MCQ
$\int_0^{\pi /2} {\frac{{1 + 2\cos x}}{{{{(2 + \cos x)}^2}}} = } $
  • A
    $\frac{\pi }{2}$
  • B
    $\pi $
  • $\frac{1}{2}$
  • D
    None of these

Answer

Correct option: C.
$\frac{1}{2}$
c
(c) $\int_0^{\pi /2} {\frac{{(1 + 2\cos x)}}{{{{(2 + \cos x)}^2}}}dx = \int_0^{\pi /2} {\frac{{2(\cos x + 2) - 3}}{{{{(2 + \cos x)}^2}}}dx} } $

$ = 2\int_0^{\pi /2} {\frac{{dx}}{{2 + \cos x}} - 3\int_0^{\pi /2} {\frac{{dx}}{{{{(2 + \cos x)}^2}}}} } $

$ = 4\int_0^1 {\frac{{dt}}{{3 + {t^2}}} - 6\int_0^1 {\frac{{1 + {t^2}}}{{{{(3 + {t^2})}^2}}}dt} } $,                     $\left[ {{\rm{Put}}\,\,\tan \frac{x}{2} = t} \right]$

$ = - 2\int_0^1 {\frac{{dt}}{{3 + {t^2}}} + 12\int_0^1 {\frac{{dt}}{{{{(3 + {t^2})}^2}}}} } $

$ = - 2\int_0^1 {\frac{{dt}}{{3 + {t^2}}} + 12} \left[ {\frac{1}{6}.\frac{t}{{{t^2} + 3}}} \right]_0^1 + \frac{1}{6}\int_0^1 {\frac{{dt}}{{3 + {t^2}}}} $

$ = 2\left[ {\frac{t}{{{t^2} + 3}}} \right]_0^1 = \frac{1}{2}$.

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