Question

The value of the integral $\int\limits_{0}^{\pi / 2} \frac{1}{1+(\tan x)^{101}} d x$ is equal to

• A. $1$
• B. $\frac{\pi}{6}$
• C. $\frac{\pi}{8}$
• D. $\frac{\pi}{4}$

Answer 1

Answer: (D)

$\frac{\pi}{4}$

Answer 2

Let $I =\int\limits_{0}^{\pi / 2} \frac{1}{1+(\tan x)^{101}} d x$
=\int\limits_{0}^{\pi / 2} \frac{d x}{1+\left\\{\tan \left(\frac{\pi}{2}-x\right)\right\\}^{101}}
$=\int\limits_{0}^{\pi / 2} \frac{d x}{1+(\cot x)^{101}}$
$=\int\limits_{0}^{\pi / 2} \frac{\tan x^{101}}{\tan x^{101}+1} d x$
$=\int\limits_{0}^{\pi / 2} \frac{1+\tan x^{101}-1}{1+\tan x^{101}}=[x]_{0}^{\pi / 2}-I$
$\Rightarrow I=\frac{\pi}{2}-I$
$\Rightarrow 2 I=\frac{\pi}{2}$
$\Rightarrow I=\frac{\pi}{4}$