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STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.2 definite integral1 Mark
MCQ
$\int_0^\pi {\frac{{x\,\tan x}}{{\sec x + \cos x}}} \,dx = $
  • ✓
    $\frac{{{\pi ^2}}}{4}$
  • B
    $\frac{{{\pi ^2}}}{2}$
  • C
    $\frac{{3{\pi ^2}}}{2}$
  • D
    $\frac{{{\pi ^2}}}{3}$
✓

Answer

Correct option: A.
$\frac{{{\pi ^2}}}{4}$
a
(a) Let $I  = \int_0^\pi {\frac{{x\tan x}}{{\sec x + \cos x}}dx} = \int_0^\pi {\frac{{(\pi - x)\tan (\pi - x)}}{{\sec (\pi - x) + \cos (\pi - x)}}dx} $

It gives $I = \frac{\pi }{2}\int_0^\pi {\frac{{\sin x}}{{1 + {{\cos }^2}x}}} dx$

Now put $\cos x = t$ and solve, we get 

$I = \frac{\pi }{2} \times \frac{\pi }{2} = \frac{{{\pi ^2}}}{4}$.

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