Question
$\int_0^\pi {\frac{{x\tan x}}{{\sec x + \tan x}}} \,dx = $
==> $2I = \frac{\pi }{2}\int_0^\pi {\frac{{\tan x}}{{\sec x + \tan x}}dx = \frac{\pi }{2}\int_0^\pi {\frac{{\sin x}}{{1 + \sin x}}dx} } $
$=\frac{\pi }{2}\left[ {\int_0^\pi {1dx - \int_0^\pi {\frac{{dx}}{{1 + \sin x}}} } } \right]$
सरल करने पर, $I = \frac{{{\pi ^2}}}{2} - \pi = \pi \left( {\frac{\pi }{2} - 1} \right)$.
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