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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
The value of $\int_{\pi /4}^{3\pi /4} {\frac{\phi  }{{1 + \sin \phi  }}\,d\phi   ,} $ is
  • ✓
    $\pi \tan \frac{\pi }{8}$
  • B
    $\log \tan \frac{\pi }{8}$
  • C
    $\tan \frac{\pi }{8}$
  • D
    None of these
✓

Answer

Correct option: A.
$\pi \tan \frac{\pi }{8}$
a
(a) $I = \int_{\pi /4}^{3\pi /4} {\frac{\phi }{{1 + \sin \phi }}d\phi } = \int_{\pi /4}^{3\pi /4} {\frac{{\pi - \phi  }}{{1 + \sin (\pi - \phi )}}d\phi } $

$\left\{ \because \frac{\pi }{4}+\frac{3\pi }{4}=\pi  \right\}$

==> $2I = \int_{\pi /4}^{3\pi /4} {\frac{\pi }{{1 + \sin \phi }}d\phi } $

On simplification, we get 

$I = \pi (\sqrt 2 - 1) = \pi \tan \frac{\pi }{8}.$

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