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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 /4} {\frac{{\sec x}}{{1 + 2{{\sin }^2}x}}} $ is equal to
  • ✓
    $\frac{1}{3}\left[ {\log (\sqrt 2 + 1) + \frac{\pi }{{2\sqrt 2 }}} \right]$
  • B
    $\frac{1}{3}\left[ {\log (\sqrt 2 + 1) - \frac{\pi }{{2\sqrt 2 }}} \right]$
  • C
    $3\left[ {\log (\sqrt 2 + 1) - \frac{\pi }{{2\sqrt 2 }}} \right]$
  • D
    $3\left[ {\log (\sqrt 2 + 1) + \frac{\pi }{{2\sqrt 2 }}} \right]$
✓

Answer

Correct option: A.
$\frac{1}{3}\left[ {\log (\sqrt 2 + 1) + \frac{\pi }{{2\sqrt 2 }}} \right]$
a
(a) Let $I = \int_0^{\pi /4} {\frac{{\cos x}}{{{{\cos }^2}x(1 + 2{{\sin }^2}x)}}} {\rm{ }}dx$

$ = \int_0^{\pi /4} {\frac{{\cos x\,dx}}{{(1 - {{\sin }^2}x)(1 + 2{{\sin }^2}x)}}} $

$ = \frac{1}{3}\int_0^{1/\sqrt 2 } {\left( {\frac{1}{{1 - {t^2}}} + \frac{2}{{1 + 2{t^2}}}} \right)} \,dt$

By partial fractions, where $t = \sin x$

$ = \frac{1}{3}\left[ {\frac{1}{{2.1}}\log \frac{{1 + t}}{{1 - t}} + \frac{2}{{\sqrt 2 }}{{\tan }^{ - 1}}t\sqrt 2 } \right]_0^{1/\sqrt 2 }$

$ = \frac{1}{3}\left[ {\frac{1}{2}\log \frac{{(\sqrt 2 + 1)}}{{(\sqrt 2 - 1)}} + \sqrt 2 {{\tan }^{ - 1}}1} \right]$

$ = \frac{1}{3}\left[ {\frac{1}{2}\log {{(\sqrt 2 + 1)}^2} + \sqrt 2 .\frac{\pi }{4}} \right]$

$ = \frac{1}{3}\left[ {\log (\sqrt 2 + 1) + \frac{\pi }{{2\sqrt 2 }}} \right]$.

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