_ ^ x ^2 x 1 - ^2 x ;dx = - Vidyadip

MCQ

$\int_{}^{} {\tan x} {\sec ^2}x\sqrt {1 - {{\tan }^2}x} \;dx = $

✓
$ - \frac{1}{3}{(1 - {\tan ^2}x)^{3/2}} + c$
B
$\frac{1}{3}{(1 - {\tan ^2}x)^{3/2}} + c$
C
$ - \frac{2}{3}{(1 - {\tan ^2}x)^{2/3}} + c$
D
None of these

✓

Answer

Correct option: A.

$ - \frac{1}{3}{(1 - {\tan ^2}x)^{3/2}} + c$

a
(a)$\int_{}^{} {\tan x\,.\,{{\sec }^2}x\sqrt {1 - {{\tan }^2}x} \,dx} $
Put $\tan x = t \Rightarrow {\sec ^2}x\,dx = dt,$ then it reduces to $\int_{}^{} {t\sqrt {1 - {t^2}} \,dt} $
Now again, put $1 - {t^2} = u,$ then its reduced form is $ - 2tdt = du$
$ - \frac{1}{2}\int_{}^{} {\sqrt u \,du} = - \frac{1}{3}{u^{3/2}} + c = - \frac{1}{3}{(1 - {\tan ^2}x)^{3/2}} + c.$

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