Question

SOLVE IT .......ⒸⒶⓃ Ⓤ ⓈⓄⓁⓋⒺ ⒾⓉ.....​

Answer 1

Given,

$\displaystyle\longrightarrow I=\dfrac{2}{\pi}\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\dfrac{1}{\left(1+e^{\sin x}\right)(2-\cos(2x))}\ dx\quad\quad\dots(1)$

We're familiar with the integral identity,

• $\displaystyle\int\limits_a^bf(x)\ dx=\int\limits_a^bf(a+b-x)\ dx$

Here,

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

• $a=-\dfrac{\pi}{4}$

• $b=\dfrac{\pi}{4}$

So,

$\longrightarrow f(a+b-x)=\dfrac{1}{\left(1+e^{\sin\left(-\frac{\pi}{4}+\frac{\pi}{4}-x\right)}\right)\big(2-\cos\left(2\left(-\frac{\pi}{4}+\frac{\pi}{4}-x\right)\right)\big)}$

$\longrightarrow f(a+b-x)=\dfrac{1}{\left(1+e^{\sin\left(-x\right)}\right)\big(2-\cos\left(-2x\right)\big)}$

$\longrightarrow f(a+b-x)=\dfrac{1}{\left(1+e^{-\sin x}\right)\big(2-\cos\left(2x\right)\big)}$

Applying the identity we get,

$\displaystyle\longrightarrow I=\dfrac{2}{\pi}\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\dfrac{1}{\left(1+e^{-\sin x}\right)(2-\cos(2x))}\ dx$

$\displaystyle\longrightarrow I=\dfrac{2}{\pi}\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\dfrac{1}{\left(1+\frac{1}{e^{\sin x}}\right)(2-\cos(2x))}\ dx$

$\displaystyle\longrightarrow I=\dfrac{2}{\pi}\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\dfrac{e^{\sin x}}{\left(1+e^{\sin x}}\right)(2-\cos(2x))}\ dx\quad\quad\dots(2)$

Adding (1) and (2),

$\displaystyle\longrightarrow 2I=\dfrac{2}{\pi}\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\dfrac{1+e^{\sin x}}{\left(1+e^{\sin x}}\right)(2-\cos(2x))}\ dx$

$\displaystyle\longrightarrow I=\dfrac{1}{\pi}\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\dfrac{1}{2-\cos(2x)}\ dx$

Since $\cos(2x)=2\cos^2x-1,$

$\displaystyle\longrightarrow I=\dfrac{1}{\pi}\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\dfrac{1}{2-(2\cos^2x-1)}\ dx$

$\displaystyle\longrightarrow I=\dfrac{1}{\pi}\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\dfrac{1}{3-2\cos^2x}\ dx$

$\displaystyle\longrightarrow I=\dfrac{1}{\pi}\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\dfrac{1}{3-\frac{2}{\sec^2x}}\ dx$

$\displaystyle\longrightarrow I=\dfrac{1}{\pi}\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\dfrac{\sec^2x}{3\sec^2x-2}\ dx$

Since $\sec^2x=1+\tan^2x,$

$\displaystyle\longrightarrow I=\dfrac{1}{\pi}\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\dfrac{\sec^2x}{3(1+\tan^2x)-2}\ dx$

$\displaystyle\longrightarrow I=\dfrac{1}{\pi}\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\dfrac{\sec^2x}{1+3\tan^2x}\ dx$

$\displaystyle\longrightarrow I=\dfrac{1}{\pi\sqrt3}\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\dfrac{\sqrt3\,\sec^2x}{1+3\tan^2x}\ dx\quad\quad\dots(3)$

Let,

$\longrightarrow u=\sqrt3\,\tan x$

$\longrightarrow du=\sqrt3\,\sec^2x\ dx$

At $x=-\dfrac{\pi}{4},$

$\longrightarrow u=\sqrt3\,\tan\left(-\dfrac{\pi}{4}\right)$

$\longrightarrow u=-\sqrt3$

At $x=\dfrac{\pi}{4},$

$\longrightarrow u=\sqrt3\,\tan\left(\dfrac{\pi}{4}\right)$

$\longrightarrow u=\sqrt3$

Then (3) becomes,

$\displaystyle\longrightarrow I=\dfrac{1}{\pi\sqrt3}\int\limits_{-\sqrt3}^{\sqrt3}\dfrac{1}{1+u^2}\ du$

$\displaystyle\longrightarrow I=\dfrac{1}{\pi\sqrt3}\Big[\tan^{-1}u\Big]_{-\sqrt3}^{\sqrt3}$

$\displaystyle\longrightarrow I=\dfrac{\dfrac{\pi}{3}-\left(-\dfrac{\pi}{3}\right)}{\pi\sqrt3}$

$\displaystyle\longrightarrow I=\dfrac{\left(\dfrac{2\pi}{3}\right)}{\pi\sqrt3}$

$\displaystyle\longrightarrow I=\dfrac{2}{3\sqrt3}$

$\displaystyle\longrightarrow I^2=\dfrac{4}{27}$

$\displaystyle\longrightarrow\underline{\underline{27I^2=4}}$

Hence 4 is the answer.