Question

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

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

Answer 1

Answer: (C)

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