_ ^ 2xdx 1 + ^2 x =

MCQ

$\int_{}^{} {\frac{{\sin 2xdx}}{{1 + {{\cos }^2}x}}} = $

A
$\frac{1}{2}\log (1 + {\cos ^2}x) + c$
B
$2\log (1 + {\cos ^2}x) + c$
C
$\frac{1}{2}\log (1 + \cos 2x) + c$
✓
$ - \log (1 + {\cos ^2}x) + c$

✓

Answer

Correct option: D.

$ - \log (1 + {\cos ^2}x) + c$

d
(d) $I = \int {\frac{{\sin 2x}}{{1 + {{\cos }^2}x}}dx = \int {\frac{{2\sin x\cos x}}{{1 + {{\cos }^2}x}}dx} } $
Put $1 + {\cos ^2}x = t$ ==> $ - 2\sin x\cos x\,\,dx = dt$
==> $\sin 2x = - dt$. Hence

$I = \int {^ - \left( {\frac{{dt}}{t}} \right)} = - \log t + c$
$ = - \log (1 + {\cos ^2}x) + c$.

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