_ ^ 1 x ^4 x ^2 x ;dx =

MCQ

$\int_{}^{} {\frac{1}{{\sqrt x }}{{\tan }^4}\sqrt x } {\sec ^2}\sqrt x \;dx = $

A
$2{\tan ^5}\sqrt x + c$
B
$\frac{1}{5}{\tan ^5}\sqrt x + c$
✓
$\frac{2}{5}{\tan ^5}\sqrt x + c$
D
None of these

✓

Answer

Correct option: C.

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

c
(c)$\int_{}^{} {\frac{1}{{\sqrt x }}{{\tan }^4}\sqrt x \,.\,{{\sec }^2}\sqrt x \,dx} $
Put $\tan \sqrt x = t \Rightarrow \frac{{{{\sec }^2}\sqrt x }}{{2\sqrt x }}\,dx = dt,$ then it reduces to
$2\int_{}^{} {{t^4}dt} = \frac{2}{5}{(\tan \sqrt x )^5} + c = \frac{2}{5}{\tan ^5}\sqrt x + c$.

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