Question

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

विकल्प

• \[\frac{ \pi}{4}\]
• \[\frac{\pi}{3}\]
• \[\frac{\pi}{2}\]

•  π

Answer 1

\[\frac{\pi}{4}\]

\[Let\, I = \int_0^\frac{\pi}{2} \frac{1}{1 + \tan x} d x ...............(1)\]
\[ = \int_0^\frac{\pi}{2} \frac{1}{1 + \tan\left( \frac{\pi}{2} - x \right)} d x \]
\[ = \int_0^\frac{\pi}{2} \frac{1}{1 + cot x} d x ................(2)\]
\[\text{Adding (1) and (2) we get}\]
\[2I = \int_0^\frac{\pi}{2} \left[ \frac{1}{1 + \tan x} + \frac{1}{1 + cotx} \right] d x\]
\[ = \int_0^\frac{\pi}{2} \left[ \frac{\left( 1 + cotx \right) + \left( 1 + \tan x \right)}{\left( 1 + \tan x \right)\left( 1 + cotx \right)} \right] d x\]
\[ = \int_0^\frac{\pi}{2} \left[ \frac{2 + \tan x + \cot x}{1 + \tan x + cotx + \tan x \cot x} \right] d x\]
\[ = \int_0^\frac{\pi}{2} \left[ \frac{2 + \tan x + \cot x}{2 + \tan x + \cot x} \right] d x\]
\[ = \int_0^\frac{\pi}{2} dx\]
\[ = \left[ x \right]_0^\frac{\pi}{2} = \frac{\pi}{2}\]
\[Hence\, I = \frac{\pi}{4}\]