Mathematics - Definite Integration Question with Solution | TestHub

$I={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{dx}{\left(1+e^{x\cos x}\right)\left({\sin }^{4}x+{\cos }^{4}x\right)} \dots \left(1\right)$

Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx$

$I={\int }_{-\pi /4}^{\pi /4}\frac{dx}{\left(1+e^{-x\cos x}\right)\left({\sin }^{4}x+{\cos }^{4}x\right)} ...\left(2\right)$

Add $\left(1\right)$ and $\left(2\right)$

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

$\Rightarrow 2I=2{\int }_0^{\frac{\pi }{4}}\frac{dx}{{\sin }^{4}x+{\cos }^{4}x}$

$\Rightarrow I={\int }_0^{\frac{\pi }{4}}\frac{\left(1+{\tan }^{2}x\right){\sec }^{2}x}{{\tan }^{4}x+1}dx$

$\Rightarrow I={\int }_0^{\frac{\pi }{4}}\frac{\left(1+\frac{1}{{\tan }^{2}x}\right){\sec }^{2}x}{{\left(\tan x-\frac{1}{\tan x}\right)}^2+2}dx$

Put $\tan x-\frac{1}{\tan x}=t$

$\Rightarrow \left(1+\frac{1}{{\tan }^{2}x}\right){\sec }^{2}xdx=dt$

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

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