dx ^3 x2 2x is equal to - Vidyadip

MCQ

$\int {\frac{{dx}}{{{{\cos }^3}x\sqrt {2\sin 2x} }}} $ is equal to

✓
$\sqrt {\tan x} \, + \frac{{{{\tan }^{5/2}}x}}{5} + c$
B
$\sqrt {\tan x} \, + \frac{2}{5}{\tan ^{5/2}}x + c$
C
$2\sqrt {\tan x} \, + \frac{2}{5}{\tan ^{5/2}}x + c$
D
None of these

✓

Answer

Correct option: A.

$\sqrt {\tan x} \, + \frac{{{{\tan }^{5/2}}x}}{5} + c$

a
(a) $\int {\frac{{dx}}{{{{\cos }^3}x\sqrt {2\sin 2x} }}} $ =$\int {\frac{{dx}}{{{{\cos }^3}x\sqrt {4\sin x\cos x} }}} $
$ = \frac{1}{2}\int {\frac{{dx}}{{{{\cos }^{7/2}}x\,\,{{\sin }^{1/2}}x}}} $
$ = \frac{1}{2}\int {\frac{{{{\sec }^4}x}}{{\sqrt {\tan x} }}dx = \frac{1}{2}\int {\frac{{(1 + {{\tan }^2}x){{\sec }^2}x}}{{\sqrt {\tan x} }}dx} } $
$ = \frac{1}{2}\int {\frac{{1 + {t^2}}}{{\sqrt t }}dt} $ (Put $\tan x = t$, $\therefore {\sec ^2}x{\rm{ }}dx = dt$)
$ = \frac{1}{2}\int {{t^{ - 1/2}}dt + \frac{1}{2}\int {{t^{3/2}}dt} } $$ = {t^{1/2}} + \frac{{{t^{5/2}}}}{5} + c$
$ = \sqrt {\tan x} + \frac{1}{5}{\tan ^{5/2}}x + c$.

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