Question

\[\int\limits_0^\pi \frac{x \tan x}{\sec x \ cosec x} dx\]

Answer 1

\[Let I = \int_0^\pi \frac{x \tan x}{secx \cos ecx} d x .............(1)\]
\[ = \int_0^\pi \frac{\left( \pi - x \right) \tan\left( \pi - x \right)}{sec\left( \pi - x \right) \cos ec\left( \pi - x \right)} dx .............\left[\text{Using }\int_0^a f\left( x \right)dx = \int_0^a f\left( a - x \right)dx \right]\]
\[ = \int_0^\pi \frac{- \left( \pi - x \right)\tan x}{- sec\ x \ cosec\ x}dx\]
\[ = \int_0^\pi \frac{\left( \pi - x \right)\tan x}{secx \cos ecx}dx ................(2)\]
\[\text{Adding (1) and (2)}\]
\[2I = \int_0^\pi \frac{x \tan x}{secx \cos ecx} + \frac{\left( \pi - x \right)\tan x}{secx \ cosec\ x} d x\]
\[ = \int_0^\pi \left( x + \pi - x \right)\frac{\tan x}{secx \ cosec\ x}dx\]
\[ = \int_0^\pi \frac{\pi\ tanx}{secx \ cosec\ x}dx\]
\[ = \int_0^\pi \pi\ sin^2 x dx\]
\[ = \pi \int_0^\pi \left( 1 - \cos^2 x \right)dx\]
\[ = \pi \left[ x \right]_0^\pi - \frac{\pi}{2} \int_0^\pi \left( 1 + \cos2x \right) dx\]
\[ = \frac{\pi}{2} \left[ x \right]_0^\pi - \frac{\pi}{2} \left[ \frac{\sin2x}{2} \right]_0^\pi \]
\[ = \frac{\pi^2}{2}\]
\[Hence\, I = \frac{\pi^2}{4}\]