_0^ x x x + x ,dx = - Vidyadip

MCQ

$\int_0^\pi {\frac{{x\tan x}}{{\sec x + \tan x}}} \,dx = $

A
$\frac{\pi }{2} - 1$
B
$\pi \left( {\frac{\pi }{2} + 1} \right)$
C
$\frac{\pi }{2} + 1$
✓
$\pi \left( {\frac{\pi }{2} - 1} \right)$

✓

Answer

Correct option: D.

$\pi \left( {\frac{\pi }{2} - 1} \right)$

d
(d) $I = \int_0^\pi {\frac{{x\tan x}}{{\sec x + \tan x}}dx = \int_0^\pi {\frac{{(\pi - x)\tan (\pi - x)}}{{\sec (\pi - x) + \tan (\pi - 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} } $

$=\frac{\pi }{2}\left[ {\int_0^\pi {1dx - \int_0^\pi {\frac{{dx}}{{1 + \sin x}}} } } \right]$

On solving, we get $I = \frac{{{\pi ^2}}}{2} - \pi = \pi \left( {\frac{\pi }{2} - 1} \right)$.

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