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7.2 definite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.2 definite integral1 Mark
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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