Question

The value of the integral $\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{dx}{\left(1+e^x\right)\left({\sin }^{6}x+{\cos }^{6}x\right)}$ is equal to

• A. $2\pi$
• B. 0
• C. $\pi$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (C)

$I=\underset{-\pi /2}{\overset{0}{\int }}\frac{dx}{\left(1+e^x\right)\left({\sin }^{6}x+{\cos }^{6}x\right)}+\underset{0}{\overset{\pi /2}{\int }}\frac{dx}{\left(1+e^x\right)\left({\sin }^{6}x+{\cos }^{6}x\right)}$
Put $x=-t$
$=\underset{\pi /2}{\overset{0}{\int }}\frac{-dt}{\left(1+e^{-t}\right)\left({\sin }^{6}t+{\cos }^{6}t\right)}+\underset{0}{\overset{\pi /2}{\int }}\frac{dx}{\left(1+e^x\right)\left({\sin }^{6}x+{\cos }^{6}x\right)}$
$=\underset{0}{\overset{\pi /2}{\int }}\frac{\left(e^x+1\right)dx}{\left(1+e^x\right)\left({\sin }^{6}x+{\cos }^{6}x\right)}$
$=\underset{0}{\overset{\pi /2}{\int }}\frac{dx}{\left({\sin }^{2}x+{\cos }^{2}x\right)\left({\sin }^{4}x-{\sin }^{2}x{\cos }^{2}x+{\cos }^{4}x\right)}$
$=\underset{0}{\overset{\pi /2}{\int }}\frac{\left(1+{\tan }^{2}x\right){\sec }^{2}xdx}{\left({\tan }^{4}x-{\tan }^{2}x+1\right)}$
Put $\tan x=t$
$=\underset{0}{\overset{\infty }{\int }}\frac{\left(1+t^2\right)dt}{\left(t^4-t^2+1\right)}$
$=\underset{0}{\overset{\infty }{\int }}\frac{\left(1+\frac{1}{t^2}\right)dt}{t^2-1+\frac{1}{t^2}}=\underset{0}{\overset{\infty }{\int }}\frac{\left(1+\frac{1}{t^2}\right)dt}{{t-\frac{1}{t})}^2+1}$
Put $t-\frac{1}{t}=z$
$\left(1+\frac{1}{t^2}\right)dt=dz$
$=\underset{-\infty }{\overset{\infty }{\int }}\frac{dz}{1+z^2}={\left({\tan }^{-1}z\right)}_{-\infty }^{\infty }$
$=\frac{\pi }{2}-\left(-\frac{\pi }{2}\right)=\pi$