Question

The value of the integral ${\int }_0^{\pi /2}\frac{1}{1+(\tan x{)}^{101}}dx$ is equal to

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

Answer 1

Answer: (D)

$\frac{\pi }{4}$

Let

$I={\int }_0^{\frac{\pi }{2}}\frac{1}{1+(\tan x{)}^{101}}dx$

$={\int }_0^{\frac{\pi }{2}}\frac{dx}{1+{\left\{\tan (\frac{x}{2}-x)\right\}}^{101}}$

$={\int }_0^{\frac{\pi }{2}}\frac{dx}{1+(\cot x{)}^{101}}$

$={\int }_0^{\frac{\pi }{2}}\frac{\tan x^{101}}{\tan x^{101}+1}dx$

$={\int }_0^{\frac{\pi }{2}}\frac{1+\tan x^{101}-1}{1+\tan x^{101}}=[x{]}_0^{\frac{x}{2}}-I$

$\Rightarrow I=\frac{\pi }{2}-I$

$\Rightarrow 2I=\frac{\pi }{2}$

$\Rightarrow I=\frac{\pi }{4}$