Question

The value of the definite integral $\underset{-\pi /4}{\overset{\pi /4}{\int }}\frac{dx}{(1+e^{x\cos x})({\sin }^{4}x+{\cos }^{4}x)}$ is equal to

• A. $-\frac{\pi }{4}$
• B. $\frac{\pi }{2\sqrt{2}}$
• C. $-\frac{\pi }{2}$
• D. $\frac{\pi }{\sqrt{2}}$

Answer 1

The correct option is B $\frac{\pi }{2\sqrt{2}}$

Let $I=\underset{-\pi /4}{\overset{\pi /4}{\int }}\frac{dx}{(1+e^{x\cos x})({\sin }^{4}x+{\cos }^{4}x)}\cdot \left(1\right)$
Applying $\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$
$\therefore I=\underset{-\pi /4}{\overset{\pi /4}{\int }}\frac{dx}{(1+e^{-x\cos x})({\sin }^{4}x+{\cos }^{4}x)}$
$\Rightarrow I=\underset{-\pi /4}{\overset{\pi /4}{\int }}\frac{e^{x\cos x}dx}{(1+e^{x\cos x})({\sin }^{4}x+{\cos }^{4}x)}\cdot \left(2\right)$
Adding $\left(1\right)$ and $\left(2\right),$ we get
$2I=\underset{-\pi /4}{\overset{\pi /4}{\int }}\frac{dx}{({\sin }^{4}x+{\cos }^{4}x)}$
$2I=\underset{-\pi /4}{\overset{\pi /4}{\int }}\frac{{\sec }^{2}x(1+{\tan }^{2}x)dx}{{\tan }^{4}x+1}$
Put $\tan x=t$
$\Rightarrow {\sec }^{2}xdx=dt$
$\therefore 2I=\underset{-1}{\overset{1}{\int }}\left(\frac{1+t^2}{1+t^4}\right)dt$
$\Rightarrow I=\underset{0}{\overset{1}{\int }}\frac{1+t^2}{1+t^4}dt$
$\Rightarrow I=\underset{0}{\overset{1}{\int }}\frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt$
$\Rightarrow I=\underset{0}{\overset{1}{\int }}\frac{1+\frac{1}{t^2}}{{(t-\frac{1}{t})}^2+2}dt$
Put $t-\frac{1}{t}=K$
$\Rightarrow (1+\frac{1}{t^2})dt=dK$

$\therefore I=\underset{-\infty }{\overset{0}{\int }}\frac{dK}{K^2+2}$
$\Rightarrow I=\frac{1}{\sqrt{2}}{\left[{\tan }^{-1}\frac{k}{\sqrt{2}}\right]}_{-\infty }^0$
$\Rightarrow I=\frac{1}{\sqrt{2}}(0+\frac{\pi }{2})$
$\therefore I=\frac{\pi }{2\sqrt{2}}$