_ ^ dx (2 x + x) ^2 = - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{{{(2\sin x + \cos x)}^2}}}} = $

A
$\frac{1}{2}\left( {\frac{1}{{2\tan x + 1}}} \right) + c$
B
$\frac{1}{2}\log (2\tan x + 1) + c$
✓
$\frac{1}{{2 + \cot x}} + c$
D
$ - \frac{1}{2}\left( {\frac{1}{{2\tan x - 1}}} \right) + c$

✓

Answer

Correct option: C.

$\frac{1}{{2 + \cot x}} + c$

c
(c) $\int_{}^{} {\frac{{dx}}{{{{(2\sin x + \cos x)}^2}}} = \int_{}^{} {\frac{{dx}}{{{{\sin }^2}x{{(2 + \cot x)}^2}}}} } = \int_{}^{} {\frac{{{\rm{cose}}{{\rm{c}}^{\rm{2}}}x\,dx}}{{{{(2 + \cot x)}^2}}}} $
Put $(2 + \cot x) = t \Rightarrow - {\rm{cose}}{{\rm{c}}^{\rm{2}}}x\,dx = dt$
$ = \int_{}^{} {\frac{{ - dt}}{{{t^2}}}} = \frac{1}{t} + c = \frac{1}{{2 + \cot x}} + c.$

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