If $I=\frac{2}{\pi} \int_{\pi / 4}^{\pi / 4} \frac{d x}{\left(1+e^{\sin x}\right)(2-\cos 2 x)}$ then $27 I^2=$ ?

Solution:

Recall: $\int_{-a}^a f(x) d x=\int_0^a[f(x)+f(-x)] d x$

Pf: $\int_{-a}^a f(x) d x=\int_{-a}^0 f(x) d x+\int_0^a f(x) d x$

Now putting $x=-y$ in the first integral, $d x=-d y$ and

when $x=-a, y=a$;

when $x=0, y_0=0$

So, $\int_{-a}^0 f(x) d x=\int_a^x f(-y)(-d y)$

$-\int_a^a f(x) d y-\int_0^a(1-y) d y$

$=\int_0^a f(-x) d x$

i.e, $\int_{-a}^a f(x) d x =\int_0^a[f(-x)+f(-x)] d x$

Recall

Special Cases:

(1) If $f$ is an old function, ie.,

$f(-x)=-f(x)$, then

$\int_{-a}^{a} f(x) d x=0$

(2) If $f$ is an even function, ie., $f(-x)=f(x)$, then

$\int_{-a}^{a} f(x) d x=2 \int_0^a f(x) d x$

Solution: We have

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

$\therefore \quad f(-x) =\frac{1}{(1+e^{\sin (-x)})(2-\cos (-2 x))}$

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

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

$\Rightarrow f(x)+f(-x) =\frac{(1+e^{\sin x})}{\left(1+e^{\sin x}\right)(2-\cos 2 x)}$

$=\frac{1}{2-\cos 2 x}$

$=\frac{1}{2-\left(2 \cos ^2 x-1\right)}$

$=\frac{1}{2-\cos 2 x}$

$=\frac{1}{2-\left(2 \cos ^2 x-1\right)}$

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

$\therefore I =\frac{2}{\pi} \int_0^{\pi / 4} \frac{1}{(3-2 \cos ^2 x)} d x$

$=\frac{2}{\pi} \int_0^{\pi / 4} \frac{\sec ^2 x}{3 \sec ^2 x-2} d x$

$=\frac{2}{\pi} \int_0^{\pi / 4} \frac{\sec ^2 x}{3(1+\tan ^2 x)-2} d x$

Put $u=\tan x$. Then $d u=\sec ^2 x d x$ When $x=0, u=0$

When $x=\frac{\pi}{4}, u=\tan \frac{\pi}{4}=1$

So, $\begin{aligned} I & =\frac{2}{\pi} \int_0^1 \frac{d u}{3 u^2+1} \ & =\frac{2}{3 \pi} \int_0^1 \frac{d u}{u^2+\left(\frac{1}{\sqrt{3}}\right)^2}\end{aligned}$

$=\left.\frac{2}{3 \pi} \frac{1}{(1 / \sqrt{3})} \tan ^{-1}\left(\frac{u}{1 / \sqrt{3}}\right)\right|_0 ^1$

$=\frac{2}{\sqrt{3} \pi}\left(\tan ^{-1} \sqrt{3}-\tan ^{-1} 0\right)$

$=\frac{2}{\sqrt{3} \pi} \times \frac{\pi}{3}=\frac{2}{3 \sqrt{3}}$

$\Rightarrow \quad I^2 =\frac{4}{27} \Rightarrow 27 I^2=4$

Find the value of the integral

$I=\int_0^{\pi / 2} \frac{3 \sqrt{\cos \theta}}{(\sqrt{\cos \theta}+\sqrt{\sin \theta})^5} d \theta$

Sol: Substitute $\theta=\frac{\pi}{2}-\varphi$

The $d \theta=-d \varphi$

When $\theta=0, \varphi=\frac{\pi}{2}$

when $\theta=\frac{\pi}{2}, \varphi=0$

Also, $\quad \cos \theta=\cos \left(\frac{\pi}{2}-\varphi\right)=\sin \varphi$

and $\frac{\sin }{\pi / 2} \theta=\cos \varphi$

$\therefore \quad I=\int_0^{\pi / 2} \frac{3 \sqrt{\sin \varphi}}{\left(\sqrt{\sin \varphi}+\sqrt{\cos \varphi)^5}\right.} d \varphi$ $=\int_0^{\pi / 2} \frac{3 \sqrt{\sin \theta}}{(\sqrt{\cos \theta}+\sqrt{\sin \theta})^5} d \theta$

Adding (i) & (ii), we get

$2 I=\int_0^{\pi / 2} \frac{3(\sqrt{\cos \theta}+\sqrt{\sin \theta})}{(\sqrt{\cos \theta}+\sqrt{\sin \theta})^5} d \theta$

$=\int_0^{\pi / 2} \frac{3}{(\sqrt{\cos \theta}+\sqrt{\sin \theta})^4} d \theta$

$=\int_0^{\pi / 2} \frac{3}{\cos ^2 \theta(1+\sqrt{\tan \theta})^4} d \theta$

$=\int_0^{\pi / 2} \frac{3 \sec \theta}{(1+\sqrt{\tan \theta})^4} d \theta$

$=\int_0^{\pi / 2} \frac{3}{(\sqrt{\cos \theta}+\sqrt{\sin -\theta})^4} d \theta$

$=\int_0^{\pi / 2} \frac{3}{\cos ^2 \theta(1+\sqrt{\tan \theta})^4} d \theta$

$=\int_0^{\pi / 2} \frac{3 \sec^2 \theta}{(1+\sqrt{\tan \theta})^4} d \theta$

$\therefore 2 I=\int_0^{\infty} \frac{3(2 t d t)}{(1+t)^4}$

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

$\operatorname{Put}$

$\tan \theta=t^2$

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

$\text { When } \theta=0, t=0$

$\text { When } \theta = \frac{\pi}{2} , t=0$

$=3 \int_0^{\infty} \frac{(t+1)-1}{(t+1)^4} d t$

$=3\left[\int_0^{\infty}(t+1)^{-3} d t-\int_0^{\infty}(t+1)^{-4} d t\right]$

$=\left.3\left(\frac{-1}{2(t+1)^2}+\frac{1}{3(t+1)^3}\right)\right|_0 ^{\infty}$

$=3\left[\int_0^{\infty}(t+1)^{-3} d t-\int_0^{\infty}(t+1)^{-4} d t\right]$

$=\left.3\left(\frac{-1}{2(t+1)^2}+\frac{1}{3(t+1)^3}\right)\right|_0 ^{\infty}$

$=3\left(0+\frac{1}{2}-\frac{1}{3}\right)=3 \times \frac{1}{6}=\frac{1}{2}$

Find the value of the integral

$I=\int_0^{1 / 4} \frac{1+\sqrt{3}}{\left[(x+1)^2(1-x)^6\right]^{1 / 4}} d x$

Solution:

$I=(1+\sqrt{3}) \int_0^{1 / 2} \frac{d x}{(x+1)^{1 / 2}(1-x)^{3 / 2}}$

$=(1+\sqrt{3}) \int_0^{1 / 2} \frac{d x}{(1-x) \sqrt{1-x^2}}$

Put $x=\sin \theta$. The $d x=\cos \theta d \theta$ When $x=0, \theta=0$

When $x=\frac{1}{2}, \theta=\frac{\pi}{6}$

$\therefore I =(\sqrt{3}+1) \int_0^{\pi / 6} \frac{\cos \theta d \theta}{(1-\sin \theta) \cos \theta}$

$=(\sqrt{3}+1) \int_0^{\pi / 6} \frac{1+\sin \theta}{\cos ^2 \theta} d \theta$

$=(\sqrt{3}+1) \int_0^{\pi /6 } {(sec ^2 \theta+\sec \theta \tan \theta) d \theta}$

$=\left.(\sqrt{3}+1)(\tan \theta+\sec \theta)\right|_0 ^{\pi / 6}$

$=(\sqrt{3}+1)\left[\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}-0-1\right]$

$=(\sqrt{3}+1)\left(\frac{3}{\sqrt{3}}-1\right)$

$=(\sqrt{3}+1)(\sqrt{3}-1)=2$

$\therefore I =2$

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f(0)=0, f\left(\frac{\pi}{2}\right)=3$ and $f^{\prime}(0)={1}$.

If $g(x)=\int_x^{\pi / 2}\left[f^{\prime}(t) \operatorname{cosec} t-\cot t \operatorname{cosec} t f(t)\right] d t$, for $x \in\left(0, \frac{\pi}{2}\right]$, the

$\lim _{x \rightarrow 0} g(x)=\text { ? }$

Soln: Note that $f^{\prime}(t) \operatorname{cosec} t-\operatorname{cosec} t \cot t f(t)$

$=\frac{d}{d t}[f(t) \text { cosect }]$

1) $g(x)=\int_x^{\pi / 2} \frac{d}{d t}[f(t) \operatorname{cosec} t] d t$

Note that cosec 0 is not defined, so we connot find $g(0)$

But $\lim _{x \rightarrow 0} g(x)=3-\lim _{x \rightarrow 0} \frac{f(x)}{\sin x}\left(\frac{0}{0}\right)$

$=3-\lim _{x \rightarrow 0} \frac{f(x) / x}{\sin x / x}$

$=3-\lim _{x \rightarrow 0} \frac{f(x) / x}{\sin x / x}$ $=3-\lim _{x \rightarrow 0} \frac{f(x)}{x} \quad\left(\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right)$ $=3-\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0} \quad(\because f(0)=0)$ $=3-f^{\prime}(0)=3-1=2$

If $I=\sum_{k=1}^{98} \int_k^{k+1} \frac{k+1}{x(x+1)} d x$, the

(A) $I>\ln 99$

(B) $I<\ln 99$

(C) $I<\frac{49}{50}$

(D) $I>\frac{49}{50}$

Note that $\ln 99=\int_1^{99} \frac{1}{x} d x=\sum_{k=1}^{98} \int_k^{k+1} \frac{1}{x} d x$

Now, if $k

$\begin{gathered}\Rightarrow \frac{k+1}{x+1}<1 \\ \therefore \frac{k+1}{x(x+1)}<\frac{1}{x+1}\end{gathered}$

So, $\quad \int_k^{k n} \frac{k n}{x(x+1)} d x<\int_k^{x+1} \frac{1}{x} d x$

$\text { 1) } I=\sum_{h=1}^{98} \int_h^{h+1} \frac{k+1}{x(x+1)} d x<\ln 99$

$\therefore(A)$ is false; (B) is true.

Similarly, $\frac{k+1}{x}>1$ if $k<x<k+1$ i. $\frac{k+1}{x(x+1)}>\frac{1}{x+1}$

$\int_{k+1}^{k+1} \frac{k+1}{x(x+1)} d x> \int_k^{k+1} \frac{1}{x+1} d x$

$\therefore=\sum_{k=1}^{98} \int_k^{k+1} \frac{k+1}{x(x+1)} d x >\int_1^{99} \frac{1}{x+1} d x$ $=\left.\ln (x+1)\right|_1 ^{99}$ $=\ln 100-\ln 2$ $=\ln 50$

Clearly, $\ln 50>1>\frac{49}{50}, \operatorname{sin} I>\frac{49}{50}$

$\therefore$ (D) is true; (C) is false

Remark, Note that $\frac{49}{50}=\frac{98}{100}$

So if we show that $\int_h^{\frac{k+1}{50}} \frac{k+1}{x(x+1)} d x>\frac{1}{100}$ for each $k$, the $I>\frac{98}{100}=\frac{49}{50}$

Put $x=k+y$. When $x=k, y=0$ the $x=k+1, y=1$

$\int_k^{k+1} \frac{k+1}{x(x+1)} d x =\int_0^1 \frac{k+1}{(k+y)(k+1+y)} d y$

$>\int_0^1 \frac{1}{(k+1+y)} d y \quad\left(\because \frac{k+1}{k+y}>1\right)$

$>\frac{1}{k+2} \geqslant \frac{1}{100} \quad \text { if } 1 \leqslant k \leqslant 98$

$\therefore \quad \int_k^{k+1} \frac{k+1}{x(x+1)} d x >\frac{1}{100}$

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be defined by

$f(x)=[x], x \leq 2$

$0, x>2$

If $I=\int_{-1}^2 \frac{x f\left(x^2\right)}{2+f(x+1)} d x$, then $I=$ ?

$=n$, if $n \leqslant x<n+1$ for any integer $n$

let $g(x)=\frac{x f\left(x^2\right)}{2+f(x+1)}$

Now :

$f\left(x^2\right) = \begin{cases}{[x^2]} & \text { if } x^2 \leq 2 \\0 & \text { if } x^2>2\end{cases}$

$= \begin{cases}0 & \text { if }-1<x<1 \\ 1 & \text { if } 1 \leq x<\sqrt{2} \\ 0 & \text { if } x>\sqrt{2}\end{cases}$

$\text { Also, } f(x+1) =\begin{cases}{[x+1] \qquad \text { if } x+1 \leq 2} \\ 0 \qquad \qquad \text { if } x+1>2\end{cases}$

$= \begin{cases}\frac{x}{2} & \text { if } 1 \leq x<\sqrt{2} \\ 0 & \text { otherwise }\end{cases}$

$\therefore I=\int_{-1}^2 g(x) d x$

$=\int_{-1}^1 g(x) d x+\int_{1}^{\sqrt{2}} g(x) d x +\int_2^2 g(x) d x$

$\begin{aligned}=\int_1^{\sqrt{2}} \frac{x}{2} d x=\left.\frac{x^2}{4}\right|_1 ^{\sqrt{2}} & =\frac{1}{4}(2-1) \& =\frac{1}{4}\end{aligned}$

So, $I=\frac{1}{4}$