Question

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

Answer 1

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

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

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

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

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

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

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

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

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

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

\[Hence\ , I = \frac{\pi}{4}\]