Question

\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan x}\]

 

Answer 1

\[\text{Let I }= \int_0^\frac{\pi}{2} \frac{1}{1 + \tan x} dx ................(1)\]

\[ = \int_0^\frac{\pi}{2} \frac{1}{1 + \tan\left( \frac{\pi}{2} - x \right)} dx ...............\left[\text{Using }\int_0^a f\left( x \right) d x = \int_0^a f\left( a - x \right) d x \right]\]

\[ = \int_0^\frac{\pi}{2} \frac{1}{1 + cotx} d x ..............(2)\]

\[\text{Adding (1) and (2)}\]

\[ 2I = \int_0^\frac{\pi}{2} \left( \frac{1}{1 + \tan x} + \frac{1}{1 + cotx} \right) d x\]

\[ = \int_0^\frac{\pi}{2} \left[ \frac{1 + cotx + 1 + \tan x}{\left( 1 + \tan x \right)\left( 1 + cotx \right)} \right] dx\]

\[ = \int_0^\frac{\pi}{2} \frac{2 + \tan x + cotx}{1 + \tan x + cotx + \tan xcotx}dx\]

\[ = \int_0^\frac{\pi}{2} \frac{2 + \tan x + cotx}{2 + \tan x + cotx}dx\]

\[ = \int_0^\frac{\pi}{2} dx\]

\[ = \left[ x \right]_0^\frac{\pi}{2} \]

\[ 2I = \frac{\pi}{2}\]

\[ \therefore I = \frac{\pi}{4}\]