# The value of the definite integral

Question:

The value of the definite integral

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\left(1+e^{x \cos x}\right)\left(\sin ^{4} x+\cos ^{4} x\right)}$

is equal to :

1. $-\frac{\pi}{2}$

2. $\frac{\pi}{2 \sqrt{2}}$

3. $-\frac{\pi}{4}$

4. $\frac{\pi}{\sqrt{2}}$

Correct Option: , 2

Solution:

$\mathrm{I}=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\mathrm{dx}}{\left(1+\mathrm{e}^{\mathrm{x} \cos \mathrm{x}}\right)\left(\sin ^{4} \mathrm{x}+\cos ^{4} \mathrm{x}\right)}$      .....(1)

Using $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$

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

$2 \mathrm{I}=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\mathrm{dx}}{\sin ^{4} \mathrm{x}+\cos ^{4} \mathrm{x}}$

$2 I=2 \int_{0}^{\frac{\pi}{4}} \frac{d x}{\sin ^{4} x+\cos ^{4} x}$

$I=\int_{0}^{\frac{\pi}{4}} \frac{\left(1+\tan ^{2} x\right) \sec ^{2} x}{\tan ^{4} x+1} d x$

$I=\int_{0}^{\frac{\pi}{4}} \frac{\left(1+\frac{1}{\tan ^{2} x}\right) \sec ^{2} x}{\left(\tan x-\frac{1}{\tan x}\right)^{2}+2} d x$

$\tan x-\frac{1}{\tan x}=t$

$\left(1+\frac{1}{\tan ^{2} x}\right) \sec ^{2} x d x=d t$

$\mathrm{I}=\int_{-\infty}^{0} \frac{\mathrm{dt}}{\mathrm{t}^{2}+2}=\left[\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\mathrm{t}}{\sqrt{2}}\right)\right]_{-\infty}^{0}$

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