Solution

L.H.S. $=\int \frac{2 x-1}{2 x+3} d x$

$\Rightarrow \int(1-\frac{4}{2 x+3}) d x \quad \begin{aligned} & \text{ [Dividing the numerator by the } \\ & \text{ denominator] }\end{aligned}$

$\Rightarrow \int 1 . d x-4 \int \frac{1}{2 x+3} d x \Rightarrow \int 1 . d x-\frac{4}{2} \int \frac{1}{x+\frac{3}{2}} d x$

$\Rightarrow \int 1 . d x-2 \int \frac{1}{x+\frac{3}{2}} d x \Rightarrow x-2 \log |x+\frac{3}{2}|+C$

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

$\Rightarrow x-\log |(2 x+3)^{2}|-\log 2^{2}+C$

$[\because n \log m=\log m^{n}]$

$\Rightarrow x-\log |(2 x+3)^{2}|+C_1 \Rightarrow$ R.H.S. $\quad[.$ where $.C_1=C-\log 2^{2}]$

L.H.S. $=$ R.H.S.

Hence proved.

Solution

L.H.S. $=\int \frac{2 x+3}{x^{2}+3 x} d x$

Put $\quad x^{2}+3 x=t$

$\therefore \quad(2 x+3) d x=d t$

$\Rightarrow \quad \int \frac{d t}{t}=\log |t| \Rightarrow \log |x^{2}+3 x|+C=$ R.H.S.

L.H.S. $=$ R.H.S.

Hence verified.

Solution

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

$ \begin{aligned} \therefore \quad I & =\int[(x-1)+\frac{3}{x+1}] d x \\ & =\int(x-1) d x+3 \int \frac{1}{x+1} d x \\ & =\frac{x^{2}}{2}-x+3 \log |x+1|+C \end{aligned} $

$ x+1 x_2^{2}+2(x-1 $

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

$-x-1$

$\frac{(+)(+)}{3}$

Hence, the required solution is $\frac{x^{2}}{2}-x+3 \log |x+1|+C$.

Solution

Let $I=\int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}} d x=\int \frac{e^{\log x^{6}}-e^{\log x^{5}}}{e^{\log x^{4}}-e^{\log x^{3}}} d x$

$ =\int \frac{x^{6}-x^{5}}{x^{4}-x^{3}} d x=\int \frac{x^{2}(x^{4}-x^{3})}{x^{4}-x^{3}} d x=\int x^{2} d x=\frac{1}{3} x^{3}+C $

Hence, the required solution is $\frac{1}{3} x^{3}+C$.

Solution

Let

$ I=\int \frac{1+\cos x}{x+\sin x} d x $

Put $\quad x+\sin x=t \quad \Rightarrow(1+\cos x) d x=d t$

$ \therefore \quad I=\int \frac{d t}{t}=\log |t|=\log |x+\sin x|+C $

Hence, the required solution is $\log |x+\sin x|+C$.

Solution

Let $I=\int \frac{d x}{1+\cos x}=\int \frac{d x}{2 \cos ^{2} x / 2} \quad[\because 1+\cos x=2 \cos ^{2} \frac{x}{2}]$

$ =\frac{1}{2} \int \sec ^{2} \frac{x}{2} d x=\frac{1}{2} \cdot 2 \tan \frac{x}{2}+C=\tan \frac{x}{2}+C $

Hence, the required solution is $\tan \frac{x}{2}+C$.

Solution

Let $I=\int \tan ^{2} x \cdot \sec ^{4} x d x$

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

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

$ \begin{aligned} \therefore \quad I & =\int t^{2}(1+t^{2}) d t=\int(t^{2}+t^{4}) d t=\int t^{2} d t+\int t^{4} d t \\ & =\frac{1}{3} t^{3}+\frac{1}{5} t^{5}=\frac{1}{3} \tan ^{3} x+\frac{1}{5} \tan ^{5} x+C \end{aligned} $

Hence, the required solution is $\frac{1}{3} \tan ^{3} x+\frac{1}{5} \tan ^{5} x+C$.

Solution

Let

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

Hence, the required solution is $x+C$.

Solution

Let $I=\int \sqrt{1+\sin x} d x$

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

Hence, the required solution is $2(\sin \frac{x}{2}-\cos \frac{x}{2})+C$.

Solution

$\quad I=\int \frac{x}{\sqrt{x}+1} d x$

$ \begin{aligned} & \text{ Put } \begin{aligned} & \sqrt{x}=t \Rightarrow x=t^{2} \quad \therefore d x=2 t \cdot d t \\ \therefore I & =\int \frac{t^{2} \cdot 2 t \cdot d t}{t+1}=2 \int \frac{t^{3}}{t+1} d t=2 \int \frac{t^{3}+1-1}{t+1} d t \\ & =2 \int \frac{t^{3}+1}{t+1} d t-2 \int \frac{1}{t+1} d t \\ & =2 \int \frac{(t+1)(t^{2}-t+1)}{t+1} d t-2 \int \frac{1}{t+1} d t \\ & =2 \int(t^{2}-t+1) d t-2 \int \frac{1}{t+1} d t \\ & =2[\frac{t^{3}}{3}-\frac{t^{2}}{2}+t]-2 \log |t+1| \\ & =2[\frac{x^{3 / 2}}{3}-\frac{x}{2}+\sqrt{x}]-2 \log |\sqrt{x}+1|+C \\ & =2[\frac{x \sqrt{x}}{3}-\frac{x}{2}+\sqrt{x}-\log |\sqrt{x}+1|]+C \end{aligned} \end{aligned} $

Hence, $I=2[\frac{x \sqrt{x}}{3}-\frac{x}{2}+\sqrt{x}-\log |\sqrt{x}+1|]+C$

Solution

Let $I=\int \sqrt{\frac{a+x}{a-x}} d x$

$ \begin{aligned} & =\int \sqrt{\frac{a+x}{a-x} \times \frac{a+x}{a+x}} d x=\int \frac{a+x}{\sqrt{(a-x)(a+x)}} d x \\ & =\int \frac{a+x}{\sqrt{a^{2}-x^{2}}} d x=\int \frac{a}{\sqrt{a^{2}-x^{2}}} d x+\int \frac{x}{\sqrt{a^{2}-x^{2}}} d x \end{aligned} $

Let $\quad I=I_1+I_2$

Now $\quad I_1=\int \frac{a}{\sqrt{a^{2}-x^{2}}} d x=a \cdot \sin ^{-1} \frac{x}{a}+C_1$

and $\quad I_2=\int \frac{x}{\sqrt{a^{2}-x^{2}}} d x$

Put $\quad a^{2}-x^{2}=t \Rightarrow-2 x d x=d t$

$ x d x=\frac{d t}{-2} $

$\therefore \quad I_2=-\frac{1}{2} \int \frac{d t}{\sqrt{t}}=-\frac{1}{2} \times 2 \sqrt{t}=-\sqrt{a^{2}-x^{2}}+C_2$

Since $\quad I=I_1+I_2$

$ =a \sin ^{-1} \frac{x}{a}+C_1-\sqrt{a^{2}-x^{2}}+C_2 $

$\therefore \quad I=a \sin ^{-1} \frac{x}{a}-\sqrt{a^{2}-x^{2}}+(C_1+C_2)$

Hence, $I=a \sin ^{-1} \frac{x}{a}-\sqrt{a^{2}-x^{2}}+C \quad[C=C_1+C_2]$

Alternate method:

Put $\quad x=a \cos 2 \theta$

$ I=\int \sqrt{\frac{a+x}{a-x}} d x $

$\therefore \quad d x=a(-2 \sin 2 \theta) d \theta=-2 a \sin 2 \theta d \theta$

$\therefore \quad I=\int \sqrt{\frac{a+a \cos 2 \theta}{a-a \cos 2 \theta}} \cdot(-2 a \sin 2 \theta) d \theta$

$=\int \sqrt{\frac{1+\cos 2 \theta}{1-\cos 2 \theta}} \cdot(-2 a \sin 2 \theta) d \theta$

$=-2 a \int \sqrt{\frac{2 \cos ^{2} \theta}{2 \sin ^{2} \theta}} \cdot \sin 2 \theta d \theta$

$=-2 a \int \sqrt{\frac{\cos ^{2} \theta}{\sin ^{2} \theta}} \cdot 2 \sin \theta \cos \theta d \theta$

$=-2 a \int \frac{\cos \theta}{\sin \theta} \cdot 2 \sin \theta \cos \theta d \theta$

$=-4 a \int \cos \theta \cos \theta d \theta=-4 a \int \cos ^{2} \theta d \theta$

$=-4 a \int \frac{1+\cos 2 \theta}{2} d \theta=-2 a \int(1+\cos 2 \theta) d \theta$

$=-2 a[\int 1 d \theta+\int \cos 2 \theta d \theta]=-2 a[\theta+\frac{1}{2} \sin 2 \theta]$

Now $\quad x=a \cos 2 \theta$

$ \frac{x}{a}=\cos 2 \theta \Rightarrow 2 \theta=\cos ^{-1} \frac{x}{a} \Rightarrow \theta=\frac{1}{2} \cos ^{-1} \frac{x}{a} $

$\sin 2 \theta=\sqrt{1-\cos ^{2} 2 \theta}=\sqrt{1-\frac{x^{2}}{a^{2}}}=\frac{\sqrt{a^{2}-x^{2}}}{a}$

$\therefore \quad I=-2 a[\frac{1}{2} \cos ^{-1} \frac{x}{a}+\frac{1}{2} \frac{\sqrt{a^{2}-x^{2}}}{a}]+C_1$

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

Hence, $I=a \sin ^{-1} \frac{x}{a}-\sqrt{a^{2}-x^{2}}+C$

Solution

Let $\quad I=\int \frac{x^{1 / 2}}{1+x^{3 / 4}} d x$

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

$ \begin{gathered} .t^{3}+1) t^{5} \\ \frac{(-t^{5}+t^{2}.}{}(t^{2}. \\ -t^{2} \end{gathered} $

$ \begin{aligned} & =\int \frac{t^{2} \cdot 4 t^{3}}{1+t^{3}} d t=4 \int \frac{t^{5}}{1+t^{3}} d t \\ & =4 \int(t^{2}-\frac{t^{2}}{t^{3}+1}) d t=4 \int t^{2} d t-4 \int \frac{t^{2}}{t^{3}+1} \cdot d t \\ I & =I_1-I_2 \end{aligned} $

Now $\quad I_1=4 \int t^{2} d t=4 \cdot \frac{t^{3}}{3}+C_1=\frac{4}{3} x^{3 / 4}+C_1$

$ I_2=4 \int \frac{t^{2}}{t^{3}+1} d t $

Put $t^{3}+1=z \Rightarrow 3 t^{2} d t=d z$

$ \begin{aligned} t^{2} d t & =\frac{1}{3} d z \\ \therefore \quad I_2 & =\frac{4}{3} \int \frac{d z}{z}=\frac{4}{3} \log |z|+C_2=\frac{4}{3} \log |t^{3}+1|+C_2 \\ & =\frac{4}{3} \log |(x)^{3 / 4}+1|+C_2 \\ \therefore \quad I & =I_1-I_2 \\ & =\frac{4}{3} x^{3 / 4}+C_1-\frac{4}{3} \log |(x)^{3 / 4}+1|-C_2 \\ & =\frac{4}{3}[x^{3 / 4}-\log |(x)^{3 / 4}+1|]+C_1-C_2 \end{aligned} $

Hence, $\quad I=\frac{4}{3}[x^{3 / 4}-\log |(x)^{3 / 4}+1|]+C \quad[\because C=C_1-C_2]$

Solution

Let $\quad I=\int \frac{\sqrt{1+x^{2}}}{x^{4}} d x=\int \sqrt{\frac{1+x^{2}}{x^{2}}} \cdot \frac{1}{x^{3}} d x=\int \sqrt{\frac{1}{x^{2}}+1} \cdot \frac{1}{x^{3}} d x$ Put $\frac{1}{x^{2}}+1=t^{2}$

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

$\therefore \quad I=\int t(t d t)=-\int t^{2} d t=-\frac{1}{3} t^{3}+C$

Hence, $I=-\frac{1}{3}(\frac{1}{x^{2}}+1)^{3 / 2}+C$

Solution

Let $\quad I=\int \frac{d x}{\sqrt{16-9 x^{2}}}$

$ \begin{aligned} & =\frac{1}{3} \int \frac{d x}{\sqrt{\frac{16}{9}-x^{2}}}=\frac{1}{3} \int \frac{d x}{\sqrt{(\frac{4}{3})^{2}-x^{2}}} \\ & =\frac{1}{3} \sin ^{-1} \frac{x}{4 / 3}+C[\because \int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} \frac{x}{a}+C] \\ & =\frac{1}{3} \sin ^{-1} \frac{3 x}{4}+C \end{aligned} $

Hence, $I=\frac{1}{3} \sin ^{-1} \frac{3 x}{4}+C$.

Solution

Let

$ \begin{aligned} & I=\int \frac{d t}{\sqrt{3 t-2 t^{2}}}=\int \frac{d t}{\sqrt{-2(t^{2}-\frac{3}{2} t)}} \\ &=\frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{-(t^{2}-\frac{3}{2} t+\frac{9}{16}-\frac{9}{16})}} \quad \text{ [Making perfect } \\ & \text{ square] } \end{aligned} $

$ \begin{aligned} & =\frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{-[(t-\frac{3}{4})^{2}-\frac{9}{16}]}}=\frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{\frac{9}{16}-(t-\frac{3}{4})^{2}}} \\ & =\frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{(\frac{3}{4})^{2}-(t-\frac{3}{4})^{2}}}=\frac{1}{\sqrt{2}} \cdot \sin ^{-1} \frac{t-\frac{3}{4}}{\frac{3}{4}}+C \\ & =\frac{1}{\sqrt{2}} \sin ^{-1} \frac{4 t-3}{3}+C \end{aligned} $

Hence, $I=\frac{1}{\sqrt{2}} \sin ^{-1}(\frac{4 t-3}{3})+C$.

Solution

Let $I=\int \frac{3 x-1}{\sqrt{x^{2}+9}} d x=\int \frac{3 x}{\sqrt{x^{2}+9}} d x-\int \frac{1}{\sqrt{x^{2}+9}} d x$

$ I=I_1-I_2 $

Now $I_1=\int \frac{3 x}{\sqrt{x^{2}+9}} d x$

Put $\quad x^{2}+9=t \Rightarrow 2 x d x=d t$

$ x d x=-d t $

$\therefore I_1=\frac{3}{2} \int \frac{d t}{\sqrt{t}}=\frac{3}{2} \cdot 2 \sqrt{t}+C_1=3 \sqrt{x^{2}+9}+C_1$

$I_2=\int \frac{1}{\sqrt{x^{2}+9}} d x=\int \frac{1}{\sqrt{x^{2}+(3)^{2}}} d x=\log |x+\sqrt{x^{2}+(3)^{2}}|+C_2$

$[\because \int \frac{1}{\sqrt{x^{2}+a^{2}}} d x=\log |x+\sqrt{x^{2}+a^{2}}|+C]$

$ \begin{aligned} & =\log |x+\sqrt{x^{2}+9}|+C_2 \\ \therefore \quad I & =I_1-I_2 \\ & =3 \sqrt{x^{2}+9}+C_1-\log |x+\sqrt{x^{2}+9}|-C_2 \\ & =3 \sqrt{x^{2}+9}-\log |x+\sqrt{x^{2}+9}|+(C_1-C_2) \end{aligned} $

Hence, $\quad I=3 \sqrt{x^{2}+9}-\log |x+\sqrt{x^{2}+9}|+C$

Solution

Let $I=\int \sqrt{5-2 x+x^{2}} d x=\int \sqrt{x^{2}-2 x+5} d x$

$ \begin{aligned} & =\int \sqrt{x^{2}-2 x+1-1+5} d x \quad \text{ (Making perfect square) } \\ & =\int \sqrt{(x-1)^{2}+4} d x=\int \sqrt{(x-1)^{2}+(2)^{2}} d x \\ & =\frac{x-1}{2} \sqrt{(x-1)^{2}+(2)^{2}}+\frac{4}{2} \log |(x-1)+\sqrt{(x-1)^{2}+(2)^{2}}|+C \\ {[\because} & .\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] \\ & =\frac{x-1}{2} \sqrt{x^{2}+1-2 x+4}+2 \log |(x-1)+\sqrt{x^{2}+1-2 x+4}|+C \\ & =\frac{x-1}{2} \sqrt{x^{2}-2 x+5}+2 \log |(x-1)+\sqrt{x^{2}-2 x+5}|+C \end{aligned} $

Hence,

$ I=\frac{x-1}{2} \sqrt{x^{2}-2 x+5}+2 \log |(x-1)+\sqrt{x^{2}-2 x+5}|+C $

Solution

Let $\quad I=\int \frac{x}{x^{4}-1} d x$

Put $x^{2}=t \Rightarrow 2 x d x=d t \Rightarrow x d x=\frac{d t}{2}$

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

Hence, $I=\frac{1}{4} \log |\frac{x^{2}-1}{x^{2}+1}|+C$.

Solution

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

Put $x^{2}=t$ for the purpose of partial fractions.

We get $\frac{t}{(1-t)(1+t)}$

Resolving into partial fractions we put

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

[where $A$ and $B$ are arbitrary constants]

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

$\Rightarrow \quad t=A+A t+B-B t$

Comparing the like terms, we get $A-B=1$ and $A+B=0$

Solving the above equations, we have $A=\frac{1}{2}$ and $B=-\frac{1}{2}$

$ \begin{aligned} \therefore \quad I & =\int \frac{1 / 2}{1-x^{2}} d x+\int \frac{-1 / 2}{1+x^{2}} d x \quad(\text{ Putting } t=x^{2}) \\ & =\frac{1}{2} \cdot \frac{1}{2 \cdot 1} \log |\frac{1+x}{1-x}|-\frac{1}{2} \tan ^{-1} x+C \\ & =\frac{1}{4} \log |\frac{1+x}{1-x}|-\frac{1}{2} \tan ^{-1} x+C \end{aligned} $

Hence, $\quad I=\frac{1}{4} \log |\frac{1+x}{1-x}|-\frac{1}{2} \tan ^{-1} x+C$.

Solution

Let $I=\int \sqrt{2 a x-x^{2}} d x$

$ =\int \sqrt{-(x^{2}-2 a x)} d x=\int \sqrt{-(x^{2}-2 a x+a^{2}-a^{2})} d x $

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

Hence, $I=\frac{x-a}{2} \sqrt{2 a x-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}(\frac{x-a}{a})+C$.

Solution

Let $\quad I=\int \frac{\sin ^{-1} x}{(1-x^{2})^{3 / 2}} d x$

Put $x=\sin \theta \Rightarrow d x=\cos \theta d \theta$

$I=\int \frac{\sin ^{-1}(\sin \theta)}{(1-\sin ^{2} \theta)^{3 / 2}} \cdot \cos \theta d \theta$

$=\int \frac{\theta \cdot \cos \theta d \theta}{(\cos ^{2} \theta)^{3 / 2}}=\int \frac{\theta \cdot \cos \theta}{\cos ^{3} \theta} d \theta$

$=\int \frac{\theta}{\cos ^{2} \theta} d \theta=\int _{\text{I }}^{\theta} \sec ^{2} \theta d \theta$

$=\theta \cdot \int \sec ^{2} \theta d \theta-\int(D(\theta) \cdot \int \sec ^{2} \theta d \theta) d \theta$

$[\because \int_I^{u . v} II d d x=u \cdot \int d x-\int(D(u) \int v d v) d v+C]$

$=\theta \cdot \tan \theta-\int 1 \cdot \tan \theta d \theta$

$=\theta \cdot \tan \theta-\log \sec \theta+C$

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

$ \begin{bmatrix} \text{ when } x=\sin \theta \\ \therefore \tan \theta=\frac{x}{\sqrt{1-x^{2}}} \text{ and } \sec \theta=\sqrt{1-x^{2}} \end{bmatrix} $

Hence, $I=\frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}}-\log |\sqrt{1-x^{2}}|+C$

Solution

Let $I=\int \frac{\cos 5 x+\cos 4 x}{1-2 \cos 3 x} d x=\int \frac{2 \cos \frac{5 x+4 x}{2} \cdot \cos \frac{5 x-4 x}{2}}{1-2(2 \cos ^{2} \frac{3 x}{2}-1)} d x$

$=\int \frac{2 \cos \frac{9 x}{2} \cdot \cos \frac{x}{2}}{1-4 \cos ^{2} \frac{3 x}{2}+2} d x=\int \frac{2 \cos \frac{9 x}{2} \cdot \cos \frac{x}{2}}{3-4 \cos ^{2} \frac{3 x}{2}} d x$

$ \begin{aligned} & =-\int \frac{2 \cos \frac{9 x}{2} \cdot \cos \frac{x}{2}}{4 \cos ^{2} \frac{3 x}{2}-3} d x=-\int \frac{2 \cos \frac{9 x}{2} \cdot \cos \frac{x}{2} \cdot \cos \frac{3 x}{2}}{4 \cos ^{3} \frac{3 x}{2}-3 \cos \frac{3 x}{2}} d x \\ & {[\text{ Multiplying and dividing by } \cos \frac{3 x}{2}]} \\ & =-\int \frac{2 \cos \frac{9 x}{2} \cdot \cos \frac{x}{2} \cdot \cos \frac{3 x}{2}}{\cos 3 \cdot \frac{3 x}{2}} d x \\ & =-\int \frac{.2 \cos 3 x=4 \cos ^{3} x-3 \cos x]}{2} \cdot \cos \frac{x}{2} \cdot \cos \frac{3 x}{2} d x=-\int 2 \cos ^{3 x} \cdot \cos \frac{x}{2} d x \\ & =-\int[\cos (\frac{3 x}{2}+\frac{x}{2})+\cos (\frac{3 x}{2}-\frac{x}{2})] d x \\ & =-\int(\cos 2 x+\cos x) d x \\ & {[\because 2 \cos A \cos B=\cos (A+B)+\cos (A-B)]} \\ & =-\int \cos 2 x d x-\int \cos x d x=-\frac{1}{2} \sin 2 x-\sin x+C \end{aligned} $

Hence, $I=-[\frac{1}{2} \sin 2 x+\sin x]+C$.

Solution

Let $I=\int \frac{\sin ^{6} x+\cos ^{6} x}{\sin ^{2} x \cdot \cos ^{2} x} d x=\int \frac{(\sin ^{2} x)^{3}+(\cos ^{2} x)^{3}}{\sin ^{2} x \cdot \cos ^{2} x} d x$

$ \begin{matrix} =\int \frac{(\sin ^{2} x+\cos ^{2} x)^{3}-3 \sin ^{2} x \cos ^{2} x(\sin ^{2} x+\cos ^{2} x)}{\sin ^{2} x \cdot \cos ^{2} x} d x \\ {[\because a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b)]} \end{matrix} $

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

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

Hence, $I=\tan x-\cot x-3 x+C$.

Solution

Let

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

Put $x^{3 / 2}=t \Rightarrow \frac{3}{2} x^{1 / 2} d x=d t \Rightarrow x^{1 / 2} d x=\frac{2}{3} d t$

$ \begin{aligned} \therefore \quad I & =\frac{2}{3} \int \frac{d t}{\sqrt{(a^{3 / 2})^{2}-(t)^{2}}} \\ & =\frac{2}{3} \sin ^{-1} \frac{t}{a^{3 / 2}}+C=\frac{2}{3} \sin ^{-1}(\frac{x^{3 / 2}}{a^{3 / 2}})+C \end{aligned} $

Hence, $I=\frac{2}{3} \sin ^{-1}(\frac{x}{a})^{3 / 2}+C$.

Solution

Let $\quad I=\int \frac{\cos x-\cos 2 x}{1-\cos x} d x$

$ \begin{aligned} &=\int \frac{2 \sin \frac{x+2 x}{2} \cdot \sin (\frac{2 x-x}{2})}{2 \sin ^{2} x / 2} d x \\ & {[\because \cos C-\cos D=2 \sin \frac{C+D}{2} \cdot \sin \frac{D-C}{2}] } \end{aligned} $

$ \begin{aligned} & =\int \frac{2 \sin \frac{3 x}{2} \cdot \sin \frac{x}{2}}{2 \sin ^{2} \frac{x}{2}} d x=\int \frac{\sin \frac{3 x}{2}}{\sin \frac{x}{2}} d x=\int \frac{\sin 3(\frac{x}{2})}{\sin \frac{x}{2}} d x \\ & =\int \frac{3 \sin \frac{x}{2}-4 \sin ^{3} \frac{x}{2}}{\sin \frac{x}{2}} d x \quad[\sin 3 x=3 \sin x-4 \sin ^{3} x] \end{aligned} $

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

Hence, $I=x+2 \sin x+C$.

Solution

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

Put $x^{2}=\sec \theta$

$\therefore \quad 2 x d x=\sec \theta \tan \theta d \theta$

$ x d x=\frac{1}{2} \sec \theta \tan \theta d \theta $

$\therefore \quad I=\frac{1}{2} \int \frac{\sec \theta \tan \theta}{\sec \theta \sqrt{\sec ^{2} \theta-1}} d \theta$

$ =\frac{1}{2} \int \frac{\sec \theta \tan \theta}{\sec \theta \cdot \tan \theta} d \theta=\frac{1}{2} \int 1 d \theta=\frac{1}{2} \theta+C $

So

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

Hence, $I=\frac{1}{2} \sec ^{-1} x^{2}+C$.

Solution

Let $\quad I=\int_0^{2}(x^{2}+3) d x$

Using the formula,

$\int_a^{b} f(x) d x=\lim _{h \to 0} h[f(a)+f(a+h)+f(a+2 h)+\ldots+f(a+\overline{n-1} h)]$

where $h=\frac{b-a}{n}$

Here, $a=0$ and $b=2$

$\therefore h=\frac{2-0}{n} \quad \therefore n h=2$

$\text{ Here, } f(x) =x^{2}+3$

$f(0) =0+3=3$

$f(0+h) =(0+h)^{2}+3=h^{2}+3$

$f(0+2 h) =(0+2 h)^{2}+3=4 h^{2}+3$

$ f(0+\overline{n-1} h)=(0+\overline{n-1} h)^{2}+3(n-1)^{2} h^{2}+3 $

Now

$ \begin{aligned} & \int_0^{2}(x^{2}+3) d x \\ & =\lim _{h \to 0} h[3+h^{2}+3+4 h^{2}+3+\ldots+(n-1)^{2} h^{2}+3] \\ & =\lim _{h \to 0} h[(3+3+3+\ldots+n)+{h^{2}+4 h^{2}+\ldots+(n-1)^{2} h^{2}}] \\ & =\lim _{h \to 0} h[3 n+h^{2}{1+4+\ldots+(n-1)^{2}}] \\ & =\lim _{h \to 0} h[3 n+h^{2} \frac{n(n-1)(2 n-1)}{6}] \\ & {[\because 1+4+9+\ldots .+(n-1)^{2}=\frac{n(n-1)(2 n-1)}{6}]} \\ & =\lim _{h \to 0}[3 n h+\frac{h^{3} n(n-1)(2 n-1)}{6}] \\ & =\lim _{h \to 0}[3 n h+\frac{n h(n h-h)(2 n h-h)}{6}] \\ & =[3 \times 2+\frac{2(2-0)(2 \times 2-0)}{6}] \quad \begin{bmatrix} \because n h & =2 \\ h & =0 \end{bmatrix} \\ & =[6+\frac{2 \times 2 \times 4}{6}]=6+\frac{8}{3}=\frac{26}{3} \end{aligned} $

Hence, $\int_0^{2}(x^{2}+3) d x=\frac{26}{3}$.

Solution

Let

$ I=\int_0^{2} e^{x} d x $

Here, $a=0$ and $b=2 \therefore h=\frac{b-a}{n} \Rightarrow h=\frac{2-0}{n} \quad \therefore n h=2$

Here

$ \begin{aligned} f(x) & =e^{x} \\ f(0) & =e^{0}=1 \\ f(0+h) & =e^{0+h}=e^{h} \\ f(0+2 h) & =e^{0+2 h}=e^{2 h} \end{aligned} $

$ \begin{aligned} & f(0+\overline{n-1} h)=e^{0+(n-1) h}=e^{(n-1) h} \\ & \text{ Using } \begin{aligned} & \int_a^{b} f(x) d x \\ &=\lim _{h \to 0} h[f(a)+f(a+h)+f(a+2 h)+\ldots+f(a+\overline{n-1} h)] \\ & \therefore \int_0^{2} e^{x} d x=\lim _{h \to 0} h[1+e^{h}+e^{2 h}+\ldots+e^{(n-1) h}] \\ &= \lim _{h \to 0} h[\frac{1(e^{n h}-1)}{e^{h}-1}] \\ &=\lim _{h \to 0} \frac{e^{n h}-1}{\frac{e^{h}-1}{h}}=\frac{e^{2}-1}{1}=e^{2}-1[\because \lim _{x \to 0} \frac{e^{x}-1}{x}=1] \end{aligned} \end{aligned} $

Hence, $I=e^{2}-1$.

Solution

Let $I=\int_0^{1} \frac{d x}{e^{x}+e^{-x}}$

$ =\int_0^{1} \frac{d x}{e^{x}+\frac{1}{e^{x}}}=\int_0^{1} \frac{d x}{\frac{e^{2 x}+1}{e^{x}}}=\int_0^{1} \frac{e^{x} d x}{e^{2 x}+1} $

Put $\quad e^{x}=t \Rightarrow e^{x} d x=d t$

Changing the limit, we have

When $x=0 \quad \therefore t=e^{0}=1$

When $x=1 \quad \therefore t=e^{1}=e$

$ \therefore I=\int_1^{e} \frac{d t}{t^{2}+1}=[\tan ^{-1} t]_1^{e}=[\tan ^{-1} e-\tan ^{-1}(1)]=\tan ^{-1} e-\frac{\pi}{4} $

Hence, $I=\tan ^{-1} e-\frac{\pi}{4}$.

Solution

Let $I=\int_0^{\pi / 2} \frac{\tan x}{1+m^{2} \tan ^{2} x} d x$

$ \begin{aligned} & =\int_0^{\pi / 2} \frac{\frac{\sin x}{\cos x}}{1+m^{2} \frac{\sin ^{2} x}{\cos ^{2} x}} d x=\int_0^{\pi / 2} \frac{\frac{\sin x}{\cos x}}{\frac{\cos ^{2} x+m^{2} \sin ^{2} x}{\cos ^{2} x}} d x \\ & =\int_0^{\pi / 2} \frac{\sin x \cos x}{\cos ^{2} x+m^{2} \sin ^{2} x} d x=\int_0^{\pi / 2} \frac{\sin x \cos x}{1-\sin ^{2} x+m^{2} \sin ^{2} x} d x \\ & =\int_0^{\pi / 2} \frac{\sin x \cos x}{1-\sin ^{2} x(1-m^{2})} d x \end{aligned} $

Put $\quad \sin ^{2} x=t$

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

$ \sin x \cos x d x=\frac{d t}{2} $

Changing the limits we get,

$ \text{ When } x=0 \quad \therefore t=\sin ^{2} 0=0 \text{; When } x=\frac{\pi}{2} \quad \therefore t=\sin ^{2} \frac{\pi}{2}=1 $

$\therefore I=\frac{1}{2} \int_0^{1} \frac{d t}{1-(1-m^{2}) t}$

$ \begin{aligned} I & =\frac{1}{2} \int_0^{1} \frac{d t}{1+(m^{2}-1) t}=\frac{1}{2}[\frac{\log [1+(m^{2}-1) t]}{m^{2}-1}]_0^{1} \\ & =\frac{1}{2(m^{2}-1)}[\log (1+m^{2}-1)-\log (1)]=\frac{\log |m^{2}|}{2(m^{2}-1)} \end{aligned} $

Hence, $I=\frac{\log |m^{2}|}{2(m^{2}-1)}=\frac{\log |m|}{m^{2}-1}$.

Solution

Let $\quad I=\int_1^{2} \frac{d x}{\sqrt{(x-1)(2-x)}}$

$ \begin{aligned} & =\int_1^{2} \frac{d x}{\sqrt{2 x-x^{2}-2+x}}=\int_1^{2} \frac{d x}{\sqrt{-x^{2}+3 x-2}} \\ & =\int_1^{2} \frac{d x}{\sqrt{-(x^{2}-3 x+2)}} \\ & =\int_1^{2} \frac{d x}{\sqrt{-(x^{2}-3 x+\frac{9}{4}-\frac{9}{4}+2)}} \quad \text{ [Making perfect } \\ & \text{ square] } \\ & =\int_1^{2} \frac{d x}{.\sqrt{-[(x-\frac{3}{2})^{2}-\frac{1}{4}.}]}=\int_1^{2} \frac{d x}{\frac{1}{4}-(x-\frac{3}{2})^{2}} \\ & =\int_1^{2} \frac{d x}{\sqrt{(\frac{1}{2})^{2}-(x-\frac{3}{2})^{2}}}=[\sin ^{-1}(\frac{x-\frac{3}{2}}{\frac{1}{2}})]_1^{2} \\ & =[\sin ^{-1}(\frac{2 x-3}{1})_1^{2}=\sin ^{-1}(4-3)-\sin ^{-1}(2-3). \\ & =\sin ^{-1}(1)-\sin ^{-1}(-1)=\sin ^{-1}(1)+\sin ^{-1}(1) \\ & =2 \sin ^{-1}(1)=2 \times \frac{\pi}{2}=\pi \\ & -\pi \end{aligned} $

Hence, $\quad I=\pi$.

Solution

Let $\quad I=\int_0^{1} \frac{x d x}{\sqrt{1+x^{2}}}$

Put $1+x^{2}=t \stackrel{0}{1+x^{2}} \Rightarrow 2 x d x=d t \Rightarrow x d x=\frac{d t}{2}$

Changing the limits, we have

When $x=0 \quad \therefore t=1$

When $x=1 \quad \therefore t=2$

$\therefore \quad I=\frac{1}{2} \int_1^{2} \frac{d t}{\sqrt{t}}=\frac{1}{2} \cdot 2[t^{1 / 2}]_1^{2}=\sqrt{2}-1$

Hence, $I=\sqrt{2}-1$.

Solution

Let $\quad I=\int_0^{\pi} x \sin x \cos ^{2} x d x$

$$ \begin{align*} & I=\int_0^{\pi}(\pi-x) \sin (\pi-x) \cos ^{2}(\pi-x) d x \tag{i}\\ & I=\int_0^{\pi}(\pi-x) \sin x \cos ^{2} x d x \tag{ii} \end{align*} $$

Adding (i) and (ii) we get,

$ \begin{aligned} & 2 I=\int_0^{\pi}[x \sin x \cos ^{2} x+(\pi-x) \sin x \cos ^{2} x] d x \\ & 2 I=\int_0^{\pi} \sin x \cos ^{2} x \cdot(x+\pi-x) d x \\ & 2 I=\int_0^{\pi} \pi \sin x \cos ^{2} x d x=\pi \int_0^{\pi} \sin x \cos ^{2} x d x \end{aligned} $

Put $\quad \cos ^{0} x=t \Rightarrow-\sin x d x=d t \Rightarrow \sin x d x=-d t$

Changing the limits, we have

When $x=0, \quad t=\cos 0=1$; When $x=\pi, \quad t=\cos \pi=-1$

$ \begin{aligned} & 2 I=\pi \int_1^{-1}-t^{2} d t=-\pi \int_1^{-1} t^{2} d t \\ & 2 I=\pi \int _{-1}^{1} t^{2} d t \quad[\int_a^{b} f(x) d x=-\int_b^{a} f(x) d x] \\ & 2 I=\pi[\frac{t^{3}}{3}] _{-1}^{1}=\pi[\frac{1}{3}+\frac{1}{3}]=\pi(\frac{2}{3}) \end{aligned} $

$ \therefore \quad I=\frac{\pi}{3} $

Solution

Let $I=\int_0^{1 / 2} \frac{d x}{(1+x^{2}) \sqrt{1-x^{2}}}$

Put $\quad x=\sin \theta$

$\therefore \quad d x=\cos \theta d \theta$

Changing the limits, we get

When $x=0 \quad \therefore \sin \theta=0 \quad \therefore \theta=0$

When $x=\frac{1}{2} \quad \therefore \sin \theta=\frac{1}{2} \quad \therefore \theta=\frac{\pi}{6}$

$ \begin{aligned} \therefore \quad I & =\int_0^{\pi / 6} \frac{\cos \theta d \theta}{(1+\sin ^{2} \theta) \sqrt{1-\sin ^{2} \theta}} \\ & =\int_0^{\pi / 6} \frac{\cos \theta d \theta}{(1+\sin ^{2} \theta) \cos \theta}=\int_0^{\pi / 6} \frac{1}{1+\sin ^{2} \theta} d \theta \end{aligned} $

Now, dividing the numerator and denominator by $\cos ^{2} \theta$, we get

$ \begin{aligned} & =\int_0^{\pi / 6} \frac{\frac{1}{\cos ^{2} \theta}}{\frac{1}{\cos ^{2} \theta}+\frac{\sin ^{2} \theta}{\cos ^{2} \theta}} d \theta=\int_0^{\pi / 6} \frac{\sec ^{2} \theta}{\sec ^{2} \theta+\tan ^{2} \theta} d \theta \\ & =\int_0^{\pi / 6} \frac{\sec ^{2} \theta}{1+\tan ^{2} \theta+\tan ^{2} \theta} d \theta=\int_0^{\pi / 6} \frac{\sec ^{2} \theta}{2 \tan ^{2} \theta+1} d \theta \end{aligned} $

Put $\tan \theta=t$

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

Changing the limits, we get

When $\theta=0 \quad \therefore t=\tan 0=0$

When $\theta=\frac{\pi}{6} \quad \therefore t=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}$

$ \begin{aligned} \therefore \quad I & =\int_0^{1 / \sqrt{3}} \frac{d t}{2 t^{2}+1}=\frac{1}{2} \int_0^{1 / \sqrt{3}} \frac{d t}{t^{2}+\frac{1}{2}}=\frac{1}{2} \int_0^{1 / \sqrt{3}} \frac{d t}{t^{2}+(\frac{1}{\sqrt{2}})^{2}} \\ & =\frac{1}{2} \times \frac{1}{1 / \sqrt{2}}[\tan ^{-1} \frac{t}{1 / \sqrt{2}}]_0^{1 / \sqrt{3}}=\frac{1}{\sqrt{2}} \tan ^{-1}[\sqrt{2} t]_0^{1 / \sqrt{3}} \\ & =\frac{1}{\sqrt{2}}[\tan ^{-1} \frac{\sqrt{2}}{\sqrt{3}}-\tan ^{-1} 0]=\frac{1}{\sqrt{2}} \tan ^{-1} \sqrt{\frac{2}{3}} \end{aligned} $

Solution

Let $I=\int \frac{x^{2}}{x^{4}-x^{2}-12} d x=\int \frac{x^{2}}{x^{4}-4 x^{2}+3 x^{2}-12} d x$

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

Put $x^{2}=t$ for the purpose of partial fraction.

We get $\frac{t}{(t-4)(t+3)}$

Let $\frac{t}{(t-4)(t+3)}=\frac{A}{t-4}+\frac{B}{t+3}$

[where $A$ and $B$ are arbitrary constants]

$\frac{t}{(t-4)(t+3)}=\frac{A(t+3)+B(t-4)}{(t-4)(t+3)}$

$\Rightarrow \quad t=A t+3 A+B t-4 B$

Comparing the like terms, we get

$ \begin{matrix} A+B=1 & \text{ and } & 3 A-4 B & =0 \\ \Rightarrow & 3 A & =4 B \\ & \therefore & A & =\frac{4}{3} B \end{matrix} $

Now $\quad \frac{4}{3} B+B=1$

$ \frac{7}{3} B=1 \quad \therefore B=\frac{3}{7} \text{ and } A=\frac{4}{3} \times \frac{3}{7}=\frac{4}{7} $

So, $\quad A=\frac{4}{7}$ and $B=\frac{3}{7}$

$\therefore \quad \int \frac{x^2}{(x^2-4)(x^2+3)}dx$

$=\frac{4}{7}\int \frac{1}{x^2-4}dx +\frac{3}{7}\int \frac{1}{x^2+3}dx$

$=\frac{4}{7} \int \frac{1}{x^{2}-(2)^{2}} d x+\frac{3}{7} \int \frac{1}{x^{2}+(\sqrt{3})^{2}} d x$

$=\frac{4}{7} \times \frac{1}{2 \times 2} \log |\frac{x-2}{x+2}|+\frac{3}{7} \times \frac{1}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}$

$=\frac{1}{7} \log |\frac{x-2}{x+2}|+\frac{\sqrt{3}}{7} \tan ^{-1} \frac{x}{\sqrt{3}}+C$

Hence, $I=\frac{1}{7} \log |\frac{x-2}{x+2}|+\frac{\sqrt{3}}{7} \tan ^{-1} \frac{x}{\sqrt{3}}+C$.

Solution

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

Put $x^{2}=t$ for the purpose of partial fraction.

We get $\frac{t}{(t+a^{2})(t+b^{2})}$

$ \begin{aligned} & \text{ Put } \frac{t}{(t+a^{2})(t+b^{2})}=\frac{A}{t+a^{2}}+\frac{B}{t+b^{2}} \\ & \Rightarrow \frac{t}{(t+a^{2})(t+b^{2})}=\frac{A(t+b^{2})+B(t+a^{2})}{(t+a^{2})(t+b^{2})} \\ & \Rightarrow \quad t=A t+A b^{2}+B t+B a^{2} \end{aligned} $

Comparing the like terms, we get

$A+B=1$ and $A b^{2}+B a^{2}=0$

$ \begin{aligned} \therefore \quad \frac{-a^{2}}{b^{2}} B+B & =1 \\ B(\frac{-a^{2}}{b^{2}}+1) & =1 \Rightarrow B(\frac{-a^{2}+b^{2}}{b^{2}})=1 \end{aligned} $

$ \begin{aligned} & A=\frac{-a^{2}}{b^{2}} B \\ & \Rightarrow \quad B=\frac{b^{2}}{b^{2}-a^{2}} \text{ and } A=\frac{-a^{2}}{b^{2}} \times \frac{b^{2}}{b^{2}-a^{2}}=\frac{a^{2}}{a^{2}-b^{2}} \\ & \text{ So } \quad A=\frac{a^{2}}{a^{2}-b^{2}} \text{ and } B=\frac{-b^{2}}{a^{2}-b^{2}} \\ & \therefore \int \frac{x^{2}}{(x^{2}+a^{2})(x^{2}+b^{2})} d x \\ & =\frac{a^{2}}{a^{2}-b^{2}} \int \frac{1}{x^{2}+a^{2}} d x-\frac{b^{2}}{a^{2}-b^{2}} \int \frac{1}{x^{2}+b^{2}} d x \\ & =\frac{a^{2}}{a^{2}-b^{2}} \times \frac{1}{a} \tan ^{-1} \frac{x}{a}-\frac{b^{2}}{a^{2}-b^{2}} \cdot \frac{1}{b} \cdot \tan ^{-1} \frac{x}{b} \\ & =\frac{a}{a^{2}-b^{2}} \tan ^{-1} \frac{x}{a}-\frac{b}{a^{2}-b^{2}} \tan ^{-1} \frac{x}{b}+C \end{aligned} $

Hence, $I=\frac{1}{a^{2}-b^{2}}[a \tan ^{-1} \frac{x}{a}-b \tan ^{-1} \frac{x}{b}]+C$

Solution

Let $I=\int_0^{\pi} \frac{x}{1+\sin x} d x$

$$ \begin{align*} & =\int_0^{\pi} \frac{\pi-x}{1+\sin (\pi-x)} d x[\text{ using } \int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x] \tag{i}\\ & =\int_0^{\pi} \frac{\pi-x}{1+\sin x} d x \tag{ii} \end{align*} $$

Adding (i) and (ii), we get

$ \begin{aligned} 2 I & =\int_0^{\pi}(\frac{x}{1+\sin x}+\frac{\pi-x}{1+\sin x}) d x \\ & =\int_0^{\pi}(\frac{x+\pi-x}{1+\sin x}) d x=\int_0^{\pi} \frac{\pi}{1+\sin x} d x \\ & =\pi \int_0^{\pi} \frac{1}{1+\sin x} d x=\pi \int_0^{\pi} \frac{1 \cdot(1-\sin x)}{(1+\sin x)(1-\sin x)} d x \\ & =\pi \int_0^{\pi} \frac{1-\sin x}{1-\sin ^{2} x} d x=\pi \int_0^{\pi} \frac{1-\sin x}{\cos ^{2} x} d x \\ & =\pi \int_0^{\pi}(\frac{1}{\cos ^{2} x}-\frac{\sin x}{\cos ^{2} x}) d x=\pi \int_0^{\pi}(\sec ^{2} x-\sec x \tan x) d x \\ & =\pi[\tan x-\sec x]_0^{\pi}=\pi[(\tan \pi-\tan 0)-(\sec \pi-\sec 0)] \\ \therefore \quad \text{ 2I } & =\pi[0-(-1-1)]=\pi(2) \\ I & =\pi \end{aligned} $

Hence, $I=\pi$.

Solution

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

Resolving into partial fraction, we put

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

$ =-\log |x-1|^{1 / 6}-\log (x+2)^{1 / 3}+\log (x-3)^{1 / 2}+C $

Hence, $\int \frac{2 x-1}{(x-1)(x+2)(x-3)} d x=\log [\frac{\sqrt{x-3}}{(x-1)^{1 / 6}(x+2)^{1 / 3}}]+C$.

Solution

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

Put $\tan ^{-1} x=t \Rightarrow \frac{1}{1+x^{2}} \cdot d x=d t$

$ =\int e^{t}(1+\tan t+\tan ^{2} t) d t=\int e^{t}(\sec ^{2} t+\tan t) d t $

Here $f(t)=\tan t$

$\therefore \quad f^{\prime}(t)=\sec ^{2} t$

$=e^{t} \cdot f(t)=e^{t} \tan t=e^{\tan ^{-1} x} \cdot x+C$

$[\because \int e^{x}[f(x)+f^{\prime}(x)] d x=e^{x} f(x)+C]$

Hence, $\quad I=e^{\tan ^{-1} x} \cdot x+C$.

Solution

Let $\quad I=\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x$

Put $\quad x=a \tan ^{2} \theta$

$d x=2 a \tan \theta \cdot \sec ^{2} \theta \cdot d \theta$

$\therefore \quad I=\int \sin ^{-1} \sqrt{\frac{a \tan ^{2} \theta}{a+a \tan ^{2} \theta}} \cdot 2 a \tan \theta \cdot \sec ^{2} \theta d \theta$

$=\int \sin ^{-1} \frac{\sqrt{a} \tan \theta}{\sqrt{a} \sec \theta} \cdot 2 a \tan \theta \cdot \sec \theta d \theta$

$=\int \sin ^{-1}(\frac{\frac{\sin \theta}{\cos \theta}}{\frac{1}{\cos \theta}}) \cdot 2 a \tan \theta \cdot \sec ^{2} \theta d \theta$

$=\int \sin ^{-1}(\sin \theta) \cdot 2 a \tan \theta \cdot \sec ^{2} \theta d \theta$

$=2 a \int \theta \tan \theta \cdot \sec ^{2} \theta d \theta$

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

Hence, $I=a[\frac{x}{a} \tan ^{-1} \sqrt{\frac{x}{a}}-\sqrt{\frac{x}{a}}+\tan ^{-1} \sqrt{\frac{x}{a}}]+C$.

Solution

Let $\quad I=\int _{\pi / 3}^{\pi / 2} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{5 / 2}} d x=\int _{\pi / 3}^{\pi / 2} \frac{\sqrt{2 \cos ^{2} x / 2}}{(2 \sin ^{2} x / 2)^{5 / 2}} d x$

$ =\int _{\pi / 3}^{\pi / 2} \frac{\sqrt{2} \cos x / 2}{(2)^{5 / 2} \sin ^{5} x / 2} d x=\frac{1}{4} \int _{\pi / 3}^{\pi / 2} \frac{\cos x / 2}{\sin ^{5} x / 2} d x $

Put $\quad \sin \frac{x}{2}=t \Rightarrow \frac{1}{2} \cos \frac{x}{2} d x=d t \Rightarrow \cos \frac{x}{2} d x=2 d t$

Changing the limits, we have

When $x=\frac{\pi}{3}, \quad \sin \frac{\pi}{6}=t \quad \therefore t=\frac{1}{2}$

When $x=\frac{\pi}{2}, \quad \sin \frac{\pi}{4}=t \quad \therefore t=\frac{1}{\sqrt{2}}$

$ \begin{aligned} \therefore \quad I & =\frac{1}{4} \times 2 \int _{1 / 2}^{1 / \sqrt{2}} \frac{d t}{t^{5}}=\frac{1}{2} \times(-\frac{1}{4})[t^{-4}] _{1 / 2}^{1 / \sqrt{2}}=-\frac{1}{8}[\frac{1}{t^{4}}] _{1 / 2}^{1 / \sqrt{2}} \\ & =-\frac{1}{8}[\frac{1}{(1 / \sqrt{2})^{4}}-\frac{1}{(1 / 2)^{4}}]=-\frac{1}{8}[4-16] \end{aligned} $

$ =-\frac{1}{8} \times(-12)=\frac{3}{2} $

Hence, $I=\frac{3}{2}$.

Solution

Let $I=\int e_II^{-3 x} \cos _I^{3} x d x$

$ \begin{aligned} & =\cos ^{3} x \cdot \int e^{-3 x} d x-\int(D(\cos ^{3} x) \cdot \int e^{-3 x} d x) d x \\ & =\cos ^{3} x \cdot \frac{e^{-3 x}}{-3}-\int(3 \cos ^{2} x \cdot(-\sin x) \cdot \frac{e^{-3 x}}{-3}) d x \\ & =-\frac{1}{3} e^{-3 x} \cos ^{3} x-\int \cos ^{2} x \sin x \cdot e^{-3 x} d x \\ & =-\frac{1}{3} e^{-3 x} \cos ^{3} x-\int(1-\sin ^{2} x) \sin x \cdot e^{-3 x} d x \\ & =-\frac{1}{3} e^{-3 x} \cos ^{3} x-\int \sin x \cdot e^{-3 x} d x+\int \sin _I^{3} x \cdot e_II^{-3 x} d x \\ & =-\frac{1}{3} e^{-3 x} \cos ^{3} x-\int \sin x \cdot e^{-3 x} d x+\sin ^{3} x \int e^{-3 x} d x \\ & -\int(D(\sin ^{3} x) \int e^{-3 x} d x) d x \\ & =-\frac{1}{3} e^{-3 x} \cos ^{3} x-\int \sin x \cdot e^{-3 x} d x+\sin ^{3} x \cdot \frac{e^{-3 x}}{-3} \\ & -\int 3 \sin ^{2} x \cdot \cos x \cdot \frac{e^{-3 x}}{-3} d x \\ & =-\frac{1}{3} e^{-3 x} \cos ^{3} x-\int \sin x \cdot e^{-3 x} d x-\frac{1}{3} e^{-3 x} \cdot \sin ^{3} x \\ & +\int \sin ^{2} x \cos x e^{-3 x} d x \\ & =-\frac{1}{3} e^{-3 x} \cos ^{3} x-\int \sin x \cdot e^{-3 x} d x-\frac{1}{3} e^{-3 x} \sin ^{3} x \\ & +\int(1-\cos ^{2} x) \cos x \cdot e^{-3 x} d x \\ & I=-\frac{1}{3} e^{-3 x} \cos ^{3} x-[\sin x \cdot \frac{e^{-3 x}}{-3}-\int \cos x \cdot \frac{e^{-3 x}}{-3}] \\ & -\frac{1}{3} e^{-3 x} \cdot \sin ^{3} x+\int \cos x \cdot e^{-3 x}-\int \cos ^{3} x \cdot e^{-3 x} d x \\ & I=-\frac{1}{3} e^{-3 x} \cos ^{3} x+\sin x \cdot \frac{e^{-3 x}}{3}-\int \cos x \cdot \frac{e^{-3 x}}{3} d x \\ & -\frac{1}{3} e^{-3 x} \sin ^{3} x+\int \cos x \cdot e^{-3 x}-I \end{aligned} $

$ \begin{aligned} 2 I=\frac{e^{-3 x}}{-3}[\cos ^{3} x+\sin ^{3} x]-[\sin x \cdot \frac{e^{-3 x}}{-3}-\int \cos x \cdot \frac{e^{-3 x}}{-3} d x] \\ +\int \cos x \cdot e^{-3 x} d x \\ =\frac{e^{-3 x}}{-3}[\cos ^{3} x+\sin ^{3} x]+\frac{1}{3} \sin x \cdot e^{-3 x}- \frac{1}{3} \int \cos x \cdot e^{-3 x} d x \\ +\int \cos x \cdot e^{-3 x} d x \\ \therefore \quad 2 I= \frac{e^{-3 x}}{-3}[\cos ^{3} x+\sin ^{3} x]+\frac{1}{3} \sin x \cdot e^{-3 x}+\frac{2}{3} \int \cos x \cdot e^{-3 x} d x \end{aligned} $

Now, put

$ \begin{aligned} & I_1=\frac{2}{3} \int \cos x \cdot e_II^{-3 x} d x \\ & =\frac{2}{3}[\cos x \cdot \int e^{-3 x} d x-\int(D(\cos x) \cdot \int e^{-3 x} d x) d x] \\ & =\frac{2}{3}[\cos x \cdot \frac{e^{-3 x}}{-3}-\int-\sin x \cdot \frac{e^{-3 x}}{-3} d x] \\ & =\frac{2}{3}[\cos x \cdot \frac{e^{-3 x}}{-3}-\frac{1}{3} \int \sin x \cdot e^{-3 x} d x] \\ & =-\frac{2}{9} \cos x \cdot e^{-3 x}-\frac{2}{9} \int \sin x \cdot e^{-3 x} d x \\ & =-\frac{2}{9} \cos x \cdot e^{-3 x}-\frac{2}{9}[\sin x \cdot \frac{e^{-3 x}}{-3}-\int \cos x \cdot \frac{e^{-3 x}}{-3} d x] \\ & I_1=-\frac{2}{9} \cos x \cdot e^{-3 x}+\frac{2}{27} \sin x \cdot e^{-3 x}-\frac{2}{27} \int \cos x \cdot e^{-3 x} d x \\ & I_1=-\frac{2}{9} \cos x \cdot e^{-3 x}+\frac{2}{27} \sin x \cdot e^{-3 x}-\frac{1}{9} \cdot \frac{2}{3} \int \cos x \cdot e^{-3 x} d x \\ & I_1=-\frac{2}{9} \cos x \cdot e^{-3 x}+\frac{2}{27} \sin x \cdot e^{-3 x}-\frac{1}{9} \cdot I_1 \\ & I_1+\frac{1}{9} I_1=-\frac{2}{9} \cos x \cdot e^{-3 x}+\frac{2}{27} \sin x \cdot e^{-3 x} \\ & \Rightarrow \frac{10 I_1}{9}=-\frac{2}{9} \cos x \cdot e^{-3 x}+\frac{2}{27} \sin x \cdot e^{-3 x} \\ & \therefore \quad I_1=-\frac{1}{10} \cos x \cdot e^{-3 x}+\frac{1}{15} \sin x \cdot e^{-3 x} \\ & \text{ So } \quad 2 I=-\frac{1}{3} e^{-3 x}[\sin ^{3} x+\cos ^{3} x]+\frac{1}{3} \sin x \cdot e^{-3 x}-\frac{1}{10} \cos x \cdot e^{-3 x} \end{aligned} $

$ \begin{aligned} & +\frac{1}{15} \sin x \cdot e^{-3 x} \\ & \therefore \quad I=-\frac{1}{6} e^{-3 x}[\sin ^{3} x+\cos ^{3} x]+\frac{1}{6} \sin x \cdot e^{-3 x}-\frac{1}{20} \cos x \cdot e^{-3 x} \\ & +\frac{1}{30} \sin x \cdot e^{-3 x} \\ & =-\frac{1}{6} e^{-3 x}[\sin ^{3} x+\cos ^{3} x]+\frac{1}{5} \sin x \cdot e^{-3 x}-\frac{1}{20} \cos x \cdot e^{-3 x} \\ & =\frac{e^{-3 x}}{24}[\sin 3 x-\cos 3 x]+\frac{3 e^{-3 x}}{40}[\sin x-3 \cos x]+C \\ & { \begin{bmatrix} \because \sin 3 x & =3 \sin x-4 \sin ^{3} x \\ \cos 3 x & =4 \cos ^{3} x-3 \cos x \end{bmatrix} } \end{aligned} $

Hence, $I=\frac{e^{-3 x}}{24}[\sin 3 x-\cos 3 x]+\frac{3 e^{-3 x}}{40}[\sin x-3 \cos x]+C$.

Solution

Let $I=\int \sqrt{\tan x} d x$

Put $\tan x=t^{2}$

$ \begin{aligned} & \sec ^{2} x d x=2 t d t \Rightarrow d x=\frac{2 t d t}{\sec ^{2} x}=\frac{2 t d t}{1+\tan ^{2} x} \Rightarrow d x=\frac{2 t d t}{1+t^{4}} \\ & \therefore \quad I=\int \frac{t \cdot 2 t}{1+t^{4}} d t=\int \frac{2 t^{2}}{1+t^{4}} d t=\int \frac{2}{t^{2}+\frac{1}{t^{2}}} d t \end{aligned} $

[Dividing the numerator and denominator by $t^{2}$ ]

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

Put $I_1=\int \frac{1+\frac{1}{t^{2}}}{(t-\frac{1}{t})^{2}+(\sqrt{2})^{2}} d t$ and $I_2=\int \frac{1-\frac{1}{t^{2}}}{(t+\frac{1}{t})^{2}-(\sqrt{2})^{2}} d t$

$\therefore \quad I=I_1+I_2$

Now $\quad I_1=\int \frac{1+\frac{1}{t^{2}}}{(t-\frac{1}{t})^{2}+(\sqrt{2})^{2}} d t$

Put $\quad t-\frac{1}{t}=u$

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

$I_1=\int \frac{d u}{u^{2}+(\sqrt{2})^{2}}=\frac{1}{\sqrt{2}} \tan ^{-1} \frac{u}{\sqrt{2}}+C_1$

$=\frac{1}{\sqrt{2}} \tan ^{-1} \frac{t-\frac{1}{t}}{\sqrt{2}}+C_1=\frac{1}{\sqrt{2}} \tan ^{-1} \frac{t^{2}-1}{\sqrt{2} t}+C_1$

$=\frac{1}{\sqrt{2}} \tan ^{-1}(\frac{\tan x-1}{\sqrt{2} \sqrt{\tan x}})+C_1$

Now $\quad I_2=\int \frac{1-\frac{1}{t^{2}}}{(t+\frac{1}{t})^{2}-(\sqrt{2})^{2}} d t$

Put $\quad t+\frac{1}{t}=v \Rightarrow(1-\frac{1}{t^{2}}) d t=d v$

$=\int \frac{d v}{v^{2}-(\sqrt{2})^{2}}=\frac{1}{2 \cdot \sqrt{2}} \log |\frac{v-\sqrt{2}}{v+\sqrt{2}}|+C_2$

$=\frac{1}{2 \sqrt{2}} \log |\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}|+C_2$

$=\frac{1}{2 \sqrt{2}} \log |\frac{t^{2}-\sqrt{2} t+1}{t^{2}+\sqrt{2} t+1}|+C_2$

$=\frac{1}{2 \sqrt{2}} \log |\frac{\tan x-\sqrt{2} \sqrt{\tan x}+1}{\tan x+\sqrt{2} \sqrt{\tan x}+1}|+C_2$

$ \begin{aligned} & \text{ So } \quad I=I_1+I_2 \\ & \Rightarrow \quad I=\frac{1}{\sqrt{2}} \tan ^{-1}[\frac{\tan x-1}{\sqrt{2 \tan x}}]+\frac{1}{2 \sqrt{2}} \log \\ & |\frac{\tan x-\sqrt{2 \tan x}+1}{\tan x+\sqrt{2 \tan x}+1}|+C_1+C_2 \end{aligned} $

Hence, $I=\frac{1}{\sqrt{2}} \tan ^{-1}[\frac{\tan x-1}{\sqrt{2 \tan x}}]+\frac{1}{2 \sqrt{2}} \log$

$ |\frac{\tan x-\sqrt{2 \tan x}+1}{\tan x+\sqrt{2 \tan x}+1}|+C . $

Solution

Let $\quad I=\int_0^{\pi / 2} \frac{d x}{(a^{2} \cos ^{2} x+b^{2} \sin ^{2} x)^{2}}$

Dividing the numerator and denominator by $\cos ^{4} x$, we have

$ \begin{aligned} I & =\int_0^{\pi / 2} \frac{\sec ^{4} x}{(\frac{a^{2} \cos ^{2} x}{\cos ^{2} x}+\frac{b^{2} \sin ^{2} x}{\cos ^{2} x})^{2}} d x \\ & =\int_0^{\pi / 2} \frac{\sec ^{2} x \cdot \sec ^{2} x}{(a^{2}+b^{2} \tan ^{2} x)^{2}} d x=\int_0^{\pi / 2} \frac{(1+\tan ^{2} x) \sec ^{2} x}{(a^{2}+b^{2} \tan ^{2} x)^{2}} d x \end{aligned} $

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

Changing the limits, we get

When $x=0, \quad t=\tan 0=0$

When $x=\frac{\pi}{2}, \quad t=\tan \frac{\pi}{2}=\infty$

$ \therefore \quad I=\int_0^{\infty} \frac{1+t^{2}}{(a^{2}+b^{2} t^{2})^{2}} d t $

Put $t^{2}=u$ only for the purpose of partial fraction

$ \begin{aligned} \therefore \quad \frac{1+u}{(a^{2}+b^{2} u)^{2}} & =\frac{A}{(a^{2}+b^{2} u)}+\frac{B}{(a^{2}+b^{2} u)^{2}} \\ 1+u & =A(a^{2}+b^{2} u)+B \end{aligned} $

Comparing the coefficients of like terms, we get

$a^{2} A+B=1$ and $b^{2} A=1 \Rightarrow A=\frac{1}{b^{2}}$

Now $a^{2} \cdot \frac{1}{b^{2}}+B=1 \quad \Rightarrow B=1-\frac{a^{2}}{b^{2}}=\frac{b^{2}-a^{2}}{b^{2}}$

$ \begin{aligned} & \therefore \quad I=\int_0^{\infty} \frac{1+t^{2}}{(a^{2}+b^{2} t^{2})^{2}}=\frac{1}{b^{2}} \int_0^{\infty} \frac{d t}{a^{2}+b^{2} t^{2}}+\frac{b^{2}-a^{2}}{b^{2}} \int_0^{\infty} \frac{d t}{(a^{2}+b^{2} t^{2})^{2}} \\ & =\frac{1}{b^{2}} \int_0^{\infty} \frac{d t}{b^{2}(\frac{a^{2}}{b^{2}}+t^{2})}+\frac{b^{2}-a^{2}}{b^{2}} \int_0^{\infty} \frac{d t}{(a^{2}+b^{2} t^{2})^{2}} \\ & =\frac{1}{a b^{3}}[\tan ^{-1} \frac{t}{a / b}]_0^{\infty}+\frac{b^{2}-a^{2}}{b^{2}}(\frac{\pi}{4} \cdot \frac{1}{a^{3} b}) \\ & =\frac{1}{a b^{3}}[\tan ^{-1} \infty-\tan 0]+\frac{b^{2}-a^{2}}{b^{2}}(\frac{\pi}{4 a^{3} b}) \\ & =\frac{1}{a b^{3}} \cdot \frac{\pi}{2}+\frac{\pi}{4} \cdot \frac{b^{2}-a^{2}}{a^{3} b^{3}}=\frac{\pi}{2 a b^{3}}+\frac{\pi}{4} \cdot \frac{b^{2}-a^{2}}{a^{3} b^{3}} \\ & =\pi[\frac{2 a^{2}+b^{2}-a^{2}}{4 a^{3} b^{3}}]=\frac{\pi}{4}(\frac{a^{2}+b^{2}}{a^{3} b^{3}}) \\ & \text{ Hence, } I=\frac{\pi}{4}(\frac{a^{2}+b^{2}}{a^{3} b^{3}}) \end{aligned} $

Solution

Let

$ \begin{aligned} \text{ I } & =\int_0^{1} x \log |1+2 x| d x \\ & =[\log |1+2 x| \cdot(\frac{x^{2}}{2})]_0^{1}-\int_0^{1}(\frac{1.2}{1+2 x} \cdot \frac{x^{2}}{2}) d x \\ & =\frac{1}{2}[x^{2} \log (1+2 x)]_0^{1}-\int_0^{1} \frac{x^{2}}{1+2 x} d x \\ & =\frac{1}{2}[\log 3-0]-\int_0^{1}(\frac{x}{2}-\frac{x / 2}{1+2 x}) d x \\ & =\frac{1}{2} \log 3-\frac{1}{2} \int_0^{1} x d x+\frac{1}{2} \int_0^{1} \frac{x}{1+2 x} d x \\ & =\frac{1}{2} \log 3-\frac{1}{2}[\frac{x^{2}}{2}]_0^{1}+\frac{1}{2} \cdot \frac{1}{2} \int_0^{1} \frac{(2 x+1-1)}{2 x+1} d x \\ & =\frac{1}{2} \log 3-\frac{1}{4}[1-0]+\frac{1}{4} \int_0^{1} 1 d x-\frac{1}{4} \int_0^{1} \frac{1}{2 x+1} d x \\ & =\frac{1}{2} \log 3-\frac{1}{4}+\frac{1}{4}[x]_0^{1}-\frac{1}{4} \cdot \frac{1}{2}[\log |2 x+1|]_0^{1} \end{aligned} $

$ \begin{aligned} & =\frac{1}{2} \log 3-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}[\log 3-0] \\ & =\frac{1}{2} \log 3-\frac{1}{8} \log 3=\frac{3}{8} \log 3 \end{aligned} $

Hence, $I=\frac{3}{8} \log 3$.

Solution

Let $\quad I=\int_0^{\pi} x \log \sin x d x$

$$ \begin{equation*} =\int_0^{\pi}(\pi-x) \log \sin (\pi-x) d x \tag{i} \end{equation*} $$

$ [\text{ using } \int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x] $

$$ \begin{equation*} I=\int_0^{\pi}(\pi-x) \log \sin x d x \tag{ii} \end{equation*} $$

Adding (i) and (ii), we get

$$ \begin{align*} 2 I & =\int_0^{\pi}[(\pi-x) \log \sin x+x \log \sin x] d x \\ 2 I & =\int_0^{\pi} \pi \log \sin x d x \\ 2 I & =2 \pi \int_0^{\pi / 2} \log \sin x d x \quad[\because \int_0^{a} f(x) d x=2 \int_0^{a / 2} f(x) d x] \\ \therefore \quad I & =\pi \int_0^{\pi / 2} \log \sin x d x \tag{iii}\\ I & =\pi \int_0^{\pi / 2} \log \sin (\frac{\pi}{2}-x) d x \\ I & =\pi \int_0^{\pi / 2} \log \cos x d x \tag{iv} \end{align*} $$

On adding (iii) and (iv), we get

$ \begin{aligned} & 2 I=\pi \int_0^{\pi / 2}(\log \sin x+\log \cos x) d x \\ & 2 I=\pi \int_0^{\pi / 2} \log \sin x \cos x d x=\pi \int_0^{\pi / 2} \frac{\log 2 \sin x \cos x}{2} d x \end{aligned} $

$ 2 \mathrm{I}=\pi \int_0^{\pi / 2} \log \sin 2 x d x-\pi \int_0^{\pi / 2} \log 2 d x $

Put $ \quad 2 x=t \Rightarrow 2 d x=d t \Rightarrow d x=\frac{d t}{2} $

$2 \mathrm{I}=\pi \int_0^\pi \log \sin t d t-\pi \cdot \log 2 \int_0^{\pi / 2} 1 d x$ [Changing the limit]

$2 \mathrm{I}=\mathrm{I}-\pi \cdot \log 2[x]_0^{\pi / 2}$[from eqn. (iii)]

$ 2 \mathrm{I}-\mathrm{I}=-\frac{\pi^2}{2} \log 2 $

So $\quad \mathrm{I}=\frac{\pi^2}{2} \log \left(\frac{1}{2}\right)$

Solution

Let $$ \begin{aligned} I & =\int_{-\pi / 4}^{\pi / 4} \log |\sin x+\cos x| d x \\ & =\int_{-\pi / 4}^{\pi / 4} \log \left|\sin \left(\frac{\pi}{4}-\frac{\pi}{4}-x\right)+\cos \left(\frac{\pi}{4}-\frac{\pi}{4}-x\right)\right| d x \\ & \quad\left[\because \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right] \\ & =\int_{-\pi / 4}^{\pi / 4} \log |\sin (-x)+\cos x| d x \\ & =\int_{-\pi / 4}^{\pi / 4} \log |\cos x-\sin x| d x \end{aligned} $$

Adding (i) and (ii), we get

$ \begin{aligned} 2 I & =\int _{-\pi / 4}^{\pi / 4} \log |\cos x+\sin x| d x+\int _{-\pi / 4}^{\pi / 4} \log |\cos x-\sin x| d x \\ & =\int _{-\pi / 4}^{\pi / 4} \log |(\cos x+\sin x)(\cos x-\sin x)| d x \\ & =\int _{-\pi / 4}^{\pi / 4} \log |\cos ^{2} x-\sin ^{2} x| d x \\ \therefore \quad 2 I & =\int _{-\pi / 4}^{\pi / 4} \log \cos 2 x d x \end{aligned} $

$ \begin{aligned} & 2 I=2 \int_0^{\pi / 4} \log \cos 2 x d x \\ & \therefore \quad I=\int_0^{\pi / 4} \log \cos 2 x d x \\ & \text{ Put } \quad 2 x=t \quad \Rightarrow d x=\frac{d t}{2} \\ & \text{ When } x=0 \quad \therefore t=0 \text{; When } x=\frac{\pi}{4} \quad \therefore t=\frac{\pi}{2} \\ & I=\frac{1}{2} \int_0^{\pi / 2} \log \cos t d t \\ & I=\frac{1}{2} \int_0^{\pi / 2} \log \cos (\frac{\pi}{2}-t) d t \\ & I=\frac{1}{2} \int_0^{\pi / 2} \log \sin t d t \end{aligned} $

$ [\because \int _{-a}^{a} f(x) d x=2 \int_0^{a} f(x) d x \text{ if } f(-x)=f(x)] $

On adding (iii) and (iv), we get,

$ \begin{aligned} & 2 I=\frac{1}{2} \int_0^{\pi / 2}(\log \cos t+\log \sin t) d t \\ & \Rightarrow \quad 2 I=\frac{1}{2} \int_0^{\pi / 2} \log \sin t \cos t d t \\ & \Rightarrow \quad 2 I=\frac{1}{2} \int_0^{\pi / 2} \frac{\log 2 \sin t \cos t}{2} d t \\ & \Rightarrow \quad 2 I=\frac{1}{2} \int_0^{\pi / 2}(\log \sin 2 t-\log 2) d t \\ & \Rightarrow \quad 4 I=\int_0^{\pi / 2} \log \sin 2 t d t-\int_0^{\pi / 2} \log 2 d t \\ & \text{ Put } 2 t=u \Rightarrow 2 d t=d u \Rightarrow d t=\frac{d u}{2} \\ & \therefore \quad 4 I=\frac{1}{2} \int_0^{\pi} \log \sin u d u-\int_0^{\pi / 2} \log 2 . d t \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 4 I=\frac{1}{2} \times 2 \int_0^{\pi / 2} \log \sin u d u-\log 2[t]_0^{\pi / 2} \\ & \Rightarrow \quad 4 I=\int_0^{\pi / 2} \log \sin u d u-\log 2 \cdot \frac{\pi}{2} \\ & \Rightarrow \quad 4 I=2 I-\frac{\pi}{2} \log 2 \\ & \Rightarrow \quad 2 I=-\frac{\pi}{2} \log 2 \Rightarrow I=\frac{\pi}{4} \log \frac{1}{2} \end{aligned} $

Hence, $I=\frac{\pi}{4} \log \frac{1}{2}$.

Solution

Let $I=\int \frac{\cos 2 x-\cos 2 \theta}{\cos x-\cos \theta} d x$

$ \begin{aligned} & =\int \frac{(2 \cos ^{2} x-1)-(2 \cos ^{2} \theta-1)}{\cos x-\cos \theta} d x \\ & =\int \frac{2 \cos ^{2} x-1-2 \cos ^{2} \theta+1}{\cos x-\cos \theta} d x \\ & =\int \frac{2 \cos ^{2} x-2 \cos ^{2} \theta}{\cos x-\cos \theta} d x=2 \int \frac{\cos ^{2} x-\cos ^{2} \theta}{\cos x-\cos \theta} d x \\ & =2 \int \frac{(\cos x+\cos \theta)(\cos x-\cos \theta)}{(\cos x-\cos \theta)} d x \\ & =2 \int(\cos x+\cos \theta) d x \\ \therefore \quad I & =2(\sin x+\cos \theta \cdot x)+C . \end{aligned} $

Hence, correct option is $(a)$.

Solution

Let

$ I=\int \frac{d x}{\sin (x-a) \cdot \sin (x-b)} $

Multiplying and dividing by $\sin (b-a)$ we get,

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

Hence, the correct option is (c).

Solution

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

Put $\sqrt{x}=\tan \theta \Rightarrow x=\tan ^{2} \theta \Rightarrow d x=2 \tan \theta \sec ^{2} \theta d \theta$

$\therefore \quad I=\int \tan ^{-1}(\tan \theta) \cdot 2 \tan \theta \sec ^{2} \theta d \theta=2 \int \theta \cdot \tan \theta \cdot \sec ^{2} \theta d \theta$

$=2[\theta \cdot \int \tan \theta \cdot \sec ^{2} \theta d \theta-\int(D(\theta) \cdot \int \tan \theta \sec ^{2} \theta d \theta) d \theta]$

Let us take

$ I_1=\int \tan \theta \sec ^{2} \theta d \theta $

Put $\tan \theta=t \Rightarrow \sec ^{2} \theta d \theta=d t$

$ \begin{matrix} \therefore \quad I_1 =\int t d t=\frac{1}{2} t^{2}=\frac{1}{2} \tan ^{2} \theta \\ \therefore \quad \quad I =2[\theta \cdot \frac{1}{2} \tan ^{2} \theta-\int(1 \cdot \frac{1}{2} \tan ^{2} \theta) d \theta] \\ =\theta \tan ^{2} \theta-\int \tan ^{2} \theta d \theta=\theta \tan ^{2} \theta-\int(\sec ^{2} \theta-1) d \theta \\ =\theta \tan ^{2} \theta-(\tan \theta-\theta)+C=\theta \tan ^{2} \theta-\tan \theta+\theta+C \\ \therefore \quad I =\tan ^{-1} \sqrt{x} \cdot x-\sqrt{x}+\tan ^{-1} \sqrt{x}+C \\ =(x+1) \tan ^{-1} \sqrt{x}-\sqrt{x}+C \end{matrix} $

Hence, the correct option is $(a)$.

Solution

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

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

Here $f(x)=\frac{1}{1+x^{2}} \quad \therefore f^{\prime}(x)=\frac{-2 x}{(1+x^{2})^{2}}$

Using $\int e^{x}[f(x)+f^{\prime}(x)] d x=e^{x} . f(x)+C$

$\therefore \quad I=e^{x} \cdot \frac{1}{(1+x^{2})}+C=\frac{e^{x}}{1+x^{2}}+C$

Hence, the correct option is (a).

Solution

Let $I=\int \frac{x^{9}}{(4 x^{2}+1)^{6}} d x=\int \frac{x^{9}}{x^{12}(4+\frac{1}{x^{2}})^{6}} d x=\int \frac{1}{x^{3}(4+\frac{1}{x^{2}})^{6}} d x$

$ \begin{aligned} & \text{ Put }(4+\frac{1}{x^{2}})=t \Rightarrow \frac{-2}{x^{3}} d x=d t \Rightarrow \frac{d x}{x^{3}}=-\frac{1}{2} d t \\ & \therefore \quad I=-\frac{1}{2} \int \frac{d t}{t^{6}} \\ & \quad=-\frac{1}{2} \times-\frac{1}{5} t^{-5}+C=\frac{1}{10} t^{-5}+C=\frac{1}{10}(4+\frac{1}{x^{2}})^{-5}+C \end{aligned} $

Hence, the correct option is $(d)$.

Solution

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

Let us resolve the given integrand into partial fractions

Put

$ \begin{aligned} & \frac{1}{(x+2)(x^{2}+1)}=\frac{A}{(x+2)}+\frac{B x+C}{(x^{2}+1)} \\ & 1=A(x^{2}+1)+(x+2)(B x+C) \\ & 1=A x^{2}+A+B x^{2}+C x+2 B x+2 C \\ & 1=(A+B) x^{2}+(C+2 B) x+(A+2 C) \end{aligned} $

Comparing the like terms, we have

$$ \begin{align*} A+B & =0 \tag{i}\\ 2 B+C & =0 \tag{ii}\\ A+2 C & =1 \tag{iii} \end{align*} $$

Subtracting (i) from (iii) we get

$ 2 C-B=1 \quad \therefore B=2 C-1 $

Putting the value of $B$ in eqn. (ii) we have

$ \begin{aligned} & 2(2 C-1)+C=0 \Rightarrow 4 C-2+C=0 \\ & 5 C=2 \quad \therefore C=\frac{2}{5} \\ & \therefore \quad B=2(\frac{2}{5})-1=-\frac{1}{5} \text{ and } A=\frac{1}{5} \\ & \therefore \int \frac{1}{(x+2)(x^{2}+1)} d x=\int \frac{\frac{1}{5}}{(x+2)} d x+\int \frac{-\frac{1}{5} x+\frac{2}{5}}{(x^{2}+1)} d x \\ & =\frac{1}{5} \int \frac{1}{(x+2)} d x-\frac{1}{5} \int \frac{x-2}{(x^{2}+1)} d x \\ & =\frac{1}{5} \int \frac{1}{x+2} d x-\frac{1}{5} \int \frac{x}{x^{2}+1} d x+\frac{2}{5} \int \frac{1}{x^{2}+1} d x \\ & =\frac{1}{5} \int \frac{1}{x+2} d x-\frac{1}{10} \int \frac{2 x}{x^{2}+1} d x+\frac{2}{5} \int \frac{1}{x^{2}+1} d x \\ & \therefore \quad I=\frac{1}{5} \log |x+2|-\frac{1}{10} \log |x^{2}+1|+\frac{2}{5} \tan ^{-1} x+C \end{aligned} $

Putting the given value of I

$ \begin{aligned} & \therefore a \log |1+x^{2}|+b \tan ^{-1} x+\frac{1}{5} \log |x+2|+C \\ & \quad=\frac{1}{5} \log |x+2|-\frac{1}{10} \log |x^{2}+1|+\frac{2}{5} \tan ^{-1} x+C \\ & \therefore a=-\frac{1}{10} \quad \text{ and } b=\frac{2}{5} \end{aligned} $

Hence, the correct option is (c).

Solution

Let $I=\int \frac{x^{3}}{x+1} d x$

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

Hence, the correct option is $(d)$.

Solution

Let $I=\int \frac{x+\sin x}{1+\cos x} d x$

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

Hence, the correct option is $(d)$.

Solution

Let $I=\int \frac{x^{3}}{\sqrt{1+x^{2}}} d x$

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

$ \begin{aligned} \therefore I & =\frac{1}{2} \int \frac{(t-1)}{\sqrt{t}} d t \\ & =\frac{1}{2} \int \frac{t}{\sqrt{t}} d t-\frac{1}{2} \int \frac{1}{\sqrt{t}} d t=\frac{1}{2} \int \sqrt{t} d t-\frac{1}{2} \int t^{-1 / 2} d t \end{aligned} $

$ =\frac{1}{2} \times \frac{2}{3}(t)^{3 / 2}-\frac{1}{2} \cdot 2 \sqrt{t}+C=\frac{1}{3}(1+x^{2})^{3 / 2}-\sqrt{1+x^{2}}+C $

But $I=a(1+x^{2})^{3 / 2}+b \sqrt{1+x^{2}}+C$

Comparing the like terms we get, $a=\frac{1}{3}$ and $b=-1$

Hence, the correct option is (d).

Solution

Let $I=\int _{-\pi / 4}^{\pi / 4} \frac{d x}{1+\cos 2 x}$

$ \begin{aligned} & =\int _{-\pi / 4}^{\pi / 4} \frac{d x}{2 \cos ^{2} x}=\frac{1}{2} \int _{-\pi / 4}^{\pi / 4} \sec ^{2} x d x=\frac{1}{2}[\tan x] _{-\pi / 4}^{\pi / 4} \\ & =\frac{1}{2}[\tan \frac{\pi}{4}-\tan (-\frac{\pi}{4})]=\frac{1}{2}[1+1]=\frac{1}{2} \times 2=1 \end{aligned} $

Hence, the correct option is $(a)$.

Solution

Let $I=\int_0^{\pi / 2} \sqrt{1-\sin 2 x} d x=\int_0^{\pi / 2} \sqrt{(\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x)} d x$

$ \begin{aligned} & =\int_0^{\pi / 2} \sqrt{(\sin x-\cos x)^{2}} d x=\int_0^{\pi / 2} \pm(\sin x-\cos x) d x \\ & =\int_0^{\pi / 4}-(\sin x-\cos x) d x+\int _{\pi / 4}^{\pi / 2}(\sin x-\cos x) d x \\ & =\int_0^{\pi / 4}(\cos x-\sin x) d x+\int _{\pi / 4}^{\pi / 2}(\sin x-\cos x) d x \\ & =[\sin x+\cos x]_0^{\pi / 4}+[-\cos x-\sin x] _{\pi / 4}^{\pi / 2} \\ & =[(\sin \frac{\pi}{4}+\cos \frac{\pi}{4})-(\sin 0-\cos 0)]-[(\cos \frac{\pi}{2}+\sin \frac{\pi}{2}). \\ & .-(\cos \frac{\pi}{4}+\sin \frac{\pi}{4})] \end{aligned} $

$ \begin{aligned} & =[(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})-(+1)]-[(0+1)-(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})] \\ & =(\frac{2}{\sqrt{2}}-1)-(1-\frac{2}{\sqrt{2}})=\frac{2}{\sqrt{2}}-1-1+\frac{2}{\sqrt{2}} \\ & =\frac{4}{\sqrt{2}}-2=2 \sqrt{2}-2=2(\sqrt{2}-1) \end{aligned} $

Hence, the correct option is $(d)$.

Solution

Let $\quad I=\int_0^{\pi / 2} \cos x \cdot e^{\sin x} d x$

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

When $x=0$ then $t=\sin 0=0$; When $x=\frac{\pi}{2}$ then $t=\sin \frac{\pi}{2}=1$

$ \therefore \quad I=\int_0^{1} e^{t} d t=[e^{t}]_0^{1}=(e^{1}-e^{0})=e-1 $

Hence, $I=e-1$.

Solution

Let $\quad I=\int \frac{x+3}{(x+4)^{2}} \cdot e^{x} d x=\int \frac{x+4-1}{(x+4)^{2}} \cdot e^{x} d x$

$=\int[\frac{x+4}{(x+4)^{2}}-\frac{1}{(x+4)^{2}}] e^{x} d x=\int[\frac{1}{x+4}-\frac{1}{(x+4)^{2}}] e^{x} d x$

Put $\quad \frac{1}{x+4}=t \Rightarrow-\frac{1}{(x+4)^{2}} d x=d t$

Let $\quad f(x)=\frac{1}{x+4} \quad \therefore f^{\prime}(x)=-\frac{1}{(x+4)^{2}}$

Using $\int e^{x}[f(x)+f^{\prime}(x)] d x=e^{x} f(x)+C$

$\therefore \quad I=e^{x} \cdot \frac{1}{x+4}+C$

Hence, $I=\frac{e^{x}}{x+4}+C$.

Solution

Given that: $\int_0^{a} \frac{1}{1+4 x^{2}} d x=\frac{\pi}{8}$

$ \begin{aligned} & \Rightarrow \quad \frac{1}{4} \int_0^{a} \frac{1}{(\frac{1}{4}+x^{2})} d x=\frac{\pi}{8} \Rightarrow \int_0^{a} \frac{1}{[(\frac{1}{2})^{2}+x^{2}]} d x=\frac{\pi}{2} \\ & \Rightarrow \quad \frac{1}{1 / 2}[\tan ^{-1} \frac{x}{1 / 2}]_0^{a}=\frac{\pi}{2} \Rightarrow 2[\tan ^{-1} 2 a-\tan ^{-1} 0]=\frac{\pi}{2} \\ & \Rightarrow \tan ^{-1} 2 a=\frac{\pi}{4} \Rightarrow 2 a=\tan \frac{\pi}{4} \Rightarrow 2 a=1 \Rightarrow a=\frac{1}{2} \end{aligned} $

Hence, the value of $a=\frac{1}{2}$.

Solution

Let $I=\int \frac{\sin x}{3+4 \cos ^{2} x} d x$

Put $\quad \cos x=t$

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

$\therefore I=-\int \frac{d t}{3+4 t^{2}}=-\frac{1}{4} \int \frac{d t}{\frac{3}{4}+t^{2}}=-\frac{1}{4} \int \frac{d t}{(\frac{\sqrt{3}}{2})^{2}+t^{2}}$

$ \begin{aligned} & =-\frac{1}{4} \times \frac{1}{\sqrt{3} / 2} \tan ^{-1}(\frac{t}{\sqrt{3} / 2})+C \\ & =-\frac{1}{2 \sqrt{3}} \tan ^{-1}(\frac{2 t}{\sqrt{3}})+C=-\frac{1}{2 \sqrt{3}} \tan ^{-1}(\frac{2 \cos x}{\sqrt{3}})+C \end{aligned} $

Hence, $I=-\frac{1}{2 \sqrt{3}} \tan ^{-1}(\frac{2}{\sqrt{3}} \cos x)+C$.

Solution

Let $I=\int _{-\pi}^{\pi} \sin ^{3} x \cdot \cos ^{2} x d x$

Let $f(x)=\sin ^{3} x \cos ^{2} x$

$ f(-x)=\sin ^{3}(-x) \cdot \cos ^{2}(-x)=-\sin ^{3} x \cos ^{2} x=-f(x) $

$\therefore \int^{\pi} \sin ^{3} x \cdot \cos ^{2} x d x$ is an odd function

$\therefore{ }^{-\pi} \quad I=0$