x , ,dx 3 + 4 ^2 x = - Vidyadip

MCQ

$\int {\frac{{\sin x\,\,dx}}{{3 + 4{{\cos }^2}x}} = } $

A
$\log (3 + 4{\cos ^2}x) + c$
B
$\frac{{ - 1}}{{2\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{\cos x}}{{\sqrt 3 }}} \right) + c$
✓
$\frac{{ - 1}}{{2\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{2\cos x}}{{\sqrt 3 }}} \right) + c$
D
$\frac{1}{{2\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{2\cos x}}{{\sqrt 3 }}} \right) + c$

✓

Answer

Correct option: C.

$\frac{{ - 1}}{{2\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{2\cos x}}{{\sqrt 3 }}} \right) + c$

c
(c) $I = \int {\frac{{\sin x}}{{3 + 4{{\cos }^2}x}}dx} $
Put $\cos x = t$==> $ - \sin x\,\,dx = dt$
$\therefore \,\,I = \int {\frac{{ - dt}}{{3 + 4{t^2}}}} $$ = \int {dt\over{t^2 + (\sqrt{3\over2})^2}} $
==> $I = - \frac{1}{{4.\frac{{\sqrt 3 }}{2}}}.{\tan ^{ - 1}}\frac{t}{{\left( {\frac{{\sqrt 3 }}{2}} \right)}} + c = \frac{{ - 1}}{{2\sqrt 3 }}{\tan ^{ - 1}}\frac{{2\,t}}{{\sqrt 3 }} + c$
$ \Rightarrow I = \frac{{ - 1}}{{2\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{2\cos x}}{{\sqrt 3 }}} \right) + c$.

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