1)}$

Let $\frac{3 x+5}{(x-1)^{2}(x+1)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^{2}}+\frac{C}{(x+1)}$

$3 x+5=A(x-1)(x+1)+B(x+1)+C(x-1)^{2}$

$3 x+5=A(x^{2}-1)+B(x+1)+C(x^{2}+1-2 x)$

Substituting $x=1$ in equation (1), we obtain

$B=4$

Equating the coefficients of $x^{2}$ and $x$, we obtain

$A+C=0$

$B-2 C=3$

On solving, we obtain

$A=-\frac{1}{2}$ and $C=\frac{1}{2}$ $\therefore \frac{3 x+5}{(x-1)^{2}(x+1)}=\frac{-1}{2(x-1)}+\frac{4}{(x-1)^{2}}+\frac{1}{2(x+1)}$

$\Rightarrow \int \frac{3 x+5}{(x-1)^{2}(x+1)} d x=-\frac{1}{2} \int \frac{1}{x-1} d x+4 \int \frac{1}{(x-1)^{2}} d x+\frac{1}{2} \int \frac{1}{(x+1)} d x$

$=-\frac{1}{2} \log |x-1|+4(\frac{-1}{x-1})+\frac{1}{2} \log |x+1|+C$

$=\frac{1}{2} \log |\frac{x+1}{x-1}|-\frac{4}{(x-1)}+C$

Solution

$\frac{2 x-3}{(x^{2}-1)(2 x+3)}=\frac{2 x-3}{(x+1)(x-1)(2 x+3)}$

Let $\frac{2 x-3}{(x+1)(x-1)(2 x+3)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{C}{(2 x+3)}$

$\Rightarrow(2 x-3)=A(x-1)(2 x+3)+B(x+1)(2 x+3)+C(x+1)(x-1)$

$\Rightarrow(2 x-3)=A(2 x^{2}+x-3)+B(2 x^{2}+5 x+3)+C(x^{2}-1)$

$\Rightarrow(2 x-3)=(2 A+2 B+C) x^{2}+(A+5 B) x+(-3 A+3 B-C)$

Equating the coefficients of $x^{2}$ and $x$, we obtain

$B=-\frac{1}{10}, A=\frac{5}{2}$, and $C=-\frac{24}{5}$ $\therefore \frac{2 x-3}{(x+1)(x-1)(2 x+3)}=\frac{5}{2(x+1)}-\frac{1}{10(x-1)}-\frac{24}{5(2 x+3)}$

$\Rightarrow \int \frac{2 x-3}{(x^{2}-1)(2 x+3)} d x=\frac{5}{2} \int \frac{1}{(x+1)} d x-\frac{1}{10} \int \frac{1}{x-1} d x-\frac{24}{5} \int \frac{1}{(2 x+3)} d x$

$=\frac{5}{2} \log |x+1|-\frac{1}{10} \log |x-1|-\frac{24}{5 \times 2} \log |2 x+3|$

$=\frac{5}{2} \log |x+1|-\frac{1}{10} \log |x-1|-\frac{12}{5} \log |2 x+3|+C$

Solution

$\frac{5 x}{(x+1)(x^{2}-4)}=\frac{5 x}{(x+1)(x+2)(x-2)}$

Let $\frac{5 x}{(x+1)(x+2)(x-2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}+\frac{C}{(x-2)}$

$5 x=A(x+2)(x-2)+B(x+1)(x-2)+C(x+1)(x+2)$

Substituting $x=-1,-2$, and 2 respectively in equation (1), we obtain

$ \begin{aligned} & A=\frac{5}{3}, B=-\frac{5}{2}, \text{ and } C=\frac{5}{6} \\ & \begin{aligned} \therefore \frac{5 x}{(x+1)(x+2)(x-2)} & =\frac{5}{3(x+1)}-\frac{5}{2(x+2)}+\frac{5}{6(x-2)} \\ \Rightarrow \int \frac{5 x}{(x+1)(x^{2}-4)} d x & =\frac{5}{3} \int \frac{1}{(x+1)} d x-\frac{5}{2} \int \frac{1}{(x+2)} d x+\frac{5}{6} \int \frac{1}{(x-2)} d x \\ & =\frac{5}{3} \log |x+1|-\frac{5}{2} \log |x+2|+\frac{5}{6} \log |x-2|+C \end{aligned} \end{aligned} $

Solution

It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing $(x^{3}+x+1)$ by $x^{2}-1$, we obtain

$\frac{x^{3}+x+1}{x^{2}-1}=x+\frac{2 x+1}{x^{2}-1}$

Let $\frac{2 x+1}{x^{2}-1}=\frac{A}{(x+1)}+\frac{B}{(x-1)}$

$2 x+1=A(x-1)+B(x+1)$

Substituting $x=1$ and -1 in equation (1), we obtain

$ A=\frac{1}{2} \text{ and } B=\frac{3}{2} $

$\therefore \frac{x^{3}+x+1}{x^{2}-1}=x+\frac{1}{2(x+1)}+\frac{3}{2(x-1)}$

$\Rightarrow \int \frac{x^{3}+x+1}{x^{2}-1} d x=\int x d x+\frac{1}{2} \int \frac{1}{(x+1)} d x+\frac{3}{2} \int \frac{1}{(x-1)} d x$

$ =\frac{x^{2}}{2}+\frac{1}{2} \log |x+1|+\frac{3}{2} \log |x-1|+C $

Solution

Let $\frac{2}{(1-x)(1+x^{2})}=\frac{A}{(1-x)}+\frac{B x+C}{(1+x^{2})}$

$2=A(1+x^{2})+(B x+C)(1-x)$

$2=A+A x^{2}+B x-B x^{2}+C-C x$

Equating the coefficient of $x^{2}, x$, and constant term, we obtain

$A-B=0$

$B-C=0$

$A+C=2$

On solving these equations, we obtain

$A=1, B=1$, and $C=1$

$\therefore \frac{2}{(1-x)(1+x^{2})}=\frac{1}{1-x}+\frac{x+1}{1+x^{2}}$

$\Rightarrow \int \frac{2}{(1-x)(1+x^{2})} d x=\int \frac{1}{1-x} d x+\int \frac{x}{1+x^{2}} d x+\int \frac{1}{1+x^{2}} d x$

$=-\int \frac{1}{x-1} d x+\frac{1}{2} \int \frac{2 x}{1+x^{2}} d x+\int \frac{1}{1+x^{2}} d x$

$=-\log |x-1|+\frac{1}{2} \log |1+x^{2}|+\tan ^{-1} x+C$

Solution

Let $\frac{3 x-1}{(x+2)^{2}}=\frac{A}{(x+2)}+\frac{B}{(x+2)^{2}}$

$\Rightarrow 3 x-1=A(x+2)+B$

Equating the coefficient of $x$ and constant term, we obtain $A=3$

$2 A+B=-1 \Rightarrow B=-7$ $\therefore \frac{3 x-1}{(x+2)^{2}}=\frac{3}{(x+2)}-\frac{7}{(x+2)^{2}}$

$\Rightarrow \int \frac{3 x-1}{(x+2)^{2}} d x=3 \int \frac{1}{(x+2)} d x-7 \int \frac{x}{(x+2)^{2}} d x$

$=3 \log |x+2|-7(\frac{-1}{(x+2)})+C$

$=3 \log |x+2|+\frac{7}{(x+2)}+C$

Solution

$\frac{1}{(x^{4}-1)}=\frac{1}{(x^{2}-1)(x^{2}+1)}=\frac{1}{(x+1)(x-1)(1+x^{2})}$

Let $\frac{1}{(x+1)(x-1)(1+x^{2})}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{C x+D}{(x^{2}+1)}$

$1=A(x-1)(x^{2}+1)+B(x+1)(x^{2}+1)+(C x+D)(x^{2}-1)$

$1=A(x^{3}+x-x^{2}-1)+B(x^{3}+x+x^{2}+1)+C x^{3}+D x^{2}-C x-D$

$1=(A+B+C) x^{3}+(-A+B+D) x^{2}+(A+B-C) x+(-A+B-D)$

Equating the coefficient of $x^{3}, x^{2}, x$, and constant term, we obtain

$A+B+C=0$

$-A+B+D=0$

$A+B-C=0$

$-A+B-D=1$

On solving these equations, we obtain

$A=-\frac{1}{4}, B=\frac{1}{4}, C=0$, and $D=-\frac{1}{2}$ $\therefore \frac{1}{x^{4}-1}=\frac{-1}{4(x+1)}+\frac{1}{4(x-1)}-\frac{1}{2(x^{2}+1)}$

$\Rightarrow \int \frac{1}{x^{4}-1} d x=-\frac{1}{4} \log |x-1|+\frac{1}{4} \log |x-1|-\frac{1}{2} \tan ^{-1} x+C$

$ =\frac{1}{4} \log |\frac{x-1}{x+1}|-\frac{1}{2} \tan ^{-1} x+C $

Solution

$\frac{1}{x(x^{n}+1)}$

Multiplying numerator and denominator by $x^{n-1}$, we obtain

$\frac{1}{x(x^{n}+1)}=\frac{x^{n-1}}{x^{n-1} x(x^{n}+1)}=\frac{x^{n-1}}{x^{n}(x^{n}+1)}$

Let $x^{n}=t \Rightarrow x^{n-1} d x=d t$

$\therefore \int \frac{1}{x(x^{n}+1)} d x=\int \frac{x^{n-1}}{x^{n}(x^{n}+1)} d x=\frac{1}{n} \int \frac{1}{t(t+1)} d t$

Let $\frac{1}{t(t+1)}=\frac{A}{t}+\frac{B}{(t+1)}$

$1=A(1+t)+B t$

Substituting $t=0,-1$ in equation (1), we obtain

$A=1$ and $B=-1$

$\therefore \frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{(1+t)}$ $\Rightarrow \int \frac{1}{x(x^{n}+1)} d x=\frac{1}{n} \int{\frac{1}{t}-\frac{1}{(t+1)}} d x$

$=\frac{1}{n}[\log |t|-\log |t+1|]+C$

$=-\frac{1}{n}[\log |x^{n}|-\log |x^{n}+1|]+C$

$=\frac{1}{n} \log |\frac{x^{n}}{x^{n}+1}|+C$

Solution

$\frac{\cos x}{(1-\sin x)(2-\sin x)}$

Let $\sin x=t \Rightarrow \cos x d x=d t$

$\therefore \int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x=\int \frac{d t}{(1-t)(2-t)}$

Let $\frac{1}{(1-t)(2-t)}=\frac{A}{(1-t)}+\frac{B}{(2-t)}$

$1=A(2-t)+B(1-t)$

Substituting $t=2$ and then $t=1$ in equation (1), we obtain

$A=1$ and $B=-1$

$\therefore \frac{1}{(1-t)(2-t)}=\frac{1}{(1-t)}-\frac{1}{(2-t)}$

$ \begin{aligned} \Rightarrow \int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x & =\int{\frac{1}{1-t}-\frac{1}{(2-t)}} d t \\ & =-\log |1-t|+\log |2-t|+C \\ & =\log |\frac{2-t}{1-t}|+C \\ & =\log |\frac{2-\sin x}{1-\sin x}|+C \end{aligned} $

Solution

$\frac{(x^{2}+1)(x^{2}+2)}{(x^{2}+3)(x^{2}+4)}=1-\frac{(4 x^{2}+10)}{(x^{2}+3)(x^{2}+4)}$

Let $\frac{4 x^{2}+10}{(x^{2}+3)(x^{2}+4)}=\frac{A x+B}{(x^{2}+3)}+\frac{C x+D}{(x^{2}+4)}$

$4 x^{2}+10=(A x+B)(x^{2}+4)+(C x+D)(x^{2}+3)$

$4 x^{2}+10=A x^{3}+4 A x+B x^{2}+4 B+C x^{3}+3 C x+D x^{2}+3 D$

$4 x^{2}+10=(A+C) x^{3}+(B+D) x^{2}+(4 A+3 C) x+(4 B+3 D)$

Equating the coefficients of $x^{3}, x^{2}, x$, and constant term, we obtain

$A+C=0$

$B+D=4$

$4 A+3 C=0$

$4 B+3 D=10$

On solving these equations, we obtain

$A=0, B=-2, C=0$, and $D=6$

$\therefore \frac{4 x^{2}+10}{(x^{2}+3)(x^{2}+4)}=\frac{-2}{(x^{2}+3)}+\frac{6}{(x^{2}+4)}$ $\frac{(x^{2}+1)(x^{2}+2)}{(x^{2}+3)(x^{2}+4)}=1-(\frac{-2}{(x^{2}+3)}+\frac{6}{(x^{2}+4)})$

$\Rightarrow \int \frac{(x^{2}+1)(x^{2}+2)}{(x^{2}+3)(x^{2}+4)} d x=\int{1+\frac{2}{(x^{2}+3)}-\frac{6}{(x^{2}+4)}} d x$

$={1+\frac{2}{x^{2}+(\sqrt{3})^{2}}-\frac{6}{x^{2}+2^{2}}}$

$=x+2(\frac{1}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}})-6(\frac{1}{2} \tan ^{-1} \frac{x}{2})+C$

$=x+\frac{2}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}-3 \tan ^{-1} \frac{x}{2}+C$

Solution

$\frac{2 x}{(x^{2}+1)(x^{2}+3)}$

Let $x^{2}=t \Rightarrow 2 x d x=d t$

$\therefore \int \frac{2 x}{(x^{2}+1)(x^{2}+3)} d x=\int \frac{d t}{(t+1)(t+3)}$

Let $\frac{1}{(t+1)(t+3)}=\frac{A}{(t+1)}+\frac{B}{(t+3)}$

$1=A(t+3)+B(t+1)$

Substituting $t=-3$ and $t=-1$ in equation (1), we obtain $A=\frac{1}{2}$ and $B=-\frac{1}{2}$

$\therefore \frac{1}{(t+1)(t+3)}=\frac{1}{2(t+1)}-\frac{1}{2(t+3)}$

$\Rightarrow \int \frac{2 x}{(x^{2}+1)(x^{2}+3)} d x=\int{\frac{1}{2(t+1)}-\frac{1}{2(t+3)}} d t$

$=\frac{1}{2} \log |(t+1)|-\frac{1}{2} \log |t+3|+C$

$=\frac{1}{2} \log |\frac{t+1}{t+3}|+C$

$=\frac{1}{2} \log |\frac{x^{2}+1}{x^{2}+3}|+C$

Solution

$\frac{1}{x(x^{4}-1)}$

Multiplying numerator and denominator by $x^{3}$, we obtain

$ \begin{aligned} & \frac{1}{x(x^{4}-1)}=\frac{x^{3}}{x^{4}(x^{4}-1)} \\ & \therefore \int \frac{1}{x(x^{4}-1)} d x=\int \frac{x^{3}}{x^{4}(x^{4}-1)} d x \end{aligned} $

Let $x^{4}=t \Rightarrow 4 x^{3} d x=d t$

$\therefore \int \frac{1}{x(x^{4}-1)} d x=\frac{1}{4} \int \frac{d t}{t(t-1)}$

Let $\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{(t-1)}$

$1=A(t-1)+B t$

Substituting $t=0$ and 1 in (1), we obtain

$A=-1$ and $B=1$

$\Rightarrow \frac{1}{t(t+1)}=\frac{-1}{t}+\frac{1}{t-1}$

$\Rightarrow \int \frac{1}{x(x^{4}-1)} d x=\frac{1}{4} \int{\frac{-1}{t}+\frac{1}{t-1}} d t$

$=\frac{1}{4}[-\log |t|+\log |t-1|]+C$

$=\frac{1}{4} \log |\frac{t-1}{t}|+C$

$=\frac{1}{4} \log |\frac{x^{4}-1}{x^{4}}|+C$

Solution

$\frac{1}{(e^{x}-1)}$

Let $e^{x}=t \Rightarrow e^{x} d x=d t$

$\Rightarrow \int \frac{1}{e^{x}-1} d x=\int \frac{1}{t-1} \times \frac{d t}{t}=\int \frac{1}{t(t-1)} d t$

Let $\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}$

$1=A(t-1)+B t$

Substituting $t=1$ and $t=0$ in equation (1), we obtain

$A=-1$ and $B=1$

$\therefore \frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{t-1}$

$\Rightarrow \int \frac{1}{t(t-1)} d t=\log |\frac{t-1}{t}|+C$

$ =\log |\frac{e^{x}-1}{e^{x}}|+C $

Solution

Let $\frac{x}{(x-1)(x-2)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}$

$x=A(x-2)+B(x-1)$

Substituting $x=1$ and 2 in (1), we obtain

$A=-1$ and $B=2$ $\therefore \frac{x}{(x-1)(x-2)}=-\frac{1}{(x-1)}+\frac{2}{(x-2)}$

$\Rightarrow \int \frac{x}{(x-1)(x-2)} d x=\int{\frac{-1}{(x-1)}+\frac{2}{(x-2)}} d x$

$ =-\log |x-1|+2 \log |x-2|+C $

$ =\log |\frac{(x-2)^{2}}{x-1}|+C $

Hence, the correct Answer is B.

Solution

Let $\frac{1}{x(x^{2}+1)}=\frac{A}{x}+\frac{B x+C}{x^{2}+1}$

$1=A(x^{2}+1)+(B x+C) x$

Equating the coefficients of $x^{2}, x$, and constant term, we obtain

$A+B=0$

$C=0$

$A=1$

On solving these equations, we obtain

$A=1, B=-1$, and $C=0$ $\therefore \frac{1}{x(x^{2}+1)}=\frac{1}{x}+\frac{-x}{x^{2}+1}$

$\Rightarrow \int \frac{1}{x(x^{2}+1)} d x=\int{\frac{1}{x}-\frac{x}{x^{2}+1}} d x$

$ =\log |x|-\frac{1}{2} \log |x^{2}+1|+C $

Hence, the correct Answer is A.

Solution

Let $I=\int x \sin x d x$

Taking $x$ as first function and $\sin x$ as second function and integrating by parts, we obtain

$ \begin{aligned} I & =x \int \sin x d x-\int\{(\frac{d}{d x} x) \int \sin x d x\} d x \\ & =x(-\cos x)-\int 1 \cdot(-\cos x) d x \\ & =-x \cos x+\sin x+C \end{aligned} $

Solution

Let $I=\int x \sin 3 x d x$

Taking $x$ as first function and $\sin 3 x$ as second function and integrating by parts, we obtain

$ \begin{aligned} I & =x \int \sin 3 x d x-\int\{(\frac{d}{d x} x) \int \sin 3 x d x\} \\ & =x(\frac{-\cos 3 x}{3})-\int 1 \cdot(\frac{-\cos 3 x}{3}) d x \\ & =\frac{-x \cos 3 x}{3}+\frac{1}{3} \int \cos 3 x d x \\ & =\frac{-x \cos 3 x}{3}+\frac{1}{9} \sin 3 x+C \end{aligned} $

Solution

$x^{2} e^{x}$

Answer

Let $I=\int x^{2} e^{x} d x$

Taking $x^{2}$ as first function and $e^{x}$ as second function and integrating by parts, we obtain

$ \begin{aligned} I & =x^{2} \int e^{x} d x-\int\{(\frac{d}{d x} x^{2}) \int e^{x} d x\} d x \\ & =x^{2} e^{x}-\int 2 x \cdot e^{x} d x \\ & =x^{2} e^{x}-2 \int x \cdot e^{x} d x \end{aligned} $

Again integrating by parts, we obtain

$ \begin{aligned} & =x^{2} e^{x}-2[x \cdot \int e^{x} d x-\int\{(\frac{d}{d x} x) \cdot \int e^{x} d x\} d x] \\ & =x^{2} e^{x}-2[x e^{x}-\int e^{x} d x] \\ & =x^{2} e^{x}-2[x e^{x}-e^{x}] \\ & =x^{2} e^{x}-2 x e^{x}+2 e^{x}+C \\ & =e^{x}(x^{2}-2 x+2)+C \end{aligned} $

Solution

Let $I=\int x \log x d x$

Taking $\log x$ as first function and $x$ as second function and integrating by parts, we obtain

$ \begin{aligned} I & =\log x \int x d x-\int\{(\frac{d}{d x} \log x) \int x d x\} d x \\ & =\log x \cdot \frac{x^{2}}{2}-\int \frac{1}{x} \cdot \frac{x^{2}}{2} d x \\ & =\frac{x^{2} \log x}{2}-\int \frac{x}{2} d x \\ & =\frac{x^{2} \log x}{2}-\frac{x^{2}}{4}+C \end{aligned} $

Solution

Let $I=\int x \log 2 x d x$

Taking $\log 2 x$ as first function and $x$ as second function and integrating by parts, we obtain

$ \begin{aligned} I & =\log 2 x \int x d x-\int\{(\frac{d}{d x} 2 \log x) \int x d x\} d x \\ & =\log 2 x \cdot \frac{x^{2}}{2}-\int \frac{2}{2 x} \cdot \frac{x^{2}}{2} d x \\ & =\frac{x^{2} \log 2 x}{2}-\int \frac{x}{2} d x \\ & =\frac{x^{2} \log 2 x}{2}-\frac{x^{2}}{4}+C \end{aligned} $

Solution

Let $I=\int x^{2} \log x d x$

Taking $\log x$ as first function and $x^{2}$ as second function and integrating by parts, we obtain

$ \begin{aligned} I & =\log x \int x^{2} d x-\int\{(\frac{d}{d x} \log x) \int x^{2} d x\} d x \\ & =\log x(\frac{x^{3}}{3})-\int \frac{1}{x} \cdot \frac{x^{3}}{3} d x \\ & =\frac{x^{3} \log x}{3}-\int \frac{x^{2}}{3} d x \\ & =\frac{x^{3} \log x}{3}-\frac{x^{3}}{9}+C \end{aligned} $

Solution

Let $I=\int x \sin ^{-1} x d x$

Taking $\sin ^{-1} x$ as first function and $x$ as second function and integrating by parts, we obtain

$ \begin{aligned} I & =\sin ^{-1} x \int x d x-\int\{(\frac{d}{d x} \sin ^{-1} x) \int x d x\} d x \\ & =\sin ^{-1} x(\frac{x^{2}}{2})-\int \frac{1}{\sqrt{1-x^{2}}} \cdot \frac{x^{2}}{2} d x \\ & =\frac{x^{2} \sin ^{-1} x}{2}+\frac{1}{2} \int \frac{-x^{2}}{\sqrt{1-x^{2}}} d x \\ & =\frac{x^2 \sin ^{-1} x}{2}+\frac{1}{2} \int [\frac{1-x^2}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}} ]d x \\ & =\frac{x^2 \sin ^{-1} x}{2}+\frac{1}{2} \int [\sqrt{1-x^2}-\frac{1}{\sqrt{1-x^2}} ] d x\\ & =\frac{x^{2} \sin ^{-1} x}{2}+\frac{1}{2}\{\int \sqrt{1-x^{2}} d x-\int \frac{1}{\sqrt{1-x^{2}}} d x\} \\ & =\frac{x^{2} \sin ^{-1} x}{2}+\frac{1}{2}\{\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x-\sin ^{-1} x\}+C \\ & =\frac{x^{2} \sin ^{-1} x}{2}+\frac{x}{4} \sqrt{1-x^{2}}+\frac{1}{4} \sin ^{-1} x-\frac{1}{2} \sin ^{-1} x+C \\ & =\frac{1}{4}(2 x^{2}-1) \sin ^{-1} x+\frac{x}{4} \sqrt{1-x^{2}}+C \end{aligned} $

Solution

Let $I=\int x \tan ^{-1} x d x$

Taking $\tan ^{-1} x$ as first function and $x$ as second function and integrating by parts, we obtain

$ \begin{aligned} I & =\tan ^{-1} x \int x d x-\int\{(\frac{d}{d x} \tan ^{-1} x) \int x d x\} d x \\ & =\tan ^{-1} x(\frac{x^{2}}{2})-\int \frac{1}{1+x^{2}} \cdot \frac{x^{2}}{2} d x \\ & =\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2} \int \frac{x^{2}}{1+x^{2}} d x \\ & =\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2} \int(\frac{x^{2}+1}{1+x^{2}}-\frac{1}{1+x^{2}}) d x \\ & =\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2} \int(1-\frac{1}{1+x^{2}}) d x \\ & =\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}(x-\tan ^{-1} x)+C \\ & =\frac{x^{2}}{2} \tan ^{-1} x-\frac{x}{2}+\frac{1}{2} \tan ^{-1} x+C \end{aligned} $

Solution

Let $I=\int x \cos ^{-1} x d x$

Taking $\cos ^{-1} x$ as first function and $x$ as second function and integrating by parts, we obtain

$$ \begin{align*} I & =\cos ^{-1} x \int x d x-\int\{(\frac{d}{d x} \cos ^{-1} x) \int x d x\} d x \\ & =\cos ^{-1} x \frac{x^{2}}{2}-\int \frac{-1}{\sqrt{1-x^{2}}} \cdot \frac{x^{2}}{2} d x \\ & =\frac{x^{2} \cos ^{-1} x}{2}-\frac{1}{2} \int \frac{1-x^{2}-1}{\sqrt{1-x^{2}}} d x \\ & =\frac{x^{2} \cos ^{-1} x}{2}-\frac{1}{2} \int\{\sqrt{1-x^{2}}+(\frac{-1}{\sqrt{1-x^{2}}})\} d x \\ & =\frac{x^{2} \cos ^{-1} x}{2}-\frac{1}{2} \int \sqrt{1-x^{2}} d x-\frac{1}{2} \int(\frac{-1}{\sqrt{1-x^{2}}}) d x \\ & =\frac{x^{2} \cos ^{-1} x}{2}-\frac{1}{2} I_1-\frac{1}{2} \cos ^{-1} x \tag{1} \end{align*} $$

where, $I_1=\int \sqrt{1-x^{2}} d x$

$\Rightarrow I_1=x \sqrt{1-x^{2}}-\int \frac{d}{d x} \sqrt{1-x^{2}} \int x d x$

$\Rightarrow I_1=x \sqrt{1-x^{2}}-\int \frac{-2 x}{2 \sqrt{1-x^{2}}} \cdot x d x$

$\Rightarrow I_1=x \sqrt{1-x^{2}}-\int \frac{-x^{2}}{\sqrt{1-x^{2}}} d x$

$\Rightarrow I_1=x \sqrt{1-x^{2}}-\int \frac{1-x^{2}-1}{\sqrt{1-x^{2}}} d x$

$\Rightarrow I_1=x \sqrt{1-x^{2}}-\{\int \sqrt{1-x^{2}} d x+\int \frac{-d x}{\sqrt{1-x^{2}}}\}$

$\Rightarrow I_1=x \sqrt{1-x^{2}}-\{I_1+\cos ^{-1} x\}$

$\Rightarrow 2 I_1=x \sqrt{1-x^{2}}-\cos ^{-1} x$

$\therefore I_1=\frac{x}{2} \sqrt{1-x^{2}}-\frac{1}{2} \cos ^{-1} x$

Substituting in (1), we obtain

$ \begin{aligned} I & =\frac{x \cos ^{-1} x}{2}-\frac{1}{2}(\frac{x}{2} \sqrt{1-x^{2}}-\frac{1}{2} \cos ^{-1} x)-\frac{1}{2} \cos ^{-1} x \\ & =\frac{(2 x^{2}-1)}{4} \cos ^{-1} x-\frac{x}{4} \sqrt{1-x^{2}}+C \end{aligned} $

Solution

Let $I=\int(\sin ^{-1} x)^{2} \cdot 1 d x$

Taking $(\sin ^{-1} x)^{2}$ as first function and 1 as second function and integrating by parts, we obtain

$ \begin{aligned} I & =(\sin ^{-1} x) \int 1 d x-\int\{\frac{d}{d x}(\sin ^{-1} x)^{2} \cdot \int 1 \cdot d x\} d x \\ & =(\sin ^{-1} x)^{2} \cdot x-\int \frac{2 \sin ^{-1} x}{\sqrt{1-x^{2}}} \cdot x d x \\ & =x(\sin ^{-1} x)^{2}+\int \sin ^{-1} x \cdot(\frac{-2 x}{\sqrt{1-x^{2}}}) d x \\ & =x(\sin ^{-1} x)^{2}+[\sin ^{-1} x \int \frac{-2 x}{\sqrt{1-x^{2}}} d x-\int\{(\frac{d}{d x} \sin ^{-1} x) \int \frac{-2 x}{\sqrt{1-x^{2}}} d x\} d x] \\ & =x(\sin ^{-1} x)^{2}+[\sin ^{-1} x \cdot 2 \sqrt{1-x^{2}}-\int \frac{1}{\sqrt{1-x^{2}}} \cdot 2 \sqrt{1-x^{2}} d x] \\ & =x(\sin ^{-1} x)^{2}+2 \sqrt{1-x^{2}} \sin ^{-1} x-\int 2 d x \\ & =x(\sin ^{-1} x)^{2}+2 \sqrt{1-x^{2}} \sin ^{-1} x-2 x+C \end{aligned} $

Solution

Let $I=\int \frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}} d x$ $I=\frac{-1}{2} \int \frac{-2 x}{\sqrt{1-x^{2}}} \cdot \cos ^{-1} x d x$

Taking $\cos ^{-1} x$ as first function and $(\frac{-2 x}{\sqrt{1-x^{2}}})$ as second function and integrating by parts, we obtain

$ \begin{aligned} I & =\frac{-1}{2}[\cos ^{-1} x \int \frac{-2 x}{\sqrt{1-x^{2}}} d x-\int\{(\frac{d}{d x} \cos ^{-1} x) \int \frac{-2 x}{\sqrt{1-x^{2}}} d x\} d x] \\ & =\frac{-1}{2}[\cos ^{-1} x \cdot 2 \sqrt{1-x^{2}}-\int \frac{-1}{\sqrt{1-x^{2}}} \cdot 2 \sqrt{1-x^{2}} d x] \\ & =\frac{-1}{2}[2 \sqrt{1-x^{2}} \cos ^{-1} x+\int 2 d x] \\ & =\frac{-1}{2}[2 \sqrt{1-x^{2}} \cos ^{-1} x+2 x]+C \\ & =-[\sqrt{1-x^{2}} \cos ^{-1} x+x]+C \end{aligned} $

Solution

Let $I=\int x \sec ^{2} x d x$

Taking $x$ as first function and $\sec ^{2} x$ as second function and integrating by parts, we obtain

$ \begin{aligned} I & =x \int \sec ^{2} x d x-\int\{\{\frac{d}{d x} x\} \int \sec ^{2} x d x\} d x \\ & =x \tan x-\int 1 \cdot \tan x d x \\ & =x \tan x+\log |\cos x|+C \end{aligned} $

Solution

Let $I=\int 1 \cdot \tan ^{-1} x d x$

Taking $\tan ^{-1} x$ as first function and 1 as second function and integrating by parts, we obtain

$ \begin{aligned} I & =\tan ^{-1} x \int 1 d x-\int\{(\frac{d}{d x} \tan ^{-1} x) \int 1 \cdot d x\} d x \\ & =\tan ^{-1} x \cdot x-\int \frac{1}{1+x^{2}} \cdot x d x \\ & =x \tan ^{-1} x-\frac{1}{2} \int \frac{2 x}{1+x^{2}} d x \\ & =x \tan ^{-1} x-\frac{1}{2} \log |1+x^{2}|+C \\ & =x \tan ^{-1} x-\frac{1}{2} \log (1+x^{2})+C \end{aligned} $

Solution

$I=\int x(\log x)^{2} d x$

Taking $(\log x)^{2}$ as first function and 1 as second function and integrating by parts, we obtain

$ \begin{aligned} I & =(\log x)^{2} \int x d x-\int[\{(\frac{d}{d x} \log x)^{2}\} \int x d x] d x \\ & =\frac{x^{2}}{2}(\log x)^{2}-[\int 2 \log x \cdot \frac{1}{x} \cdot \frac{x^{2}}{2} d x] \\ & =\frac{x^{2}}{2}(\log x)^{2}-\int x \log x d x \end{aligned} $

Again integrating by parts, we obtain

$ \begin{aligned} I & =\frac{x^{2}}{2}(\log x)^{2}-[\log x \int x d x-\int\{(\frac{d}{d x} \log x) \int x d x\} d x] \\ & =\frac{x^{2}}{2}(\log x)^{2}-[\frac{x^{2}}{2}-\log x-\int \frac{1}{x} \cdot \frac{x^{2}}{2} d x] \\ & =\frac{x^{2}}{2}(\log x)^{2}-\frac{x^{2}}{2} \log x+\frac{1}{2} \int x d x \\ & =\frac{x^{2}}{2}(\log x)^{2}-\frac{x^{2}}{2} \log x+\frac{x^{2}}{4}+C \end{aligned} $

Solution

Let $I=\int(x^{2}+1) \log x d x=\int x^{2} \log x d x+\int \log x d x$

Let $I=I_1+I_2 \ldots$ (1)

Where, $I_1=\int x^{2} \log x d x$ and $I_2=\int \log x d x$

$I_1=\int x^{2} \log x d x$

Taking $\log x$ as first function and $x^{2}$ as second function and integrating by parts, we obtain

$$ \begin{align*} I_1 & =\log x-\int x^{2} d x-\int\{(\frac{d}{d x} \log x) \int x^{2} d x\} d x \\ & =\log x \cdot \frac{x^{3}}{3}-\int \frac{1}{x} \cdot \frac{x^{3}}{3} d x \\ & =\frac{x^{3}}{3} \log x-\frac{1}{3}(\int x^{2} d x) \\ & =\frac{x^{3}}{3} \log x-\frac{x^{3}}{9}+C_1 \tag{2}\\ I_2 & =\int \log x d x \end{align*} $$

Taking $\log x$ as first function and 1 as second function and integrating by parts, we obtain

$$ \begin{align*} I_2 & =\log x \int 1 \cdot d x-\int\{(\frac{d}{d x} \log x) \int 1 \cdot d x\} \\ & =\log x \cdot x-\int \frac{1}{x} \cdot x d x \\ & =x \log x-\int 1 d x \\ & =x \log x-x+C_2 \tag{3} \end{align*} $$

Using equations (2) and (3) in (1), we obtain

$ \begin{aligned} I & =\frac{x^{3}}{3} \log x-\frac{x^{3}}{9}+C_1+x \log x-x+C_2 \\ & =\frac{x^{3}}{3} \log x-\frac{x^{3}}{9}+x \log x-x+(C_1+C_2) \\ & =(\frac{x^{3}}{3}+x) \log x-\frac{x^{3}}{9}-x+C \end{aligned} $

Solution

Let $I=\int e^{x}(\sin x+\cos x) d x$

Let $f(x)=\sin x$

$e^x\left(f(x)+f^{\prime}(x)\right)^{\prime}(x)=\cos x$

$e^x\left(f(x)+f^{\prime}(x)\right) {I=\int e^{x}}\{f(x)+f^{\prime}(x)\} d x$

It is known that, $\int e^{x}\{f(x)+f^{\prime}(x)\} d x=e^{x} f(x)+C$

$\therefore I=e^{x} \sin x+C$

Solution

Let $I=\int \frac{x e^{x}}{(1+x)^{2}} d x=\int e^{x}\{\frac{x}{(1+x)^{2}}\} d x$

$=\int e^{x}\{\frac{1+x-1}{(1+x)^{2}}\} d x$

$=\int e^{x}\{\frac{1}{1+x}-\frac{1}{(1+x)^{2}}\} d x$

Let $f(x)=\frac{1}{1+x} e^x\left(f(x)+f^{\prime}(x)\right) f^{\prime}(x)=\frac{-1}{(1+x)^{2}}$

$\Rightarrow \int \frac{x e^{x}}{(1+x)^{2}} d x=\int e^{x}\{f(x)+f^{\prime}(x)\} d x$

It is known that, $e^x\left(f(x)+f^{\prime}(x)\right)$

$\therefore \int \frac{x e^{x}}{(1+x)^{2}} d x=\frac{e^{x}}{1+x}+C$

Solution

$e^{x}(\frac{1+\sin x}{1+\cos x})$

$=e^{x}(\frac{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}})$

$=\frac{e^{x}(\sin \frac{x}{2}+\cos \frac{x}{2})^{2}}{2 \cos ^{2} \frac{x}{2}}$

$=\frac{1}{2} e^{x} \cdot(\frac{\sin \frac{x}{2}+\cos \frac{x}{2}}{\cos \frac{x}{2}})^{2}$

$=\frac{1}{2} e^{x}[\tan \frac{x}{2}+1]^{2}$

$=\frac{1}{2} e^{2}(1+\tan \frac{x}{2})^{2}$

$=\frac{1}{2} e^{x}[1+\tan ^{2} \frac{x}{2}+2 \tan \frac{x}{2}]$

$=\frac{1}{2} e^{x}[\sec ^{2} \frac{x}{2}+2 \tan \frac{x}{2}]$

$\frac{e^{x}(1+\sin x) d x}{(1+\cos x)}=e^{x}[\frac{1}{2} \sec ^{2} \frac{x}{2}+\tan \frac{x}{2}]$

Let $\tan \frac{x}{2}=f(x) \quad f^{\prime}(x)=\frac{1}{2} \sec ^{2} \frac{x}{2}$

It is known that, $\int e^{x}\{f(x)+f^{\prime}(x)\} d x=e^{x} f(x)+C$

From equation (1), we obtain

$\int \frac{e^{x}(1+\sin x)}{(1+\cos x)} d x=e^{x} \tan \frac{x}{2}+C$

Solution

Let $I=\int e^{x}[\frac{1}{x}-\frac{1}{x^{2}}] d x$

Also, let $\frac{1}{x}=f(x) \quad f^{\prime}(x)=\frac{-1}{x^{2}}$

It is known that, $\int e^{x}\{f(x)+f^{\prime}(x)\} d x=e^{x} f(x)+C$

$\therefore I=\frac{e^{x}}{x}+C$

Solution

$ \begin{aligned} & \begin{aligned} \int e^{x}\{\frac{x-3}{(x-1)^{3}}\} d x & =\int e^{x}\{\frac{x-1-2}{(x-1)^{3}}\} d x \\ & =\int e^{x}\{\frac{1}{(x-1)^{2}}-\frac{2}{(x-1)^{3}}\} d x \end{aligned} \\ & \text{ Let } f(x)=\frac{1}{(x-1)^{2}} \quad f^{\prime}(x)=\frac{-2}{(x-1)^{3}} \end{aligned} $

It is known that, $\int e^x\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)+C$

$\therefore \int e^{x}\{\frac{(x-3)}{(x-1)^{2}}\} d x=\frac{e^{x}}{(x-1)^{2}}+C$

Solution

Let $I=\int e^{2 x} \sin x d x$

Integrating by parts, we obtain

$I=\sin x \int e^{2 x} d x-\int\{(\frac{d}{d x} \sin x) \int e^{2 x} d x\} d x$

$\Rightarrow I=\sin x \cdot \frac{e^{2 x}}{2}-\int \cos x \cdot \frac{e^{2 x}}{2} d x$

$\Rightarrow I=\frac{e^{2 x} \sin x}{2}-\frac{1}{2} \int e^{2 x} \cos x d x$

Again integrating by parts, we obtain

$ \begin{aligned} & I=\frac{e^{2 x} \cdot \sin x}{2}-\frac{1}{2}[\cos x \int e^{2 x} d x-\int\{(\frac{d}{d x} \cos x) \int e^{2 x} d x\} d x] \\ & \Rightarrow I=\frac{e^{2 x} \sin x}{2}-\frac{1}{2}[\cos x \cdot \frac{e^{2 x}}{2}-\int(-\sin x) \frac{e^{2 x}}{2} d x] \\ & \Rightarrow I=\frac{e^{2 x} \cdot \sin x}{2}-\frac{1}{2}[\frac{e^{2 x} \cos x}{2}+\frac{1}{2} \int e^{2 x} \sin x d x] \\ & \Rightarrow I=\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}-\frac{1}{4} I \\ & \Rightarrow I+\frac{1}{4} I=\frac{e^{2 x} \cdot \sin x}{2}-\frac{e^{2 x} \cos x}{4} \\ & \Rightarrow \frac{5}{4} I=\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4} \\ & \Rightarrow I=\frac{4}{5}[\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}]+C \\ & \Rightarrow I=\frac{e^{2 x}}{5}[2 \sin x-\cos x]+C \end{aligned} $

Solution

Let $x=\tan \theta \quad \square d x=\sec ^{2} \theta d \theta$ $\therefore \sin ^{-1}(\frac{2 x}{1+x^{2}})=\sin ^{-1}(\frac{2 \tan \theta}{1+\tan ^{2} \theta})=\sin ^{-1}(\sin 2 \theta)=2 \theta$

$\int \sin^{-1} (\frac{2x}{1+x^2})dx=\int 2 \theta \cdot \sec^2 \theta d \theta=2 \int \theta \cdot \sec^2 \theta d \theta$

Integrating by parts, we obtain

$ \begin{aligned} & 2[\theta \cdot \int \sec ^{2} \theta d \theta-\int\{(\frac{d}{d \theta} \theta) \int \sec ^{2} \theta d \theta\} d \theta] \\ & =2[\theta \cdot \tan \theta-\int \tan \theta d \theta] \\ & =2[\theta \tan \theta+\log |\cos \theta|]+C \\ & =2[x \tan ^{-1} x+\log |\frac{1}{\sqrt{1+x^{2}}}|]+C \\ & =2 x \tan ^{-1} x+2 \log (1+x^{2})^{-\frac{1}{2}}+C \\ & =2 x \tan ^{-1} x+2[-\frac{1}{2} \log (1+x^{2})]+C \\ & =2 x \tan ^{-1} x-\log (1+x^{2})+C \end{aligned} $

Solution

Let $I=\int x^{2} e^{x^{3}} d x$

Also, let $x^{3}=t3 x^{2} d x=d t$

$ \begin{aligned} \Rightarrow I & =\frac{1}{3} \int e^{t} d t \\ & =\frac{1}{3}(e^{t})+C \\ & =\frac{1}{3} e^{x^{3}}+C \end{aligned} $

Hence, the correct Answer is A.

Solution

$\int e^{x} \sec x(1+\tan x) d x$

Let $I=\int e^{x} \sec x(1+\tan x) d x=\int e^{x}(\sec x+\sec x \tan x) d x$

Also, let $\sec x=f(x)\Rightarrow {\sec x \tan x=f^{\prime}(x)}$

It is known that, $\int e^{x}\{f(x)+f^{\prime}(x)\} d x=e^{x} f(x)+C$

$\therefore I=e^{x} \sec x+C$

Hence, the correct Answer is B.

Solution

Let $I=\int \sqrt{4-x^{2}} d x=\int \sqrt{(2)^{2}-(x)^{2}} d x$

It is known that, $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}} \frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+C$

$ \begin{aligned} \therefore I & =\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}+C \\ & =\frac{x}{2} \sqrt{4-x^{2}}+2 \sin ^{-1} \frac{x}{2}+C \end{aligned} $

Solution

Let $I=\int \sqrt{1-4 x^{2}} d x=\int \sqrt{(1)^{2}-(2 x)^{2}} d x$

Let $2 x=t \Rightarrow 2 d x=d t$

$\therefore I=\frac{1}{2} \int \sqrt{(1)^{2}-(t)^{2}} d t$

It is known that, $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+C$

$ \begin{aligned} \Rightarrow I & =\frac{1}{2}[\frac{t}{2} \sqrt{1-t^{2}}+\frac{1}{2} \sin ^{-1} t]+C \\ & =\frac{t}{4} \sqrt{1-t^{2}}+\frac{1}{4} \sin ^{-1} t+C \\ & =\frac{2 x}{4} \sqrt{1-4 x^{2}}+\frac{1}{4} \sin ^{-1} 2 x+C \\ & =\frac{x}{2} \sqrt{1-4 x^{2}}+\frac{1}{4} \sin ^{-1} 2 x+C \end{aligned} $

Solution

Let $I=\int \sqrt{x^{2}+4 x+6} d x$

$ \begin{aligned} & =\int \sqrt{x^{2}+4 x+4+2} d x \\ & =\int \sqrt{(x^{2}+4 x+4)+2} d x \\ & =\int \sqrt{(x+2)^{2}+(\sqrt{2})^{2}} d x \end{aligned} $

It is known that, $\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log |x+\sqrt{x^{2}+a^{2}}|+C$

$ \begin{aligned} \therefore I & =\frac{(x+2)}{2} \sqrt{x^{2}+4 x+6}+\frac{2}{2} \log |(x+2)+\sqrt{x^{2}+4 x+6}|+C \\ & =\frac{(x+2)}{2} \sqrt{x^{2}+4 x+6}+\log |(x+2)+\sqrt{x^{2}+4 x+6}|+C \end{aligned} $

Solution

Let $I=\int \sqrt{x^{2}+4 x+1} d x$

$ \begin{aligned} & =\int \sqrt{(x^{2}+4 x+4)-3} d x \\ & =\int \sqrt{(x+2)^{2}-(\sqrt{3})^{2}} d x \end{aligned} $

It is known that, $\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log |x+\sqrt{x^{2}-a^{2}}|+C$

$\therefore I=\frac{(x+2)}{2} \sqrt{x^{2}+4 x+1}-\frac{3}{2} \log |(x+2)+\sqrt{x^{2}+4 x+1}|+C$

Solution

Let $I=\int \sqrt{1-4 x-x^{2}} d x$

$ \begin{aligned} & =\int \sqrt{1-(x^{2}+4 x+4-4)} d x \\ & =\int \sqrt{1+4-(x+2)^{2}} d x \\ & =\int \sqrt{(\sqrt{5})^{2}-(x+2)^{2}} d x \end{aligned} $

It is known that, $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+C$

$\therefore I=\frac{(x+2)}{2} \sqrt{1-4 x-x^{2}}+\frac{5}{2} \sin ^{-1}(\frac{x+2}{\sqrt{5}})+C$

Solution

Let $I=\int \sqrt{x^{2}+4 x-5} d x$

$ \begin{aligned} & =\int \sqrt{(x^{2}+4 x+4)-9} d x \\ & =\int \sqrt{(x+2)^{2}-(3)^{2}} d x \end{aligned} $

It is known that, $\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log |x+\sqrt{x^{2}-a^{2}}|+C$

$\therefore I=\frac{(x+2)}{2} \sqrt{x^{2}+4 x-5}-\frac{9}{2} \log |(x+2)+\sqrt{x^{2}+4 x-5}|+C$

Solution

Let $I=\int \sqrt{1+3 x-x^{2}} d x$

$ \begin{aligned} & =\int \sqrt{1-(x^{2}-3 x+\frac{9}{4}-\frac{9}{4})} d x \\ & =\int \sqrt{(1+\frac{9}{4})-(x-\frac{3}{2})^{2}} d x \\ & =\int \sqrt{(\frac{\sqrt{13}}{2})^{2}-(x-\frac{3}{2})^{2}} d x \end{aligned} $

It is known that, $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+C$

$ \begin{aligned} \therefore I & =\frac{x-\frac{3}{2}}{2} \sqrt{1+3 x-x^{2}}+\frac{13}{4 \times 2} \sin ^{-1}(\frac{x-\frac{3}{2}}{\frac{\sqrt{13}}{2}})+C \\ & =\frac{2 x-3}{4} \sqrt{1+3 x-x^{2}}+\frac{13}{8} \sin ^{-1}(\frac{2 x-3}{\sqrt{13}})+C \end{aligned} $

Solution

Let $I=\int \sqrt{x^{2}+3 x} d x$

$ \begin{aligned} & =\int \sqrt{x^{2}+3 x+\frac{9}{4}-\frac{9}{4}} d x \\ & =\int \sqrt{(x+\frac{3}{2})^{2}-(\frac{3}{2})^{2}} d x \end{aligned} $

It is known that, $\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log |x+\sqrt{x^{2}-a^{2}}|+C$

$ \begin{aligned} \therefore I & =\frac{(x+\frac{3}{2})}{2} \sqrt{x^{2}+3 x}-\frac{9}{2} \log |(x+\frac{3}{2})+\sqrt{x^{2}+3 x}|+C \\ & =\frac{(2 x+3)}{4} \sqrt{x^{2}+3 x}-\frac{9}{8} \log |(x+\frac{3}{2})+\sqrt{x^{2}+3 x}|+C \end{aligned} $

Solution

Let $I=\int \sqrt{1+\frac{x^{2}}{9}} d x=\frac{1}{3} \int \sqrt{9+x^{2}} d x=\frac{1}{3} \int \sqrt{(3)^{2}+x^{2}} d x$

It is known that, $\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log |x+\sqrt{x^{2}+a^{2}}|+C$

$ \begin{aligned} \therefore I & =\frac{1}{3}[\frac{x}{2} \sqrt{x^{2}+9}+\frac{9}{2} \log |x+\sqrt{x^{2}+9}|]+C \\ & =\frac{x}{6} \sqrt{x^{2}+9}+\frac{3}{2} \log |x+\sqrt{x^{2}+9}|+C \end{aligned} $

Solution

It is known that, $\int \sqrt{a^{2}+x^{2}} d x=\frac{x}{2} \sqrt{a^{2}+x^{2}}+\frac{a^{2}}{2} \log |x+\sqrt{x^{2}+a^{2}}|+C$

$ \therefore \int \sqrt{1+x^{2}} d x=\frac{x}{2} \sqrt{1+x^{2}}+\frac{1}{2} \log |x+\sqrt{1+x^{2}}|+C $

Hence, the correct Answer is A.

Solution

Let $I=\int \sqrt{x^{2}-8 x+7} d x$

$ \begin{aligned} & =\int \sqrt{(x^{2}-8 x+16)-9} d x \\ & =\int \sqrt{(x-4)^{2}-(3)^{2}} d x \end{aligned} $

It is known that, $\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log |x+\sqrt{x^{2}-a^{2}}|+C$

$ \therefore I=\frac{(x-4)}{2} \sqrt{x^{2}-8 x+7}-\frac{9}{2} \log |(x-4)+\sqrt{x^{2}-8 x+7}|+C $

Hence, the correct Answer is D.

Solution

Let $I=\int _{-1}^{1}(x+1) d x$

$\int(x+1) d x=\frac{x^{2}}{2}+x=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(1)-F(-1) \\ & =(\frac{1}{2}+1)-(\frac{1}{2}-1) \\ & =\frac{1}{2}+1-\frac{1}{2}+1 \\ & =2 \end{aligned} $

Solution

Let $I=\int_2^{3} \frac{1}{x} d x$

$\int \frac{1}{x} d x=\log |x|=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(3)-F(2) \\ & =\log |3|-\log |2|=\log \frac{3}{2} \end{aligned} $

Solution

Let $I=\int_1^{2}(4 x^{3}-5 x^{2}+6 x+9) d x$

$\int(4 x^{3}-5 x^{2}+6 x+9) d x=4(\frac{x^{4}}{4})-5(\frac{x^{3}}{3})+6(\frac{x^{2}}{2})+9(x)$

$=x^{4}-\frac{5 x^{3}}{3}+3 x^{2}+9 x=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(2)-F(1) \\ I & ={2^{4}-\frac{5 \cdot(2)^{3}}{3}+3(2)^{2}+9(2)}-{(1)^{4}-\frac{5(1)^{3}}{3}+3(1)^{2}+9(1)} \\ & =(16-\frac{40}{3}+12+18)-(1-\frac{5}{3}+3+9) \\ & =16-\frac{40}{3}+12+18-1+\frac{5}{3}-3-9 \\ & =33-\frac{35}{3} \\ & =\frac{99-35}{3} \\ & =\frac{64}{3} \end{aligned} $

Solution

Let $I=\int_0^{\pi} \sin 2 x d x$

$\int \sin 2 x d x=(\frac{-\cos 2 x}{2})=F(x)$

By second fundamental theorem of calculus, we obtain

$ I =F(\frac{\pi}{4})-F(0) $

$ =-\frac{1}{2}[\cos 2(\frac{\pi}{4})-\cos 0] $

$ =-\frac{1}{2}[\cos (\mathrm{\pi} / 2)] $

$ =-\frac{1}{2}[0-1] $

$ =\frac{1}{2} $

Solution

Let $I=\int_0^{\frac{\pi}{2}} \cos 2 x d x$

$\int \cos 2 x d x=(\frac{\sin 2 x}{2})=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(\frac{\pi}{2})-F(0) \\ & =\frac{1}{2}[\sin 2(\frac{\pi}{2})-\sin 0] \\ & =\frac{1}{2}[\sin \pi-\sin 0] \\ & =\frac{1}{2}[0-0]=0 \end{aligned} $

Solution

Let $I=\int_4^{5} e^{x} d x$

$\int e^{x} d x=e^{x}=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(5)-F(4) \\ & =e^{5}-e^{4} \\ & =e^{4}(e-1) \end{aligned} $

Solution

Let $I=\int_0^{\frac{\pi}{4}} \tan x d x$

$\int \tan x d x=-\log |\cos x|=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(\frac{\pi}{4})-F(0) \\ & =-\log |\cos \frac{\pi}{4}|+\log |\cos 0| \\ & =-\log |\frac{1}{\sqrt{2}}|+\log |1| \\ & =-\log (2)^{-\frac{1}{2}} \\ & =\frac{1}{2} \log 2 \end{aligned} $

Solution

Let $I=\int _{\frac{\pi}{6}}^{\frac{\pi}{4}} cosec x d x$

$\int cosec x d x=\log |cosec x-\cot x|=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(\frac{\pi}{4})-F(\frac{\pi}{6}) \\ & =\log |cosec \frac{\pi}{4}-\cot \frac{\pi}{4}|-\log |cosec \frac{\pi}{6}-\cot \frac{\pi}{6}| \\ & =\log |\sqrt{2}-1|-\log |2-\sqrt{3}| \\ & =\log (\frac{\sqrt{2}-1}{2-\sqrt{3}}) \end{aligned} $

Solution

Let $I=\int_0^{1} \frac{d x}{\sqrt{1-x^{2}}}$

$\int \frac{d x}{\sqrt{1-x^{2}}}=\sin ^{-1} x=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(1)-F(0) \\ & =\sin ^{-1}(1)-\sin ^{-1}(0) \\ & =\frac{\pi}{2}-0 \\ & =\frac{\pi}{2} \end{aligned} $

Solution

Let $I=\int_0^{1} \frac{d x}{1+x^{2}}$

$\int \frac{d x}{1+x^{2}}=\tan ^{-1} x=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(1)-F(0) \\ & =\tan ^{-1}(1)-\tan ^{-1}(0) \\ & =\frac{\pi}{4} \end{aligned} $

Solution

Let $I=\int_2^{3} \frac{d x}{x^{2}-1}$

$\int \frac{d x}{x^{2}-1}=\frac{1}{2} \log |\frac{x-1}{x+1}|=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(3)-F(2) \\ & =\frac{1}{2}[\log |\frac{3-1}{3+1}|-\log |\frac{2-1}{2+1}|] \\ & =\frac{1}{2}[\log |\frac{2}{4}|-\log |\frac{1}{3}|] \\ & =\frac{1}{2}[\log \frac{1}{2}-\log \frac{1}{3}] \\ & =\frac{1}{2}[\log \frac{3}{2}] \end{aligned} $

Solution

Let $I=\int_0^{\frac{\pi}{2}} \cos ^{2} x d x$

$\int \cos ^{2} x d x=\int(\frac{1+\cos 2 x}{2}) d x=\frac{x}{2}+\frac{\sin 2 x}{4}=\frac{1}{2}(x+\frac{\sin 2 x}{2})=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =[F(\frac{\pi}{2})-F(0)] \\ & =\frac{1}{2}[(\frac{\pi}{2}-\frac{\sin \pi}{2})-(0+\frac{\sin 0}{2})] \\ & =\frac{1}{2}[\frac{\pi}{2}+0-0-0] \\ & =\frac{\pi}{4} \end{aligned} $

Solution

Let $I=\int_2^{3} \frac{x}{x^{2}+1} d x$

$\int \frac{x}{x^{2}+1} d x=\frac{1}{2} \int \frac{2 x}{x^{2}+1} d x=\frac{1}{2} \log (1+x^{2})=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(3)-F(2) \\ & =\frac{1}{2}[\log (1+(3)^{2})-\log (1+(2)^{2})] \\ & =\frac{1}{2}[\log (10)-\log (5)] \\ & =\frac{1}{2} \log (\frac{10}{5})=\frac{1}{2} \log 2 \end{aligned} $

Solution

Let $I=\int_0^{1} \frac{2 x+3}{5 x^{2}+1} d x$

$\int \frac{2 x+3}{5 x^{2}+1} d x=\frac{1}{5} \int \frac{5(2 x+3)}{5 x^{2}+1} d x$

$=\frac{1}{5} \int \frac{10 x+15}{5 x^{2}+1} d x$

$=\frac{1}{5} \int \frac{10 x}{5 x^{2}+1} d x+3 \int \frac{1}{5 x^{2}+1} d x$

$=\frac{1}{5} \int \frac{10 x}{5 x^{2}+1} d x+3 \int \frac{1}{5(x^{2}+\frac{1}{5})} d x$

$=\frac{1}{5} \log (5 x^{2}+1)+\frac{3}{5} \cdot \frac{1}{\frac{1}{\sqrt{5}}} \tan ^{-1} \frac{x}{\frac{1}{\sqrt{5}}}$

$=\frac{1}{5} \log (5 x^{2}+1)+\frac{3}{\sqrt{5}} \tan ^{-1}(\sqrt{5} x)$

$=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(1)-F(0) \\ & ={\frac{1}{5} \log (5+1)+\frac{3}{\sqrt{5}} \tan ^{-1}(\sqrt{5})}-{\frac{1}{5} \log (1)+\frac{3}{\sqrt{5}} \tan ^{-1}(0)} \\ & =\frac{1}{5} \log 6+\frac{3}{\sqrt{5}} \tan ^{-1} \sqrt{5} \end{aligned} $

Solution

Let $I=\int_0^{1} x e^{x^{2}} d x$

Put $x^{2}=t \Rightarrow 2 x d x=d t$

As $x \to 0, t \to 0$ and as $x \to 1, t \to 1$,

$\therefore I=\frac{1}{2} \int_0^{1} e^{t} d t$

$\frac{1}{2} \int e^{t} d t=\frac{1}{2} e^{t}=F(t)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(1)-F(0) \\ & =\frac{1}{2} e-\frac{1}{2} e^{0} \\ & =\frac{1}{2}(e-1) \end{aligned} $

Solution

Let $I=\int_1^{2} \frac{5 x^{2}}{x^{2}+4 x+3} d x$

Dividing $5 x^{2}$ by $x^{2}+4 x+3$, we obtain

$I=\int_1^{2}{5-\frac{20 x+15}{x^{2}+4 x+3}} d x$

$=\int_1^{2} 5 d x-\int_1^{2} \frac{20 x+15}{x^{2}+4 x+3} d x$

$=[5 x]_1^{2}-\int_1^{2} \frac{20 x+15}{x^{2}+4 x+3} d x$

$I=5-I_1$, where $I=\int_1^{2} \frac{20 x+15}{x^{2}+4 x+3} d x$

Consider $I_1=\int_1^{2} \frac{20 x+15}{x^{2}+4 x+8} d x$

Let $20 x+15=A \frac{d}{d x}(x^{2}+4 x+3)+B$

$ =2 A x+(4 A+B) $

Equating the coefficients of $x$ and constant term, we obtain

$A=10$ and $B=-25$

$\Rightarrow I_1=10 \int_1^{2} \frac{2 x+4}{x^{2}+4 x+3} d x-25 \int_1^{2} \frac{d x}{x^{2}+4 x+3}$

Let $x^{2}+4 x+3=t$

$\Rightarrow(2 x+4) d x=d t$

$\Rightarrow I_1=10 \int \frac{d t}{t}-25 \int \frac{d x}{(x+2)^{2}-1^{2}}$

$=10 \log t-25[\frac{1}{2} \log (\frac{x+2-1}{x+2+1})]$

$=[10 \log (x^{2}+4 x+3)]_1^{2}-25[\frac{1}{2} \log (\frac{x+1}{x+3})]_1^{2}$

$=[10 \log 15-10 \log 8]-25[\frac{1}{2} \log \frac{3}{5}-\frac{1}{2} \log \frac{2}{4}]$

$=[10 \log (5 \times 3)-10 \log (4 \times 2)]-\frac{25}{2}[\log 3-\log 5-\log 2+\log 4]$

$=[10 \log 5+10 \log 3-10 \log 4-10 \log 2]-\frac{25}{2}[\log 3-\log 5-\log 2+\log 4]$

$=[10+\frac{25}{2}] \log 5+[-10-\frac{25}{2}] \log 4+[10-\frac{25}{2}] \log 3+[-10+\frac{25}{2}] \log 2$

$=\frac{45}{2} \log 5-\frac{45}{2} \log 4-\frac{5}{2} \log 3+\frac{5}{2} \log 2$

$=\frac{45}{2} \log \frac{5}{4}-\frac{5}{2} \log \frac{3}{2}$

Substituting the value of $I_1$ in (1), we obtain

$ \begin{aligned} I & =5-[\frac{45}{2} \log \frac{5}{4}-\frac{5}{2} \log \frac{3}{2}] \\ & =5-\frac{5}{2}[9 \log \frac{5}{4}-\log \frac{3}{2}] \end{aligned} $

Solution

Let $I=\int_0^{\frac{\pi}{4}}(2 \sec ^{2} x+x^{3}+2) d x$

$\int(2 \sec ^{2} x+x^{3}+2) d x=2 \tan x+\frac{x^{4}}{4}+2 x=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(\frac{\pi}{4})-F(0) \\ & ={(2 \tan \frac{\pi}{4}+\frac{1}{4}(\frac{\pi}{4})^{4}+2(\frac{\pi}{4}))-(2 \tan 0+0+0)} \\ & =2 \tan \frac{\pi}{4}+\frac{\pi^{4}}{4^{5}}+\frac{\pi}{2} \\ & =2+\frac{\pi}{2}+\frac{\pi^{4}}{1024} \end{aligned} $

Solution

Let $I=\int_0^{\pi}(\sin ^{2} \frac{x}{2}-\cos ^{2} \frac{x}{2}) d x$

$ \begin{aligned} & =-\int_0^{\pi}(\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}) d x \\ & =-\int_0^{\pi} \cos x d x \end{aligned} $

$\int \cos x d x=\sin x=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(\pi)-F(0) \\ & =\sin \pi-\sin 0 \\ & =0 \end{aligned} $

Solution

Let $I=\int_0^{2} \frac{6 x+3}{x^{2}+4} d x$

$ \begin{aligned} \int \frac{6 x+3}{x^{2}+4} d x & =3 \int \frac{2 x+1}{x^{2}+4} d x \\ & =3 \int \frac{2 x}{x^{2}+4} d x+3 \int \frac{1}{x^{2}+4} d x \\ & =3 \log (x^{2}+4)+\frac{3}{2} \tan ^{-1} \frac{x}{2}=F(x) \end{aligned} $

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(2)-F(0) \\ & ={3 \log (2^{2}+4)+\frac{3}{2} \tan ^{-1}(\frac{2}{2})}-{3 \log (0+4)+\frac{3}{2} \tan ^{-1}(\frac{0}{2})} \\ & =3 \log 8+\frac{3}{2} \tan ^{-1} 1-3 \log 4-\frac{3}{2} \tan ^{-1} 0 \\ & =3 \log 8+\frac{3}{2}(\frac{\pi}{4})-3 \log 4-0 \\ & =3 \log (\frac{8}{4})+\frac{3 \pi}{8} \\ & =3 \log 2+\frac{3 \pi}{8} \end{aligned} $

Solution

Let $I=\int_0^{1}(x e^{x}+\sin \frac{\pi x}{4}) d x$

$ \begin{aligned} \int(x e^{x}+\sin \frac{\pi x}{4}) d x & =x \int e^{x} d x-\int{(\frac{d}{d x} x) \int e^{x} d x} d x+{\frac{-\cos \frac{\pi x}{4}}{\frac{\pi}{4}}} \\ & =x e^{x}-\int e^{x} d x-\frac{4} {\pi}* \cos(\pi x/4) \\ & =x e^{x}-e^{x}-\frac{4}{\pi} \cos \frac{\pi x}{4} \\ & =F(x) \end{aligned} $

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(1)-F(0) \\ & =(1 . e^{1}-e^{1}-\frac{4}{\pi} \cos \frac{\pi}{4})-(0 . e^{0}-e^{0}-\frac{4}{\pi} \cos 0) \\ & =e-e-\frac{4}{\pi}(\frac{1}{\sqrt{2}})+1+\frac{4}{\pi} \\ & =1+\frac{4}{\pi}-\frac{2 \sqrt{2}}{\pi} \end{aligned} $

Solution

$ \int \frac{d x}{1+x^{2}}=\tan ^{-1} x=F(x) $

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} & \int_1^{\sqrt{3}} \frac{d x}{1+x^{2}}=F(\sqrt{3})-F(1) \\ & =\tan ^{-1} \sqrt{3}-\tan ^{-1} 1 \\ & =\frac{\pi}{3}-\frac{\pi}{4} \\ & =\frac{\pi}{12} \end{aligned} $

Hence, the correct Answer is D.

Solution

$ \int \frac{d x}{4+9 x^{2}}=\int \frac{d x}{(2)^{2}+(3 x)^{2}} $

Put $3 x=t \Rightarrow 3 d x=d t$

$\therefore \int \frac{d x}{(2)^{2}+(3 x)^{2}}=\frac{1}{3} \int \frac{d t}{(2)^{2}+t^{2}}$

$ \begin{aligned} & =\frac{1}{3}[\frac{1}{2} \tan ^{-1} \frac{t}{2}] \\ & =\frac{1}{6} \tan ^{-1}(\frac{3 x}{2}) \\ & =F(x) \end{aligned} $

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} \int_0^{\frac{2}{3}} \frac{d x}{4+9 x^{2}} & =F(\frac{2}{3})-F(0) \\ & =\frac{1}{6} \tan ^{-1}(\frac{3}{2} \cdot \frac{2}{3})-\frac{1}{6} \tan ^{-1} 0 \\ & =\frac{1}{6} \tan ^{-1} 1-0 \\ & =\frac{1}{6} \times \frac{\pi}{4} \\ & =\frac{\pi}{24} \end{aligned} $

Hence, the correct Answer is C.

Solution

$\int_0^{1} \frac{x}{x^{2}+1} d x$

Let $x^{2}+1=t \Rightarrow 2 x d x=d t$

When $x=0, t=1$ and when $x=1, t=2$

$\therefore \int_0^{1} \frac{x}{x^{2}+1} d x=\frac{1}{2} \int_1^{2} \frac{d t}{t}$

$=\frac{1}{2}[\log |t|]_1^{2}$

$=\frac{1}{2}[\log 2-\log 1]$

$=\frac{1}{2} \log 2$

Solution

Let $I=\int_0^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{5} \phi d \phi=\int_0^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{4} \phi \cos \phi d \phi$

Also, let $\sin \phi=t \Rightarrow \cos \phi d \phi=d t$

When $\phi=0, t=0$ and when $\phi=\frac{\pi}{2}, t=1$

$\therefore I=\int_0^{1} \sqrt{t}(1-t^{2})^{2} d t$

$=\int_0^{1} t^{\frac{1}{2}}(1+t^{4}-2 t^{2}) d t$

$=\int_0^{1}[t^{\frac{1}{2}}+t^{\frac{9}{2}}-2 t^{\frac{5}{2}}] d t$

$=[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+\frac{t^{\frac{11}{2}}}{\frac{11}{2}}-\frac{2 t^{\frac{7}{2}}}{\frac{7}{2}}]_0^{1}$

$=\frac{2}{3}+\frac{2}{11}-\frac{4}{7}$

$=\frac{154+42-132}{231}$

$=\frac{64}{231}$

Solution

Let $I=\int_0^{1} \sin ^{-1}(\frac{2 x}{1+x^{2}}) d x$

Also, let $x=\tan \theta d x=\sec ^{2} \theta d \theta$

When $x=0, \theta=0$ and when $x=1, \theta=\frac{\pi}{4}$

$ \begin{aligned} I & =\int_0^{\pi/4} \sin ^{-1}(\frac{2 \tan \theta}{1+\tan ^{2} \theta}) \sec ^{2} \theta d \theta \\ & =\int_0^{\frac{\pi}{4}} \sin ^{-1}(\sin 2 \theta) \sec ^{2} \theta d \theta \\ & =\int_0^{\frac{\pi}{4}} 2 \theta \cdot \sec ^{2} \theta d \theta \\ & =2 \int_0^{\pi/4} \theta \cdot \sec ^{2} \theta d \theta \end{aligned} $

Taking $\theta$ as first function and $\sec ^{2} \theta$ as second function and integrating by parts, we obtain

$ \begin{aligned} I & =2[\theta \int \sec ^{2} \theta d \theta-\int{(\frac{d}{d x} \theta) \int \sec ^{2} \theta d \theta} d \theta]_0^{\frac{\pi}{4}} \\ & =2[\theta \tan \theta-\int \tan \theta d \theta]_0^{\frac{\pi}{4}} \\ & =2[\theta \tan \theta+\log |\cos \theta|]_0^{\frac{\pi}{4}} \\ & =2[\frac{\pi}{4} \tan \frac{\pi}{4}+\log |\cos \frac{\pi}{4}|-\log |\cos 0|] \\ & =2[\frac{\pi}{4}+\log (\frac{1}{\sqrt{2}})-\log 1] \\ & =2[\frac{\pi}{4}-\frac{1}{2} \log 2] \\ & =\frac{\pi}{2}-\log 2 \end{aligned} $

Solution

$\int_0^{2} x \sqrt{x+2} d x$

Let $x+2=t^{2} d x=2 t d t$

When $x=0, t=\sqrt{2}$ and when $x=2, t=2$ $\therefore \int_0^{2} x \sqrt{x+2} d x=\int _{\sqrt{2}}^{2}(t^{2}-2) \sqrt{t^{2}} 2 t d t$

$.=2 \int _{\sqrt{2}}^{2}(t^{2}-2))^{2} d t$

$=2 \int _{\sqrt{2}}^{2}(t^{4}-2 t^{2}) d t$

$=2[\frac{t^{5}}{5}-\frac{2 t^{3}}{3}] _{\sqrt{2}}^{2}$

$=2[\frac{32}{5}-\frac{16}{3}-\frac{4 \sqrt{2}}{5}+\frac{4 \sqrt{2}}{3}]$

$=2[\frac{96-80-12 \sqrt{2}+20 \sqrt{2}}{15}]$

$=2[\frac{16+8 \sqrt{2}}{15}]$

$=\frac{16(2+\sqrt{2})}{15}$

$=\frac{16 \sqrt{2}(\sqrt{2}+1)}{15}$

Solution

$\int_0^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} d x$

Let $\cos x=t -\sin x d x=d t$

When $x=0, t=1$ and when $x=\frac{\pi}{2}, t=0$ $\Rightarrow \int_0^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} d x=-\int_1^{0} \frac{d t}{1+t^{2}}$

$=-[\tan ^{-1} t]_1^{0}$

$=-[\tan ^{-1} 0-\tan ^{-1} 1]$

$=-[-\frac{\pi}{4}]$

$=\frac{\pi}{4}$

Solution

$\int_0^{2} \frac{d x}{x+4-x^{2}}=\int_0^{2} \frac{d x}{-(x^{2}-x-4)}$

$=\int_0^{2} \frac{d x}{-(x^{2}-x+\frac{1}{4}-\frac{1}{4}-4)}$

$=\int_0^{2} \frac{d x}{-[(x-\frac{1}{2})^{2}-\frac{17}{4}]}$

$=\int_0^{2} \frac{d x}{(\frac{\sqrt{17}}{2})^{2}-(x-\frac{1}{2})^{2}}$

Let $\quad x-\frac{1}{2}=t \quad \square d x=d t$

When $x=0, t=-\frac{1}{2}$ and when $x=2, t=\frac{3}{2}$

$\therefore \int_0^{2} \frac{d x}{(\frac{\sqrt{17}}{2})^{2}-(x-\frac{1}{2})^{2}}=\int _{-\frac{1}{2}}^{\frac{3}{2}} \frac{d t}{(\frac{\sqrt{17}}{2})^{2}-t^{2}}$

$=[\frac{1}{2(\frac{\sqrt{17}}{2})} \log \frac{\frac{\sqrt{17}}{2}+t}{\frac{\sqrt{17}}{2}-t}] _{-\frac{1}{2}}^{\frac{3}{2}}$

$=\frac{1}{\sqrt{17}}[\log \frac{\frac{\sqrt{17}}{2}+\frac{3}{2}}{\frac{\sqrt{17}}{2}-\frac{3}{2}}-\frac{\log \frac{\sqrt{17}}{2}-\frac{1}{2}}{\log \frac{\sqrt{17}}{2}+\frac{1}{2}}]$

$=\frac{1}{\sqrt{17}}[\log \frac{\sqrt{17}+3}{\sqrt{17}-3}-\log \frac{\sqrt{17}-1}{\sqrt{17}+1}]$

$=\frac{1}{\sqrt{17}} \log \frac{\sqrt{17}+3}{\sqrt{17}-3} \times \frac{\sqrt{17}+1}{\sqrt{17}-1}$

$=\frac{1}{\sqrt{17}} \log [\frac{17+3+4 \sqrt{17}}{17+3-4 \sqrt{17}}]$

$=\frac{1}{\sqrt{17}} \log [\frac{20+4 \sqrt{17}}{20-4 \sqrt{17}}]$

$=\frac{1}{\sqrt{17}} \log (\frac{5+\sqrt{17}}{5-\sqrt{17}})$

$=\frac{1}{\sqrt{17}} \log [\frac{(5+\sqrt{17})(5+\sqrt{17})}{25-17}]$

$=\frac{1}{\sqrt{17}} \log [\frac{25+17+10 \sqrt{17}}{8}]$

$=\frac{1}{\sqrt{17}} \log (\frac{42+10 \sqrt{17}}{8})$

$=\frac{1}{\sqrt{17}} \log (\frac{21+5 \sqrt{17}}{4})$

Solution

$\int _{-1}^{1} \frac{d x}{x^{2}+2 x+5}=\int _{-1}^{1} \frac{d x}{(x^{2}+2 x+1)+4}=\int _{-1}^{1} \frac{d x}{(x+1)^{2}+(2)^{2}}$

Let $x+1=t d x=d t$

When $x=-1, t=0$ and when $x=1, t=2$

$\therefore \int _{-1}^{1} \frac{d x}{(x+1)^{2}+(2)^{2}}=\int_0^{2} \frac{d t}{t^{2}+2^{2}}$

$=[\frac{1}{2} \tan ^{-1} \frac{t}{2}]_0^{2}$

$=\frac{1}{2} \tan ^{-1} 1-\frac{1}{2} \tan ^{-1} 0$

$=\frac{1}{2}(\frac{\pi}{4})=\frac{\pi}{8}$

Solution

$\int(\frac{1}{x}-\frac{1}{2 x^{2}}) e^{2 x} d x$

Let $2 x=t 2 d x=d t$

When $x=1, t=2$ and when $x=2, t=4$

$ \begin{aligned} \therefore \int_1^{2}(\frac{1}{x}-\frac{1}{2 x^{2}}) e^{2 x} d x & =\frac{1}{2} \int_2^{4}(\frac{2}{t}-\frac{2}{t^{2}}) e^{t} d t \\ & =\int_2^{4}(\frac{1}{t}-\frac{1}{t^{2}}) e^{t} d t \end{aligned} $

Let $\frac{1}{t}=f(t)$

Then, $f^{\prime}(t)=-\frac{1}{t^{2}}$

$ \begin{aligned} \Rightarrow \int_2^{4}(\frac{1}{t}-\frac{1}{t^{2}}) e^{t} d t & =\int_2^{4} e^{t}[f(t)+f^{\prime}(t)] d t \\ & =[e^{t} f(t)]_2^{4} \\ & =[e^{t} \cdot \frac{2}{t}]_2^{4} \\ & =[\frac{e^{t}}{t}]_2^{4} \\ & =\frac{e^{4}}{4}-\frac{e^{2}}{2} \\ & =\frac{e^{2}(e^{2}-2)}{4} \end{aligned} $

Solution

Let $I=\int _{\frac{1}{3}}^{1} \frac{(x-x^{3})^{\frac{1}{3}}}{x^{4}} d x$

Also, let $x=\sin \theta \Rightarrow d x=\cos \theta d \theta$

When $x=\frac{1}{3}, \theta=\sin ^{-1}(\frac{1}{3})$ and when $x=1, \theta=\frac{\pi}{2}$

$\Rightarrow I=\int _{\sin ^{-1}(\frac{1}{3})}^{\frac{\pi}{2}} \frac{(\sin \theta-\sin ^{3} \theta)^{\frac{1}{3}}}{\sin ^{4} \theta} \cos \theta d \theta$

$ =\int _{\sin ^{-1}(\frac{1}{3})}^{\frac{\pi}{2}} \frac{(\sin \theta)^{\frac{1}{3}}(1-\sin ^{2} \theta)^{\frac{1}{3}}}{\sin ^{4} \theta} \cos \theta d \theta $

$=\int _{\sin ^{-1}(\frac{1}{3})}^{\frac{\pi}{2}} \frac{(\sin \theta)^{\frac{1}{3}}(\cos \theta)^{\frac{2}{3}}}{\sin ^{4} \theta} \cos \theta d \theta$

$=\int _{\sin ^{-1}(\frac{1}{3})}^{\frac{\pi}{2}} \frac{(\sin \theta)^{\frac{1}{3}}(\cos \theta)^{\frac{2}{3}}}{\sin ^{2} \theta \sin ^{2} \theta} \cos \theta d \theta$

$=\int _{\sin ^{-1}(\frac{1}{3}.}^{\frac{\pi}{2}} \frac{(\cos \theta)^{\frac{5}{3}}}{(\sin \theta)^{\frac{5}{3}}} cosec^{2} \theta d \theta$

$=\int _{\sin ^{-1}(\frac{1}{3})}^{\frac{\pi}{2}}(\cot \theta)^{\frac{5}{3}} cosec^{2} \theta d \theta$

Let $\cot \theta=t -cosec 2 \theta d \theta=d t$

$ \begin{aligned} \therefore I & =-\int _{2 \sqrt{2}}^{0}(t)^{\frac{5}{3}} d t \\ & =-[\frac{3}{8}(t)^{\frac{8}{3}}] _{2 \sqrt{2}}^{0} \\ & =-\frac{3}{8}[(t)^{\frac{8}{3}}] _{2 \sqrt{2}}^{0} \\ & =-\frac{3}{8}[-(2 \sqrt{2})^{\frac{8}{3}}] \\ & =\frac{3}{8}[(\sqrt{8})^{\frac{8}{3}}] \\ & =\frac{3}{8}[(8)^{\frac{4}{3}}] \\ & =\frac{3}{8}[16] \\ & =3 \times 2 \\ & =6 \end{aligned} $

$ \text{ When } \theta=\sin ^{-1}(\frac{1}{3}), t=2 \sqrt{2} \text{ and when } \theta=\frac{\pi}{2}, t=0 $

Hence, the correct Answer is A.

Solution

$f(x)=\int_0^{x} t \sin t d t$

Integrating by parts, we obtain

$ \begin{aligned} f(x) & =t \int_0^{x} \sin t d t-\int_0^{x}{(\frac{d}{d t} t) \int \sin t d t} d t \\ & =[t(-\cos t)]_0^{x}-\int_0^{x}(-\cos t) d t \\ & =[-t \cos t+\sin t]_0^{x} \\ & =-x \cos x+\sin x \\ \Rightarrow f^{\prime}(x) & =-[{x(-\sin x)}+\cos x]+\cos x \\ & =x \sin x-\cos x+\cos x \\ & x \sin x \end{aligned} $

Hence, the correct Answer is B.

Solution

$I=\int_0^{\frac{\pi}{2}} \cos ^{2} x d x$

$\Rightarrow I=\int_0^{\frac{\pi}{2}} \cos ^{2}(\frac{\pi}{2}-x) d x$

$(\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x)$

$\Rightarrow I=\int_0^{\frac{\pi}{2}} \sin ^{2} x d x$

Adding (1) and (2), we obtain

$2 I=\int_0^{\frac{\pi}{2}}(\sin ^{2} x+\cos ^{2} x) d x$

$\Rightarrow 2 I=\int_0^{\frac{\pi}{2}} 1 \cdot d x$

$\Rightarrow 2 I=[x]_0^{\frac{\pi}{2}}$

$\Rightarrow 2 I=\frac{\pi}{2}$

$\Rightarrow I=\frac{\pi}{4}$

Solution

$\int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$

Let $I=\int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$

$\Rightarrow I=\int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin (\frac{\pi}{2}-x)}}{\sqrt{\sin (\frac{\pi}{2}-x)}+\sqrt{\cos (\frac{\pi}{2}-x)}} d x$

$\Rightarrow I=\int_0^{\frac{\pi}{2}} \frac{\sqrt{\cos(x) }}{\sqrt{\cos(x) }+\sqrt{\sin x}} d x$

Adding (1) and (2), we obtain

$2 I=\int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$

$\Rightarrow 2 I=\int_0^{\frac{\pi}{2}} 1 \cdot d x$

$\Rightarrow 2 I=[x]_0^{\frac{\pi}{2}}$

$\Rightarrow 2 I=\frac{\pi}{2}$

$\Rightarrow I=\frac{\pi}{4}$

Solution

$(\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x)$

Let $I=\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x$

$\Rightarrow I=\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}}(\frac{\pi}{2}-x)}{\sin ^{\frac{3}{2}}(\frac{\pi}{2}-x)+\cos ^{\frac{3}{2}}(\frac{\pi}{2}-x)} d x$

$ (\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x) $

$\Rightarrow I=\int_0^{\frac{\pi}{2}} \frac{\cos ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x$

Adding (1) and (2), we obtain

$2 I=\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x$

$\Rightarrow 2 I=\int_0^{\frac{\pi}{2}} 1 \cdot d x$

$\Rightarrow 2 I=[x]_0^{\frac{\pi}{2}}$

$\Rightarrow 2 I=\frac{\pi}{2}$

$\Rightarrow I=\frac{\pi}{4}$

Solution

Let $I=\int_0^{\frac{\pi}{2}} \frac{\cos ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x$

$\Rightarrow I=\int_0^{\frac{\pi}{2}} \frac{\cos ^{5}(\frac{\pi}{2}-x)}{\sin ^{5}(\frac{\pi}{2}-x)+\cos ^{5}(\frac{\pi}{2}-x)} d x$

$\Rightarrow I=\int_0^{\frac{\pi}{2}} \frac{\sin ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x$

Adding (1) and (2), we obtain

$2 I=\int_0^{\frac{\pi}{2}} \frac{\sin ^{5} x+\cos ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x$

$\Rightarrow 2 I=\int_0^{\frac{\pi}{2}} 1 \cdot d x$

$\Rightarrow 2 I=[x]_0^{\frac{\pi}{2}}$

$\Rightarrow 2 I=\frac{\pi}{2}$

$\Rightarrow I=\frac{\pi}{4}$

Solution

Let $I=\int _{-5}^{5}|x+2| d x$

It can be seen that $(x+2) \leq 0$ on $[-5,-2]$ and $(x+2) \geq 0$ on $[-2,5]$.

$ \begin{aligned} \therefore I & =\int _{-5}^{-2}-(x+2) d x+\int _{-2}^{5}(x+2) d x \quad(\int_a^{b} f(x)=\int_a^{c} f(x)+\int_c^{b} f(x)) \\ I & =-[\frac{x^{2}}{2}+2 x] _{-5}^{-2}+[\frac{x^{2}}{2}+2 x] _{-2}^{5} \\ & =-[\frac{(-2)^{2}}{2}+2(-2)-\frac{(-5)^{2}}{2}-2(-5)]+[\frac{(5)^{2}}{2}+2(5)-\frac{(-2)^{2}}{2}-2(-2)] \\ & =-[2-4-\frac{25}{2}+10]+[\frac{25}{2}+10-2+4] \\ & =-2+4+\frac{25}{2}-10+\frac{25}{2}+10-2+4 \\ & =29 \end{aligned} $

Solution

Let $I=\int_2^{8}|x-5| d x$

It can be seen that $(x-5) \leq 0$ on $[2,5]$ and $(x-5) \geq 0$ on $[5,8]$.

$ \begin{matrix} I & =\int_2^{5}-(x-5) d x+\int_5^{8}(x-5) d x & (\int_a^{b} f(x)=\int_a^{c} f(x)+\int_c^{b} f(x)) \\ & =-[\frac{x^{2}}{2}-5 x]_2^{5}+[\frac{x^{2}}{2}-5 x]_5^{8} \\ & =-[\frac{25}{2}-25-2+10]+[32-40-\frac{25}{2}+25] \\ & =9 \end{matrix} $

Solution

$ \begin{aligned} \text{ Let } & I=\int_0^{1} x(1-x)^{n} d x \\ \therefore I & =\int_0^{1}(1-x)(1-(1-x))^{n} d x \\ & =\int_0^{1}(1-x)(x)^{n} d x \\ & =\int_0^{1}(x^{n}-x^{n+1}) d x \\ & =[\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}]_0^{1} \quad(\int_0^{n} f(x) d x=\int_0^{n} f(a-x) d x) \\ & =[\frac{1}{n+1}-\frac{1}{n+2}] \\ & =\frac{(n+2)-(n+1)}{(n+1)(n+2)} \\ & =\frac{1}{(n+1)(n+2)} \end{aligned} $

Solution

Let $I=\int_0^{\frac{\pi}{4}} \log (1+\tan x) d x$

$\therefore I=\int_0^{\frac{\pi}{4}} \log [1+\tan (\frac{\pi}{4}-x)] d x$

$\Rightarrow I=\int_0^{\frac{\pi}{4}} \log ({1+\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \tan x}}) d x$

$\Rightarrow I=\int_0^{\frac{\pi}{4}} \log ({1+\frac{1-\tan x}{1+\tan }}) d x$

$\Rightarrow I=\int_0^{\frac{\pi}{4}} \log \frac{2}{(1+\tan x)} d x$

$\Rightarrow I=\int_0^{\frac{\pi}{4}} \log 2 d x-\int_0^{\frac{\pi}{4}} \log (1+\tan x) d x$

$\Rightarrow I=\int_0^{\frac{\pi}{4}} \log 2 d x-I$

$\Rightarrow 2 I=[x \log 2]_0^{\frac{\pi}{4}}$

$\Rightarrow 2 I=\frac{\pi}{4} \log 2$

$\Rightarrow I=\frac{\pi}{8} \log 2$

Solution

$(\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x)$

[From (1)]

Let $I=\int_0^{2} x \sqrt{2-x} d x$

$ \begin{aligned} I & =\int_0^{2}(2-x) \sqrt{x} d x \\ & =\int_0^{2}{2 x^{\frac{1}{2}}-x^{\frac{3}{2}}} d x \\ & .=[2(\frac{x^{\frac{3}{2}}}{\frac{3}{2}})-\frac{x^{\frac{5}{2}}}{\frac{5}{2}}]_0^{2} f(x)d x=\int_0^{a} f(a-x) d x) \\ & =[\frac{4}{3} x^{\frac{3}{2}}-\frac{2}{5} x^{\frac{5}{2}}]_0^{2} \\ & =\frac{4}{3}(2)^{\frac{3}{2}}-\frac{2}{5}(2)^{\frac{5}{2}} \\ & =\frac{4 \times 2 \sqrt{2}}{3}-\frac{2}{5} \times 4 \sqrt{2} \\ & =\frac{8 \sqrt{2}}{3}-\frac{8 \sqrt{2}}{5} \\ & =\frac{40 \sqrt{2}-24 \sqrt{2}}{15} \\ & =\frac{16 \sqrt{2}}{15} \end{aligned} $

Solution

Let $I=\int_0^{\frac{\pi}{2}}(2 \log \sin x-\log \sin 2 x) d x$

$\Rightarrow I=\int_0^{\frac{\pi}{2}}{2 \log \sin x-\log (2 \sin x \cos x)} d x$

$\Rightarrow I=\int_0^{\frac{\pi}{2}}{2 \log \sin x-\log \sin x-\log \cos x-\log 2} d x$

$\Rightarrow I=\int_0^{\frac{\pi}{2}}{\log \sin x-\log \cos x-\log 2} d x$

It is known that, $(\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x)$

$\Rightarrow I=\int_0^{\frac{\pi}{2}}{\log \cos x-\log \sin x-\log 2} d x$

Adding (1) and (2), we obtain

$ \begin{aligned} & 2 I=\int_0^{\frac{\pi}{2}}(-\log 2-\log 2) d x \\ & \Rightarrow 2 I=-2 \log 2 \int_0^{\frac{\pi}{2}} 1 d x \\ & \Rightarrow I=-\log 2[\frac{\pi}{2}] \\ & \Rightarrow I=\frac{\pi}{2}(-\log 2) \\ & \Rightarrow I=\frac{\pi}{2}[\log \frac{1}{2}] \\ & \Rightarrow I=\frac{\pi}{2} \log \frac{1}{2} \end{aligned} $

Solution

Let $I=\int _{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{2} x d x$

As $\sin ^{2}(-x)=(\sin (-x))^{2}=(-\sin x)^{2}=\sin ^{2} x$, therefore, $\sin ^{2} x$ is an even function.

It is known that if $f(x)$ is an even function, then $\int _{-a}^{a} f(x) d x=2 \int_0^{a} f(x) d x$

$ \begin{aligned} I & =2 \int_0^{\frac{\pi}{2}} \sin ^{2} x d x \\ & =2 \int_0^{\frac{\pi}{2}} \frac{1-\cos 2 x}{2} d x \\ & =\int_0^{\frac{\pi}{2}}(1-\cos 2 x) d x \\ & =[x-\frac{\sin 2 x}{2}]_0^{\frac{\pi}{2}} \\ & =\frac{\pi}{2} \end{aligned} $

Solution

Let $I=\int_0^{\pi} \frac{x d x}{1+\sin x}$

$\Rightarrow I=\int_0^{\pi} \frac{(\pi-x)}{1+\sin (\pi-x)} d x$

$(\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x)$

$\Rightarrow I=\int_0^{\pi} \frac{(\pi-x)}{1+\sin x} d x$

Adding (1) and (2), we obtain

$ \begin{aligned} & 2 I=\int_0^{\pi} \frac{\pi}{1+\sin x} d x \\ & \Rightarrow 2 I=\pi \int_0^{\pi} \frac{(1-\sin x)}{(1+\sin x)(1-\sin x)} d x \\ & \Rightarrow 2 I=\pi \int_0^{\pi} \frac{1-\sin x}{\cos ^{2} x} d x \\ & \Rightarrow 2 I=\pi \int_0^{\pi}{\sec ^{2} x-\tan x \sec x} d x \\ & \Rightarrow 2 I=\pi[\tan x-\sec x]_0^{\pi} \\ & \Rightarrow 2 I=\pi[2] \\ & \Rightarrow I=\pi \end{aligned} $

Solution

Let $I=\int _{\frac{\pi}{2}}^{\frac{-\pi}{2}} \sin ^{7} x d x$

As $\sin ^{7}(-x)=(\sin (-x))^{7}=(-\sin x)^{7}=-\sin ^{7} x$, therefore, $\sin ^{2} x$ is an odd function.

It is known that, if $f(x)$ is an odd function, then $\int _{-a}^{a} f(x) d x=0$

$\therefore I=\int _{\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{7} x d x=0$

Solution

Let $I=\int_0^{2 \pi} \cos ^{5} x d x$

$\cos ^{5}(2 \pi-x)=\cos ^{5} x$

It is known that,

$ \begin{aligned} \int_0^{2 a} f(x) d x & =2 \int_0^{a} f(x) d x \text{, if } f(2 a-x)=f(x) \\ & =0 \text{ if } f(2 a-x)=-f(x) \end{aligned} $

$\therefore I=2 \int_0^{\pi} \cos ^{5} x d x$

$\Rightarrow I=2(0)=0 \quad[\cos ^{5}(\pi-x)=-\cos ^{5} x]$

Solution

Let $I=\int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$

$\Rightarrow I=\int_0^{\frac{\pi}{2}} \frac{\sin (\frac{\pi}{2}-x)-\cos (\frac{\pi}{2}-x)}{1+\sin (\frac{\pi}{2}-x) \cos (\frac{\pi}{2}-x)} d x$

$ (\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x) $

$\Rightarrow I=\int_0^{\frac{\pi}{2}} \frac{\cos x-\sin x}{1+\sin x \cos x} d x$

Adding (1) and (2), we obtain

$ \begin{aligned} & 2 I=\int_0^{\frac{\pi}{2}} \frac{0}{1+\sin x \cos x} d x \\ & \Rightarrow I=0 \end{aligned} $

Solution

$$ \begin{align*} & \text{ Let } I=\int_0^{\pi} \log (1+\cos x) d x \tag{1}\\ & \Rightarrow I=\int_0^{\pi} \log (1+\cos (\pi-x)) d x \\ & \Rightarrow I=\int_0^{\pi} \log (1-\cos x) d x \end{align*} $$

$(\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x)$

Adding (1) and (2), we obtain

$2 I=\int_0^{\pi}{\log (1+\cos x)+\log (1-\cos x)} d x$

$\Rightarrow 2 I=\int_0^{\pi} \log (1-\cos ^{2} x) d x$

$\Rightarrow 2 I=\int_0^{\pi} \log \sin ^{2} x d x$

$\Rightarrow 2 I=2 \int_0^{\pi} \log \sin x d x$

$\Rightarrow I=\int_0^{\pi} \log \sin x d x$

$\sin (\pi-x)=\sin x$

$\therefore I=2 \int_0^{\frac{\pi}{2}} \log \sin x d x$

$\Rightarrow I=2 \int_0^{\frac{\pi}{2}} \log \sin (\frac{\pi}{2}-x) d x=2 \int_0^{\frac{\pi}{2}} \log \cos x d x$

Adding (4) and (5), we obtain

$2 I=2 \int_0^{\frac{\pi}{2}}(\log \sin x+\log \cos x) d x$

$\Rightarrow I=\int_0^{\frac{\pi}{2}}(\log \sin x+\log \cos x+\log 2-\log 2) d x$

$\Rightarrow I=\int_0^{\frac{\pi}{2}}(\log 2 \sin x \cos x-\log 2) d x$

$\Rightarrow I=\int_0^{\frac{\pi}{2}} \log \sin 2 x d x-\int_0^{\frac{\pi}{2}} \log 2 d x$

Let $2 x=t\Rightarrow 2 d x=d t$

When $x=0, t=0$ and when $x=\frac{\pi}{2}, \pi=t$

$\therefore I=\frac{1}{2} \int_0^{\pi} \log \sin t d t-\frac{2} \log 2$

$\Rightarrow I=\frac{1}{2} I-\frac{2} \log 2$

$\Rightarrow \frac{I}{2}=-\frac{\pi}{2} \log 2$

$\Rightarrow I=-\pi \log 2$

Solution

Let $I=\int_0^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x$

It is known that, $(\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x)$

$I=\int_0^{a} \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}} d x$

Adding (1) and (2), we obtain

$2 I=\int_0^{a} \frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}} d x$

$\Rightarrow 2 I=\int_0^{a} 1 d x$

$\Rightarrow 2 I=[x]_0^{a}$

$\Rightarrow 2 I=a$

$\Rightarrow I=\frac{a}{2}$

Solution

$I=\int_0^{4}|x-1| d x$

It can be seen that, $(x-1) \leq 0$ when $0 \leq x \leq 1$ and $(x-1) \geq 0$ when $1 \leq x \leq 4$

$ \begin{aligned} I & =\int_0^{1}|x-1| d x+\int_0^{4}|x-1| d x \quad(\int_a^{b} f(x)=\int_a^{c} f(x)+\int_a^{b} f(x)) \\ & =\int_0^{1}-(x-1) d x+\int_1^{4}(x-1) d x \\ & =[x-\frac{x^{2}}{2}]_0^{1}+[\frac{x^{2}}{2}-x]_1^{4} \\ & =1-\frac{1}{2}+\frac{(4)^{2}}{2}-4-\frac{1}{2}+1 \\ & =1-\frac{1}{2}+8-4-\frac{1}{2}+1 \\ & =5 \end{aligned} $

Solution

$$ \begin{align*} & \text{ Let } I=\int_0^{a} f(x) g(x) d x \tag{1}\\ & \Rightarrow I=\int_0^{a} f(a-x) g(a-x) d x \\ & \Rightarrow I=\int_0^{a} f(x) g(a-x) d x \end{align*} $$

Adding (1) and (2), we obtain

$ \begin{aligned} & 2 I=\int_0^{a}{f(x) g(x)+f(x) g(a-x)} d x \\ & \Rightarrow 2 I=\int_0^{a} f(x)[{g(x)+g(a-x)}] d x \\ & \Rightarrow 2 I=\int_0^{a} f(x) \times 4 d x \quad[g(x)+g(a-x)=4] \\ & \Rightarrow I=2 \int_0^{a} f(x) d x \end{aligned} $

Solution

Let $I=\int _{\frac{-\pi}{2}}^{\frac{\pi}{2}}(x^{3}+x \cos x+\tan ^{5} x+1) d x$

$\Rightarrow I=\int _{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^{3} d x+\int _{-\frac{\pi}{2}}^{\frac{\pi}{2}}x \cos x+\int _{-\frac{\pi}{2}}^{\frac{\pi}{2}} \tan ^{5} x d x+\int _{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \cdot d x$

It is known that if $f(x)$ is an even function, then $\int _{-a}^{a} f(x) d x=2 \int_0^{a} f(x) d x$ and

if $f(x)$ is an odd function, then $\int _{-a}^{a} f(x) d x=0$

$ \begin{aligned} I & =0+0+0+2 \int_0^{\frac{\pi}{2}} 1 \cdot d x \\ & =2[x]_0^{\frac{\pi}{2}} \\ & =\frac{2 \pi}{2} \\ I & =\pi \end{aligned} $

Hence, the correct Answer is C.

Solution

Let $I=\int_0^{\frac{\pi}{2}} \log (\frac{4+3 \sin x}{4+3 \cos x}) d x$

$\Rightarrow I=\int_0^{\frac{\pi}{2}} \log [\frac{4+3 \sin (\frac{\pi}{2}-x)}{4+3 \cos (\frac{\pi}{2}-x)}] d x \quad(\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x)$

$\Rightarrow I=\int_0^{\frac{\pi}{2}} \log (\frac{4+3 \cos x}{4+3 \sin x}) d x$

Adding (1) and (2), we obtain

$2 I=\int_0^{\frac{\pi}{2}}{\log (\frac{4+3 \sin x}{4+3 \cos x})+\log (\frac{4+3 \cos x}{4+3 \sin x})} d x$

$\Rightarrow 2 I=\int_0^{\frac{\pi}{2}} \log (\frac{4+3 \sin x}{4+3 \cos x} \times \frac{4+3 \cos x}{4+3 \sin x}) d x$

$\Rightarrow 2 I=\int_0^{\frac{\pi}{2}} \log 1 d x$

$\Rightarrow 2 I=\int_0^{\frac{\pi}{2}} 0 d x$

$\Rightarrow I=0$

Hence, the correct Answer is C.

Solution

$\frac{1}{x-x^{3}}=\frac{1}{x(1-x^{2})}=\frac{1}{x(1-x)(1+x)}$

Let $\frac{1}{x(1-x)(1+x)}=\frac{A}{x}+\frac{B}{(1-x)}+\frac{C}{1+x}$

$\Rightarrow 1=A(1-x^{2})+B x(1+x)+C x(1-x)$

$\Rightarrow 1=A-A x^{2}+B x+B x^{2}+C x-C x^{2}$

Equating the coefficients of $x^{2}, x$, and constant term, we obtain

$-A+B-C=0$

$B+C=0$

$A=1$

On solving these equations, we obtain

$A=1, B=\frac{1}{2}$, and $C=-\frac{1}{2}$

From equation (1), we obtain $\frac{1}{x(1-x)(1+x)}=\frac{1}{x}+\frac{1}{2(1-x)}-\frac{1}{2(1+x)}$

$\Rightarrow \int \frac{1}{x(1-x)(1+x)} d x=\int \frac{1}{x} d x+\frac{1}{2} \int \frac{1}{1-x} d x-\frac{1}{2} \int \frac{1}{1+x} d x$

$=\log |x|-\frac{1}{2} \log |(1-x)|-\frac{1}{2} \log |(1+x)|$

$=\log |x|-\log |(1-x)^{\frac{1}{2}}|-\log |(1+x)^{\frac{1}{2}}|$

$=\log |\frac{x}{(1-x)^{\frac{1}{2}}(1+x)^{\frac{1}{2}}}|+C$

$=\log |(\frac{x^{2}}{1-x^{2}})^{\frac{1}{2}}|+C$

$=\frac{1}{2} \log |\frac{x^{2}}{1-x^{2}}|+C$

Solution

$ \begin{aligned} \frac{1}{\sqrt{x+a}+\sqrt{x+b}} & =\frac{1}{\sqrt{x+a}+\sqrt{x+b}} \times \frac{\sqrt{x+a}-\sqrt{x+b}}{\sqrt{x+a}-\sqrt{x+b}} \\ & =\frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)} \\ & =\frac{(\sqrt{x+a}-\sqrt{x+b})}{a-b} \end{aligned} $

$\Rightarrow \int \frac{1}{\sqrt{x+a}-\sqrt{x+b}} d x=\frac{1}{a-b} \int(\sqrt{x+a}-\sqrt{x+b}) d x$

$ \begin{aligned} & =\frac{1}{(a-b)}[\frac{(x+a)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{(x+b)^{\frac{3}{2}}}{\frac{3}{2}}] \\ & =\frac{2}{3(a-b)}[(x+a)^{\frac{3}{2}}-(x+b)^{\frac{3}{2}}]+C \end{aligned} $

Solution

$ \begin{aligned} & \frac{1}{x \sqrt{a x-x^{2}}} \\ & \text{ Let } x=\frac{a}{t} \Rightarrow d x=-\frac{a}{t^{2}} d t \\ & \Rightarrow \int \frac{1}{x \sqrt{a x-x^{2}}} d x=\int \frac{1}{\frac{a}{t} \sqrt{a \cdot \frac{a}{t}-(\frac{a}{t})^{2}}}(-\frac{a}{t^{2}} d t) \\ &=-\int \frac{1}{a t} \cdot \frac{1}{\sqrt{\frac{1}{t}-\frac{1}{t^{2}}}} d t \\ &=-\frac{1}{a} \int \frac{1}{\sqrt{\frac{t^{2}}{t}-\frac{t^{2}}{t^{2}}} d t} \\ &=-\frac{1}{a} \int \frac{1}{\sqrt{t-1} d t} \\ &=-\frac{1}{a}[2 \sqrt{t-1}]+C \\ &=-\frac{1}{a}[2 \sqrt{\frac{a}{x}-1}]+C \\ &=-\frac{2}{a}(\frac{\sqrt{a-x}}{\sqrt{x}})+C \\ &=-\frac{2}{a}(\sqrt{\frac{a-x}{x}})+C \end{aligned} $

Solution

$\frac{1}{x^{2}(x^{4}+1)^{\frac{3}{4}}}$

Multiplying and dividing by $x^{-3}$, we obtain

$ \begin{aligned} \frac{x^{-3}}{x^{2} \cdot x^{-3}(x^{4}+1)^{\frac{3}{4}}} & =\frac{x^{-3}(x^{4}+1)^{\frac{-3}{4}}}{x^{2} \cdot x^{-3}} \\ & =\frac{(x^{4}+1)^{\frac{-3}{4}}}{x^{5} \cdot(x^{4})^{-\frac{3}{4}}} \\ & =\frac{1}{x^{5}}(\frac{x^{4}+1}{x^{4}})^{-\frac{3}{4}} \\ & =\frac{1}{x^{5}}(1+\frac{1}{x^{4}})^{-\frac{3}{4}} \end{aligned} $

Let $\frac{1}{x^{4}}=t \Rightarrow-\frac{4}{x^{5}} d x=d t \Rightarrow \frac{1}{x^{5}} d x=-\frac{d t}{4}$

$\therefore \int \frac{1}{x^{2}(x^{4}+1)^{\frac{3}{4}}} d x=\int \frac{1}{x^{5}}(1+\frac{1}{x^{4}})^{-\frac{3}{4}} d x$

$ =-\frac{1}{4} \int(1+t)^{-\frac{3}{4}} d t $

$ \begin{aligned} & =-\frac{1}{4}[\frac{(1+t)^{\frac{1}{4}}}{\frac{1}{4}}]+C \\ & =-\frac{1}{4} \frac{(1+\frac{1}{x^{4}})^{\frac{1}{4}}}{\frac{1}{4}}+C \\ & =-(1+\frac{1}{x^{4}})^{\frac{1}{4}}+C \end{aligned} $

Solution

$\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=\frac{1}{x^{\frac{1}{3}}(1+x^{\frac{1}{6}})}$

Let $x=t^{6} \Rightarrow d x=6 t^{5} d t$

$\therefore \int \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}} d x=\int \frac{1}{x^{\frac{1}{3}}(1+x^{\frac{1}{6}})} d x$

$=\int \frac{6 t^{5}}{t^{2}(1+t)} d t$

$=6 \int \frac{t^{3}}{(1+t)} d t$

On dividing, we obtain

$ \begin{aligned} \int \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}} d x & =6 \int{(t^{2}-t+1)-\frac{1}{1+t}} d t \\ & =6[(\frac{t^{3}}{3})-(\frac{t^{2}}{2})+t-\log |1+t|] \\ & =2 x^{\frac{1}{2}}-3 x^{\frac{1}{3}}+6 x^{\frac{1}{6}}-6 \log (1+x^{\frac{1}{6}})+C \\ & =2 \sqrt{x}-3 x^{\frac{1}{3}}+6 x^{\frac{1}{6}}-6 \log (1+x^{\frac{1}{6}})+C \end{aligned} $

Solution

Let $\frac{5 x}{(x+1)(x^{2}+9)}=\frac{A}{(x+1)}+\frac{B x+C}{(x^{2}+9)}$

$\Rightarrow 5 x=A(x^{2}+9)+(B x+C)(x+1)$

$\Rightarrow 5 x=A x^{2}+9 A+B x^{2}+B x+C x+C$

Equating the coefficients of $x^{2}, x$, and constant term, we obtain

$A+B=0$

$B+C=5$

$9 A+C=0$

On solving these equations, we obtain

$A=-\frac{1}{2}, B=\frac{1}{2}$, and $C=\frac{9}{2}$

From equation (1), we obtain

$ \begin{aligned} & \frac{5 x}{(x+1)(x^{2}+9)}=\frac{-1}{2(x+1)}+\frac{\frac{x}{2}+\frac{9}{2}}{(x^{2}+9)} \\ & \begin{aligned} \int \frac{5 x}{(x+1)(x^{2}+9)} d x & =\int{\frac{-1}{2(x+1)}+\frac{(x+9)}{2(x^{2}+9)}} d x \\ & =-\frac{1}{2} \log |x+1|+\frac{1}{2} \int \frac{x}{x^{2}+9} d x+\frac{9}{2} \int \frac{1}{x^{2}+9} d x \\ & =-\frac{1}{2} \log |x+1|+\frac{1}{4} \int \frac{2 x}{x^{2}+9} d x+\frac{9}{2} \int \frac{1}{x^{2}+9} d x \\ & =-\frac{1}{2} \log |x+1|+\frac{1}{4} \log |x^{2}+9|+\frac{9}{2} \cdot \frac{1}{3} \tan ^{-1} \frac{x}{3} \\ & =-\frac{1}{2} \log |x+1|+\frac{1}{4} \log (x^{2}+9)+\frac{3}{2} \tan ^{-1} \frac{x}{3}+C \end{aligned} \end{aligned} $

Solution

$ \frac{\sin x}{\sin (x-a)} $

Let $x-a=t d x=d t$

$ \begin{aligned} \int \frac{\sin x}{\sin (x-a)} d x & =\int \frac{\sin (t+a)}{\sin t} d t \\ & =\int \frac{\sin t \cos a+\cos t \sin a}{\sin t} d t \\ & =\int(\cos a+\cot t \sin a) d t \\ & =t \cos a+\sin a \log |\sin t|+C_1 \\ & =(x-a) \cos a+\sin a \log |\sin (x-a)|+C_1 \\ & =x \cos a+\sin a \log |\sin (x-a)|-a \cos a+C_1 \\ & =\sin a \log |\sin (x-a)|+x \cos a+C \end{aligned} $

Solution

$ \begin{aligned} \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} & =\frac{e^{4 \log x}(e^{\log x}-1)}{e^{2 \log x}(e^{\log x}-1)} \\ & =e^{2 \log x} \\ & =e^{\log x^{2}} \\ & =x^{2} \\ \therefore \int \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} d x & =\int x^{2} d x=\frac{x^{3}}{3}+C \end{aligned} $

Solution

$\frac{\cos x}{\sqrt{4-\sin ^{2} x}}$

Let $\sin x=t \cos x d x=d t$

$\Rightarrow \int \frac{\cos x}{\sqrt{4-\sin ^{2} x}} d x=\int \frac{d t}{\sqrt{(2)^{2}-(t)^{2}}}$

$ \begin{aligned} & =\sin ^{-1}(\frac{t}{2})+C \\ & =\sin ^{-1}(\frac{\sin x}{2})+C \end{aligned} $

Solution

$ \begin{aligned} & \frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}=\frac{(\sin ^{4} x+\cos ^{4} x)(\sin ^{4} x-\cos ^{4} x)}{\sin ^{2} x+\cos ^{2} x-\sin ^{2} x \cos ^{2} x-\sin ^{2} x \cos ^{2} x} \\ &=\frac{(\sin ^{4} x+\cos ^{4} x)(\sin ^{2} x+\cos ^{2} x)(\sin ^{2} x-\cos ^{2} x)}{(\sin ^{2} x-\sin ^{2} x \cos ^{2} x)+(\cos ^{2} x-\sin ^{2} x \cos ^{2} x)} \\ &=\frac{(\sin ^{4} x+\cos ^{4} x)(\sin ^{2} x-\cos ^{2} x)}{\sin ^{2} x(1-\cos ^{2} x)+\cos ^{2} x(1-\sin ^{2} x)} \\ &=\frac{-(\sin ^{4} x+\cos ^{4} x)(\cos ^{2} x-\sin ^{2} x)}{(\sin ^{4} x+\cos ^{4} x)} \\ &=-\cos 2 x \\ & \therefore \int \frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x} d x=\int-\cos 2 x d x=-\frac{\sin 2 x}{2}+C \end{aligned} $

Solution

$\frac{1}{\cos (x+a) \cos (x+b)}$

Multiplying and dividing by $\sin (a-b)$, we obtain

$ \begin{aligned} & \frac{1}{\sin (a-b)}[\frac{\sin (a-b)}{\cos (x+a) \cos (x+b)}] \\ & =\frac{1}{\sin (a-b)}[\frac{\sin [(x+a)-(x+b)]}{\cos (x+a) \cos (x+b)}] \\ & =\frac{1}{\sin (a-b)}[\frac{\sin (x+a) \cdot \cos (x+b)-\cos (x+a) \sin (x+b)}{\cos (x+a) \cos (x+b)}] \\ & =\frac{1}{\sin (a-b)}[\frac{\sin (x+a)}{\cos (x+a)}-\frac{\sin (x+b)}{\cos (x+b)}] \\ & =\frac{1}{\sin (a-b)}[\tan (x+a)-\tan (x+b)] \end{aligned} $

$ \begin{aligned} \int \frac{1}{\cos (x+a) \cos (x+b)} d x & =\frac{1}{\sin (a-b)} \int[\tan (x+a)-\tan (x+b)] d x \\ & =\frac{1}{\sin (a-b)}[-\log |\cos (x+a)|+\log |\cos (x+b)|]+C \\ & =\frac{1}{\sin (a-b)} \log |\frac{\cos (x+b)}{\cos (x+a)}|+C \end{aligned} $

Solution

$\frac{x^{3}}{\sqrt{1-x^{8}}}$

Let $x^{4}=t 4 x^{3} d x=d t$ $\Rightarrow \int \frac{x^{3}}{\sqrt{1-x^{8}}} d x=\frac{1}{4} \int \frac{d t}{\sqrt{1-t^{2}}}$

$ \begin{aligned} & =\frac{1}{4} \sin ^{-1} t+C \\ & =\frac{1}{4} \sin ^{-1}(x^{4})+C \end{aligned} $

Solution

$\frac{e^{x}}{(1+e^{x})(2+e^{x})}$

Let $e^{x}=t e^{x} d x=d t$

$\Rightarrow \int \frac{e^{x}}{(1+e^{x})(2+e^{x})} d x=\int \frac{d t}{(t+1)(t+2)}$

$=\int[\frac{1}{(t+1)}-\frac{1}{(t+2)}] d t$

$=\log |t+1|-\log |t+2|+C$

$=\log |\frac{t+1}{t+2}|+C$

$=\log |\frac{1+e^{x}}{2+e^{x}}|+C$

Solution

$\therefore \frac{1}{(x^{2}+1)(x^{2}+4)}=\frac{A x+B}{(x^{2}+1)}+\frac{C x+D}{(x^{2}+4)}$

$\Rightarrow 1=(A x+B)(x^{2}+4)+(C x+D)(x^{2}+1)$

$\Rightarrow 1=A x^{3}+4 A x+B x^{2}+4 B+C x^{3}+C x+D x^{2}+D$

Equating the coefficients of $x^{3}, x^{2}, x$, and constant term, we obtain

$A+C=0$

$B+D=0$

$4 A+C=0$

$4 B+D=1$

On solving these equations, we obtain

$A=0, B=\frac{1}{3}, C=0$, and $D=-\frac{1}{3}$

From equation (1), we obtain

$ \begin{aligned} & \frac{1}{(x^{2}+1)(x^{2}+4)}=\frac{1}{3(x^{2}+1)}-\frac{1}{3(x^{2}+4)} \\ & \begin{aligned} \int \frac{1}{(x^{2}+1)(x^{2}+4)} d x & =\frac{1}{3} \int \frac{1}{x^{2}+1} d x-\frac{1}{3} \int \frac{1}{x^{2}+4} d x \\ & =\frac{1}{3} \tan ^{-1} x-\frac{1}{3} \cdot \frac{1}{2} \tan ^{-1} \frac{x}{2}+C \\ & =\frac{1}{3} \tan ^{-1} x-\frac{1}{6} \tan ^{-1} \frac{x}{2}+C \end{aligned} \end{aligned} $

Solution

$\cos ^{3} x e^{\log \sin x}=\cos ^{3} x \times \sin x$

Let $\cos x=t -\sin x d x=d t$ $\Rightarrow \int \cos ^{3} x e^{\log \sin x} d x=\int \cos ^{3} x \sin x d x$

$ \begin{aligned} & =-\int t^{3} \cdot d t \\ & =-\frac{t^{4}}{4}+C \\ & =-\frac{\cos ^{4} x}{4}+C \end{aligned} $

Solution

$e^{3 \log x}(x^{4}+1)^{-1}=e^{\log x^{3}}(x^{4}+1)^{-1}=\frac{x^{3}}{(x^{4}+1)}$

Let $x^{4}+1=t \Rightarrow 4 x^{3} d x=d t$

$\Rightarrow \int e^{3 \log x}(x^{4}+1)^{-1} d x=\int \frac{x^{3}}{(x^{4}+1)} d x$

$ \begin{aligned} & =\frac{1}{4} \int \frac{d t}{t} \\ & =\frac{1}{4} \log |t|+C \\ & =\frac{1}{4} \log |x^{4}+1|+C \\ & =\frac{1}{4} \log (x^{4}+1)+C \end{aligned} $

Solution

$ f^{\prime}(a x+b)[f(a x+b)]^{n} $

Let $f(a x+b)=t \Rightarrow a f^{\prime}(a x+b) d x=d t$

$\Rightarrow \int f^{\prime}(a x+b)[f(a x+b)]^{n} d x=\frac{1}{a} \int t^{n} d t$

$ \begin{aligned} & =\frac{1}{a}[\frac{t^{n+1}}{n+1}] \\ & =\frac{1}{a(n+1)}(f(a x+b))^{n+1}+C \end{aligned} $

Solution

$ \begin{aligned} \frac{1}{\sqrt{\sin ^{3} x \sin (x+\alpha)}} & =\frac{1}{\sqrt{\sin ^{3} x(\sin x \cos \alpha+\cos x \sin \alpha)}} \\ & =\frac{1}{\sqrt{\sin ^{4} x \cos \alpha+\sin ^{3} x \cos x \sin \alpha}} \\ & =\frac{1}{\sin ^{2} x \sqrt{\cos \alpha+\cot x \sin \alpha}} \\ & =\frac{cosec^{2}x}{\sqrt{\cos \alpha+\cot x \sin \alpha}} \end{aligned} $

Let $\cos \alpha+\cot x \sin \alpha=t \Rightarrow-cosec^{2} x \sin \alpha d x=d t$

$ \begin{aligned} \therefore \int \frac{1}{\sin ^{3} x \sin (x+\alpha)} d x & =\int \frac{cosec^{2} x}{\sqrt{\cos \alpha+\cot x \sin \alpha}} d x \\ & =\frac{-1}{\sin \alpha} \int \frac{d t}{\sqrt{t}} \\ & =\frac{-1}{\sin \alpha}[2 \sqrt{t}]+C \\ & =\frac{-1}{\sin \alpha}[2 \sqrt{\cos \alpha+\cot x \sin \alpha}]+C \\ & =\frac{-2}{\sin \alpha} \sqrt{\cos \alpha+\frac{\cos x \sin \alpha}{\sin x}}+C \\ & =\frac{-2}{\sin \alpha} \sqrt{\frac{\sin x \cos \alpha+\cos x \sin \alpha}{\sin x}}+C \\ & =-\frac{2}{\sin \alpha} \sqrt{\frac{\sin (x+\alpha)}{\sin x}+C} \end{aligned} $

Solution

$ I=\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x $

Let $x=\cos ^{2} \theta \Rightarrow d x=-2 \sin \theta \cos \theta d \theta$

$ I=\int \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}(-2 \sin \theta \cos \theta) d \theta $

$ \begin{aligned} & =-\int \sqrt{\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}} \sin 2 \theta d \theta \\ & =-\int \tan \frac{\theta}{2} \cdot 2 \sin \theta \cos \theta d \theta \\ & =-2 \int \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}(2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}) \cos \theta d \theta \end{aligned} $

$=-4 \int \sin ^{2} \frac{\theta}{2} \cos \theta d \theta$

$=-4 \int \sin ^{2} \frac{\theta}{2} \cdot(2 \cos ^{2} \frac{\theta}{2}-1) d \theta$

$=-4 \int(2 \sin ^{2} \frac{\theta}{2} \cos ^{2} \frac{\theta}{2}-\sin ^{2} \frac{\theta}{2}) d \theta$

$=-8 \int \sin ^{2} \frac{\theta}{2} \cdot \cos ^{2} \frac{\theta}{2} d \theta+4 \int \sin ^{2} \frac{\theta}{2} d \theta$

$=-2 \int \sin ^{2} \theta d \theta+4 \int \sin ^{2} \frac{\theta}{2} d \theta$

$=-2 \int(\frac{1-\cos 2 \theta}{2}) d \theta+4 \int \frac{1-\cos \theta}{2} d \theta$

$=-2[\frac{\theta}{2}-\frac{\sin 2 \theta}{4}]+4[\frac{\theta}{2}-\frac{\sin \theta}{2}]+C$

$=-\theta+\frac{\sin 2 \theta}{2}+2 \theta-2 \sin \theta+C$

$=\theta+\frac{\sin 2 \theta}{2}-2 \sin \theta+C$

$=\theta+\frac{2 \sin \theta \cos \theta}{2}-2 \sin \theta+C$

$=\theta+\sqrt{1-\cos ^{2} \theta} \cdot \cos \theta-2 \sqrt{1-\cos ^{2} \theta}+C$

$=\cos ^{-1} \sqrt{x}+\sqrt{1-x} \cdot \sqrt{x}-2 \sqrt{1-x}+C$

$=-2 \sqrt{1-x}+\cos ^{-1} \sqrt{x}+\sqrt{x(1-x)}+C$

$=-2 \sqrt{1-x}+\cos ^{-1} \sqrt{x}+\sqrt{x-x^{2}}+C$

Solution

$ \begin{aligned} I & =\int(\frac{2+\sin 2 x}{1+\cos 2 x}) e^{x} \\ & =\int(\frac{2+2 \sin x \cos x}{2 \cos ^{2} x}) e^{x} \\ & =\int(\frac{1+\sin x \cos x}{\cos ^{2} x}) e^{x} \\ & =\int(\sec ^{2} x+\tan x) e^{x} \end{aligned} $

Let $f(x)=\tan x \Rightarrow f^{\prime}(x)=\sec ^{2} x$

$ \begin{aligned} \therefore I & =\int(f(x)+f^{\prime}(x)] e^{x} d x \\ & =e^{x} f(x)+C \\ & =e^{x} \tan x+C \end{aligned} $

Solution

Let $\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+1)^{2}}+\frac{C}{(x+2)}$

$\Rightarrow x^{2}+x+1=A(x+1)(x+2)+B(x+2)+C(x^{2}+2 x+1)$

$\Rightarrow x^{2}+x+1=A(x^{2}+3 x+2)+B(x+2)+C(x^{2}+2 x+1)$

$\Rightarrow x^{2}+x+1=(A+C) x^{2}+(3 A+B+2 C) x+(2 A+2 B+C)$

Equating the coefficients of $x^{2}, x$, and constant term, we obtain

$A+C=1$

$3 A+B+2 C=1$

$2 A+2 B+C=1$

On solving these equations, we obtain

$A=-2, B=1$, and $C=3$

From equation (1), we obtain

$ \begin{aligned} \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} & =\frac{-2}{(x+1)}+\frac{3}{(x+2)}+\frac{1}{(x+1)^{2}} \\ \int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} d x & =-2 \int \frac{1}{x+1} d x+3 \int \frac{1}{(x+2)} d x+\int \frac{1}{(x+1)^{2}} d x \\ & =-2 \log |x+1|+3 \log |x+2|-\frac{1}{(x+1)}+C \end{aligned} $

Solution

$I=\tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x$

Let $x=\cos \theta \Rightarrow d x=-\sin \theta d \theta$

$I=\int \tan ^{-1} \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}(-\sin \theta d \theta)$

$=-\int \tan ^{-1} \sqrt{\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}} \sin \theta d \theta$

$=-\int \tan ^{-1} \tan \frac{\theta}{2} \cdot \sin \theta d \theta$

$=-\frac{1}{2} \int \theta \cdot \sin \theta d \theta$

$=-\frac{1}{2}[\theta \cdot(-\cos \theta)-\int 1 \cdot(-\cos \theta) d \theta]$

$=-\frac{1}{2}[-\theta \cos \theta+\sin \theta]$

$=+\frac{1}{2} \theta \cos \theta-\frac{1}{2} \sin \theta$

$=\frac{1}{2} \cos ^{-1} x \cdot x-\frac{1}{2} \sqrt{1-x^{2}}+C$

$=\frac{x}{2} \cos ^{-1} x-\frac{1}{2} \sqrt{1-x^{2}}+C$

$=\frac{1}{2}(x \cos ^{-1} x-\sqrt{1-x^{2}})+C$

Solution

$ \begin{aligned} \frac{\sqrt{x^{2}+1}[\log (x^{2}+1)-2 \log x]}{x^{4}} & =\frac{\sqrt{x^{2}+1}}{x^{4}}[\log (x^{2}+1)-\log x^{2}] \\ & =\frac{\sqrt{x^{2}+1}}{x^{4}}[\log (\frac{x^{2}+1}{x^{2}})] \\ & =\frac{\sqrt{x^{2}+1}}{x^{4}} \log (1+\frac{1}{x^{2}}) \\ & =\frac{1}{x^{3}} \sqrt{\frac{x^{2}+1}{x^{2}}} \log (1+\frac{1}{x^{2}}) \\ & =\frac{1}{x^{3}} \sqrt{1+\frac{1}{x^{2}}} \log (1+\frac{1}{x^{2}}) \end{aligned} $

Let $1+\frac{1}{x^{2}}=t \Rightarrow \frac{-2}{x^{3}} d x=d t$

$ \begin{aligned} \therefore I & =\int \frac{1}{x^{3}} \sqrt{1+\frac{1}{x^{2}}} \log (1+\frac{1}{x^{2}}) d x \\ & =-\frac{1}{2} \int \sqrt{t} \log t d t \\ & =-\frac{1}{2} \int t^{\frac{1}{2}} \cdot \log t d t \end{aligned} $

Integrating by parts, we obtain

$ \begin{aligned} I & =-\frac{1}{2}[\log t \cdot \int t^{\frac{1}{2}} d t-{(\frac{d}{d t} \log t) \int t^{\frac{1}{2}} d t} d t] \\ & =-\frac{1}{2}[\log t \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}}-\int \frac{1}{t} \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}} d t] \\ & =-\frac{1}{2}[\frac{2}{3} t^{\frac{3}{2}} \log t-\frac{2}{3} \int t^{\frac{1}{2}} d t] \\ & =-\frac{1}{2}[\frac{2}{3} t^{\frac{3}{2}} \log t-\frac{4}{9} t^{\frac{3}{2}}] \\ & =-\frac{1}{3} t^{\frac{3}{2}} \log t+\frac{2}{9} t^{\frac{3}{2}} \\ & =-\frac{1}{3} t^{\frac{3}{2}}[\log t-\frac{2}{3}] \\ & =-\frac{1}{3}(1+\frac{1}{x^{2}})^{\frac{3}{2}}[\log (1+\frac{1}{x^{2}})-\frac{2}{3}]+C \end{aligned} $

Solution

$ \begin{aligned} I & =\int _{\frac{\pi}{2}}^{\pi} e^{x}(\frac{1-\sin x}{1-\cos x}) d x \\ & =\int _{\frac{\pi}{2}}^{\pi} e^{x}(\frac{1-2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin ^{2} \frac{x}{2}}) d x \\ & =\int _{\frac{\pi}{2}}^{\pi} e^{x}(\frac{cosec^{2} \frac{x}{2}}{2}-\cot \frac{x}{2}) d x \end{aligned} $

Let $f(x)=-\cot \frac{x}{2}$

$\Rightarrow f^{\prime}(x)=-(-\frac{1}{2} cosec^{2} \frac{x}{2})=\frac{1}{2} cosec^{2} \frac{x}{2}$

$\therefore I=\int _{\frac{\pi}{2}}^{\pi} e^{x}(f(x)+f^{\prime}(x)] d x$

$=[e^{x} \cdot f(x) d x] _{\frac{\pi}{2}}^{x}$

$=-[e^{x} \cdot \cot \frac{x}{2}] _{\frac{\pi}{2}}^{\pi}$

$=-[e^{\pi} \times \cot \frac{\pi}{2}-e^{\frac{\pi}{2}} \times \cot \frac{\pi}{4}]$

$=-[e^{\pi} \times 0-e^{\frac{\pi}{2}} \times 1]$

$=e^{\frac{\pi}{2}}$

Solution

Let $I=\int_0^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x$

$ \begin{aligned} & \Rightarrow I=\int_0^{\frac{\pi}{4}} \frac{\frac{(\sin x \cos x)}{\cos ^{4} x}}{\frac{(\cos ^{4} x+\sin ^{4} x)}{\cos ^{4} x}} d x \\ & \Rightarrow I=\int_0^{\frac{\pi}{4}} \frac{\tan x \sec ^{2} x}{1+\tan ^{4} x} d x \end{aligned} $

Let $\tan ^{2} x=t \Rightarrow 2 \tan x \sec ^{2} x d x=d t$

$ \begin{aligned} \therefore I & =\frac{1}{2} \int_0^{1} \frac{d t}{1+t^{2}} \\ & =\frac{1}{2}[\tan ^{-1} t]_0^{1} \\ & =\frac{1}{2}[\tan ^{-1} 1-\tan ^{-1} 0] \\ & =\frac{1}{2}[\frac{\pi}{4}] \\ & =\frac{\pi}{8} \end{aligned} $

$ \text{ When } x=0, t=0 \text{ and when } x=\frac{\pi}{4}, t=1 $

Solution

Let $I=\int_0^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4 \sin ^{2} x} d x$

$\Rightarrow I=\int_0^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4(1-\cos ^{2} x)} d x$

$\Rightarrow I=\int_0^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4-4 \cos ^{2} x} d x$

$\Rightarrow I=\frac{-1}{3} \int_0^{\frac{\pi}{2}} \frac{4-3 \cos ^{2} x-4}{4-3 \cos ^{2} x} d x$

$\Rightarrow I=\frac{-1}{3} \int_0^{\frac{\pi}{2}} \frac{4-3 \cos ^{2} x}{4-3 \cos ^{2} x} d x+\frac{1}{3} \int_0^{\frac{\pi}{2}} \frac{4}{4-3 \cos ^{2} x} d x$

$\Rightarrow I=\frac{-1}{3} \int_0^{\frac{\pi}{2}} 1 d x+\frac{1}{3} \int_0^{\frac{\pi}{2}} \frac{4 \sec ^{2} x}{4 \sec ^{2} x-3} d x$

$\Rightarrow I=\frac{-1}{3}[x]_0^{\frac{\pi}{2}}+\frac{1}{3} \int_0^{\frac{\pi}{2}} \frac{4 \sec ^{2} x}{4(1+\tan ^{2} x)-3} d x$

$\Rightarrow I=-\frac{\pi}{6}+\frac{2}{3} \int_0^{\frac{\pi}{2}} \frac{2 \sec ^{2} x}{1+4 \tan ^{2} x} d x$

Consider, $\int_0^{\frac{\pi}{2}} \frac{2 \sec ^{2} x}{1+4 \tan ^{2} x} d x$

Let $2 \tan x=t \Rightarrow 2 \sec ^{2} x d x=d t$

When $x=0, t=0$ and when $x=\frac{\pi}{2}, t=\infty$

$\Rightarrow \int_0^{\frac{\pi}{2}} \frac{2 \sec ^{2} x}{1+4 \tan ^{2} x} d x=\int_0^{\infty} \frac{d t}{1+t^{2}}$

$ \begin{aligned} & =[\tan ^{-1} t]_0^{\infty} \\ & =[\tan ^{-1}(\infty)-\tan ^{-1}(0)] \\ & =\frac{\pi}{2} \end{aligned} $

Therefore, from (1),we obtain

$I=-\frac{\pi}{6}+\frac{2}{3}[\frac{\pi}{2}]=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$

Solution

Let $I=\int _{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$

$\Rightarrow I=\int _{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x+\cos x)}{\sqrt{-(-\sin 2 x)}} d x$

$\Rightarrow I=\int _{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{-(-1+1-2 \sin x \cos x)}} d x$

$\Rightarrow I=\int _{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x+\cos x)}{\sqrt{1-(\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x)}} d x$

$\Rightarrow I=\int _{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x+\cos x) d x}{\sqrt{1-(\sin x-\cos x)^{2}}}$

Let $(\sin x-\cos x)=t \Rightarrow(\sin x+\cos x) d x=d t$

When $x=\frac{\pi}{6}, t=(\frac{1-\sqrt{3}}{2})$ and when $\quad x=\frac{\pi}{3}, t=(\frac{\sqrt{3}-1}{2})$

$I=\int _{\frac{1-\sqrt{3}}{2}}^{\frac{\sqrt{3}-1}{2}} \frac{d t}{\sqrt{1-t^{2}}}$

$\Rightarrow I=\int _{-(\frac{\sqrt{3}-1}{2})}^{.\frac{\sqrt{3}-1}{2})} \frac{d t}{\sqrt{1-t^{2}}}$

As $\frac{1}{\sqrt{1-(-t)^{2}}}=\frac{1}{\sqrt{1-t^{2}}}$, therefore, $\frac{1}{\sqrt{1-t^{2}}}$ is an even function.

It is known that if $f(x)$ is an even function, then $\int _{-a}^{a} f(x) d x=2 \int_0^{a} f(x) d x$

$ \begin{aligned} \Rightarrow I & =2 \int_0^{\sqrt{3}-1} \frac{d t}{\sqrt{1-t^{2}}} \\ & =[2 \sin ^{-1} t]_0^{\frac{\sqrt{3}-1}{2}} \\ & =2 \sin ^{-1}(\frac{\sqrt{3}-1}{2}) \end{aligned} $

Solution

$ \begin{aligned} & \text{ Let } I=\int_0^{1} \frac{d x}{\sqrt{1+x}-\sqrt{x}} \\ & \begin{aligned} I & =\int_0^{1} \frac{1}{(\sqrt{1+x}-\sqrt{x})} \times \frac{(\sqrt{1+x}+\sqrt{x})}{(\sqrt{1+x}+\sqrt{x})} d x \\ & =\int_0^{1} \frac{\sqrt{1+x}+\sqrt{x}}{1+x-x} d x \\ & =\int_0^{1} \sqrt{1+x} d x+\int_0^{1} \sqrt{x} d x \\ & =[\frac{2}{3}(1+x)^{\frac{3}{2}}]_0^{1}+[\frac{2}{3}(x)^{\frac{3}{2}}]_0^{1} \\ & =\frac{2}{3}[(2)^{\frac{3}{2}}-1]+\frac{2}{3}[1] \\ & =\frac{2}{3}(2)^{\frac{3}{2}} \\ & =\frac{2 \cdot 2 \sqrt{2}}{3} \\ & =\frac{4 \sqrt{2}}{3} \end{aligned} \\ & \text{ (1) } \end{aligned} $

Solution

Let $I=\int_0^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x$

Also, let $\sin x-\cos x=t \Rightarrow(\cos x+\sin x) d x=d t$

When $x=0, t=-1$ and when $x=\frac{\pi}{4}, t=0$

$\Rightarrow(\sin x-\cos x)^{2}=t^{2}$

$\Rightarrow \sin ^{2} x+\cos ^{2} x-2 \sin x \cos x=t^{2}$

$\Rightarrow 1-\sin 2 x=t^{2}$

$\Rightarrow \sin 2 x=1-t^{2}$

$\therefore I=\int _{-1}^{0} \frac{d t}{9+16(1-t^{2})}$

$=\int _{-1}^{0} \frac{d t}{9+16-16 t^{2}}$

$=\int _{-1}^{0} \frac{d t}{25-16 t^{2}}=\int _{-1}^{0} \frac{d t}{(5)^{2}-(4 t)^{2}}$

$=\frac{1}{4}[\frac{1}{2(5)} \log |\frac{5+4 t}{5-4 t}|] _{-1}^{0}$

$=\frac{1}{40}[\log (1)-\log |\frac{1}{9}|]$

$=\frac{1}{40} \log 9$

Solution

Let $I=\int_0^{\frac{\pi}{2}} \sin 2 x \tan ^{-1}(\sin x) d x=\int_0^{\frac{\pi}{2}} 2 \sin x \cos x \tan ^{-1}(\sin x) d x$

Also, let $\sin x=t \Rightarrow \cos x d x=d t$

When $x=0, t=0$ and when $x=\frac{\pi}{2}, t=1$

$\Rightarrow I=2 \int_0^{1} t \tan ^{-1}(t) d t$

Consider $\int t \cdot \tan ^{-1} t d t=\tan ^{-1} t \cdot \int t d t-\int{\frac{d}{d t}(\tan ^{-1} t) \int t d t} d t$

$ \begin{aligned} & =\tan ^{-1} t \cdot \frac{t^{2}}{2}-\int \frac{1}{1+t^{2}} \cdot \frac{t^{2}}{2} d t \\ & =\frac{t^{2} \tan ^{-1} t}{2}-\frac{1}{2} \int \frac{t^{2}+1-1}{1+t^{2}} d t \\ & =\frac{t^{2} \tan ^{-1} t}{2}-\frac{1}{2} \int 1 d t+\frac{1}{2} \int \frac{1}{1+t^{2}} d t \\ & =\frac{t^{2} \tan ^{-1} t}{2}-\frac{1}{2} \cdot t+\frac{1}{2} \tan ^{-1} t \end{aligned} $

$ \begin{aligned} \Rightarrow \int_0^{1} t \cdot \tan ^{-1} t d t & =[\frac{t^{2} \cdot \tan ^{-1} t}{2}-\frac{t}{2}+\frac{1}{2} \tan ^{-1} t]_0^{1} \\ & =\frac{1}{2}[\frac{\pi}{4}-1+\frac{\pi}{4}] \\ & =\frac{1}{2}[\frac{\pi}{2}-1]=\frac{\pi}{4}-\frac{1}{2} \end{aligned} $

From equation (1), we obtain

$I=2[\frac{\pi}{4}-\frac{1}{2}]=\frac{\pi}{2}-1$

Solution

Let $I=\int_1^{4}[|x-1|+|x-2|+|x-3|] d x$

$\Rightarrow I=\int^{4}|x-1| d x+\int_1^{4}|x-2| d x+\int^{4}|x-3| d x$

$I=I_1+I_2+I_3$

where, $I_1=\int_1^{4}|x-1| d x, I_2=\int_1^{4}|x-2| d x$, and $I_3=\int_1^{4}|x-3| d x$

$I_1=\int_1^{4}|x-1| d x$

$(x-1) \geq 0$ for $1 \leq x \leq 4$

$\therefore I_1=\int_1^{4}(x-1) d x$

$\Rightarrow I_1=[\frac{x^{2}}{x}-x]_1^{4}$

$\Rightarrow I_1=[8-4-\frac{1}{2}+1]=\frac{9}{2}$

$I_2=\int^{4}|x-2| d x$

$x-2 \geq 0$ for $2 \leq x \leq 4$ and $x-2 \leq 0$ for $1 \leq x \leq 2$

$\therefore I_2=\int^{2}(2-x) d x+\int_2^{4}(x-2) d x$

$\Rightarrow I_2=[2 x-\frac{x^{2}}{2}]_1^{2}+[\frac{x^{2}}{2}-2 x]_2^{4}$

$\Rightarrow I_2=[4-2-2+\frac{1}{2}]+[8-8-2+4]$

$\Rightarrow I_2=\frac{1}{2}+2=\frac{5}{2}$ $I_3=\int^{4}|x-3| d x$

$x-3 \geq 0$ for $3 \leq x \leq 4$ and $x-3 \leq 0$ for $1 \leq x \leq 3$

$\therefore I_3=\int_1^{3}(3-x) d x+\int_3^{4}(x-3) d x$

$\Rightarrow I_3=[3 x-\frac{x^{2}}{2}]_1^{3}+[\frac{x^{2}}{2}-3 x]_3^{4}$

$\Rightarrow I_3=[9-\frac{9}{2}-3+\frac{1}{2}]+[8-12-\frac{9}{2}+9]$

$\Rightarrow I_3=[6-4]+[\frac{1}{2}]=\frac{5}{2}$

From equations (1), (2), (3), and (4), we obtain

$I=\frac{9}{2}+\frac{5}{2}+\frac{5}{2}=\frac{19}{2}$

Solution

Let $I=\int_1^{3} \frac{d x}{x^{2}(x+1)}$

Also, let $\frac{1}{x^{2}(x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1}$

$\Rightarrow 1=A x(x+1)+B(x+1)+C(x^{2})$

$\Rightarrow 1=A x^{2}+A x+B x+B+C x^{2}$

Equating the coefficients of $x^{2}, x$, and constant term, we obtain

$A+C=0$

$A+B=0$

$B=1$

On solving these equations, we obtain

$A=-1, C=1$, and $B=1$ $\therefore \frac{1}{x^{2}(x+1)}=\frac{-1}{x}+\frac{1}{x^{2}}+\frac{1}{(x+1)}$

$\Rightarrow I=\int^{3}{-\frac{1}{x}+\frac{1}{x^{2}}+\frac{1}{(x+1)}} d x$

$ =[-\log x-\frac{1}{x}+\log (x+1)]_1^{3} $

$ =[\log (\frac{x+1}{x})-\frac{1}{x}]_1^{3} $

$=\log (\frac{4}{3})-\frac{1}{3}-\log (\frac{2}{1})+1$

$=\log 4-\log 3-\log 2+\frac{2}{3}$

$=\log 2-\log 3+\frac{2}{3}$

$=\log (\frac{2}{3})+\frac{2}{3}$

Hence, the given result is proved.

Solution

Let $I=\int_0^{1} x e^{x} d x$

Integrating by parts, we obtain

$ \begin{aligned} I & =x \int_0^{1} e^{x} d x-\int_0^{1}{(\frac{d}{d x}(x)) \int e^{x} d x} d x \\ & =[x e^{x}]_0^{1}-\int_0^{1} e^{x} d x \\ & =[x e^{x}]_0^{1}-[e^{x}]_0^{1} \\ & =e-e+1 \\ & =1 \end{aligned} $

Hence, the given result is proved.

Solution

Let $I=\int _{-1}^{1} x^{17} \cos ^{4} x d x$

Also, let $f(x)=x^{17} \cos ^{4} x$

$\Rightarrow f(-x)=(-x)^{17} \cos ^{4}(-x)=-x^{17} \cos ^{4} x=-f(x)$

Therefore, $f(x)$ is an odd function.

It is known that if $f(x)$ is an odd function, then $\int _{-a}^{a} f(x) d x=0$

$\therefore I=\int _{-1}^{1} x^{17} \cos ^{4} x d x=0$

Hence, the given result is proved.

Solution

Let $I=\int_0^{\frac{\pi}{2}} \sin ^{3} x d x$

$I=\int_0^{\frac{\pi}{2}} \sin ^{2} x \cdot \sin x d x$

$=\int_0^{\frac{\pi}{2}}(1-\cos ^{2} x) \sin x d x$

$=\int_0^{\frac{\pi}{2}} \sin x d x-\int_0^{\frac{\pi}{2}} \cos ^{2} x \cdot \sin x d x$

$=[-\cos x]_0^{\frac{\pi}{2}}+[\frac{\cos ^{3} x}{3}]_0^{\frac{\pi}{2}}$

$=1+\frac{1}{3}[-1]=1-\frac{1}{3}=\frac{2}{3}$

Hence, the given result is proved.

Solution

Let $I=\int_0^{\frac{\pi}{4}} 2 \tan ^{3} x d x$

$I=2 \int_0^{\frac{\pi}{4}} \tan ^{2} x \tan x d x=2 \int_0^{\frac{\pi}{4}}(\sec ^{2} x-1) \tan x d x$

$=2 \int_0^{\frac{\pi}{4}} \sec ^{2} x \tan x d x-2 \int_0^{\frac{\pi}{4}} \tan x d x$

$=2[\frac{\tan ^{2} x}{2}]_0^{\frac{\pi}{4}}+2[\log \cos x]_0^{\frac{\pi}{4}}$

$=1+2[\log \cos \frac{\pi}{4}-\log \cos 0]$

$=1+2[\log \frac{1}{\sqrt{2}}-\log 1]$

$=1-\log 2-\log 1=1-\log 2$

Hence, the given result is proved.

Solution

Let $I=\int_0^{1} \sin ^{-1} x d x$

$\Rightarrow I=\int_0^{1} \sin ^{-1} x \cdot 1 \cdot d x$

Integrating by parts, we obtain

$ \begin{aligned} I & =[\sin ^{-1} x \cdot x]_0^{1}-\int_0^{1} \frac{1}{\sqrt{1-x^{2}}} \cdot x d x \\ & =[x \sin ^{-1} x]_0^{1}+\frac{1}{2} \int_0^{1} \frac{(-2 x)}{\sqrt{1-x^{2}}} d x \end{aligned} $

Let $1-x^{2}=t \square-2 x d x=d t$

When $x=0, t=1$ and when $x=1, t=0$

$ \begin{aligned} I & =[x \sin ^{-1} x]_0^{1}+\frac{1}{2} \int_0^{0} \frac{d t}{\sqrt{t}} \\ & =[x \sin ^{-1} x]_0^{1}+\frac{1}{2}[2 \sqrt{t}]_1^{0} \\ & =\sin ^{-1}(1)+[-\sqrt{1}] \\ & =\frac{\pi}{2}-1 \end{aligned} $

Hence, the given result is proved.

Solution

Let $I=\int \frac{d x}{e^{x}+e^{-x}} d x=\int \frac{e^{x}}{e^{2 x}+1} d x$

Also, let $e^{x}=t \Rightarrow e^{x} d x=d t$

$ \begin{aligned} \therefore I & =\int \frac{d t}{1+t^{2}} \\ & =\tan ^{-1} t+C \\ & =\tan ^{-1}(e^{x})+C \end{aligned} $

Hence, the correct Answer is A.

Solution

Let $I=\frac{\cos 2 x}{(\cos x+\sin x)^{2}}$

$ \begin{aligned} I & =\int \frac{\cos ^{2} x-\sin ^{2} x}{(\cos x+\sin x)^{2}} d x \\ & =\int \frac{(\cos x+\sin x)(\cos x-\sin x)}{(\cos x+\sin x)^{2}} d x \\ & =\int \frac{\cos x-\sin x}{\cos +\sin x} d x \end{aligned} $

Let $\cos x+\sin x=t \Rightarrow(\cos x-\sin x) d x=d t$

$ \begin{aligned} \therefore I & =\int \frac{d t}{t} \\ & =\log |t|+C \\ & =\log |\cos x+\sin x|+C \end{aligned} $

Hence, the correct Answer is B.

Solution

Let $I=\int_a^{b} x f(x) d x$

$ \begin{aligned} & I=\int_a^{b}(a+b-x) f(a+b-x) d x \\ & (\int_a^{b} f(x) d x=\int_a^{b} f(a+b-x) d x) \\ & \Rightarrow I=\int_a^{b}(a+b-x) f(x) d x \\ & \Rightarrow I=(a+b) \int_a^{b} f(x) d x \quad-I \\ & \Rightarrow I+I=(a+b) \int_a^{b} f(x) d x \\ & \Rightarrow 2 I=(a+b) \int_a^{b} f(x) d x \\ & \Rightarrow I=(\frac{a+b}{2}) \int_a^{b} f(x) d x \end{aligned} $

Hence, the correct Answer is D.