Question

\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\cot}x} dx\] is

विकल्प

•  π/3

•  π/6

• π/12

• π/2

Answer 1

\[\frac{\pi}{12}\]
\[Let\, I = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{1}{1 + \sqrt{cotx}} d x .............(1)\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{1}{1 + \sqrt{cot\left( \frac{\pi}{3} + \frac{\pi}{6} - x \right)}} dx ..............\left[\text{Using }\int_a^b f\left( x \right) d x = \int_a^b f\left( a + b - x \right) d x \right]\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{1}{1 + \sqrt{\tan x}} d x .................(2)\]
Adding (1) and (2) we get
\[2I = \int_\frac{\pi}{6}^\frac{\pi}{3} \left[ \frac{1}{1 + \sqrt{cotx}} + \frac{1}{1 + \sqrt{\tan x}} \right] d x \]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{2 + \sqrt{cotx} + \sqrt{\tan x}}{\left( 1 + \sqrt{cotx} \right)\left( 1 + \sqrt{\tan x} \right)}dx\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \left[ \frac{2 + \sqrt{cotx} + \sqrt{\tan x}}{2 + \sqrt{cotx} + \sqrt{\tan x}} \right]dx \]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} dx\]
\[ = \left[ x \right]_\frac{\pi}{6}^\frac{\pi}{3} \]
\[ = \frac{\pi}{3} - \frac{\pi}{6}\]
\[ = \frac{\pi}{6}\]
\[\text{Hence, }I = \frac{\pi}{12}\]