Question

$\int \frac{{\tan }^4\sqrt{x}{\sec }^2\sqrt{x}}{\sqrt{x}}dx$

Answer 1

$I=\int \frac{{\tan }^4\sqrt{x}.{\sec }^2\sqrt{x}}{\sqrt{x}}dx$

Let $\tan \sqrt{x}=t$

${\sec }^2\sqrt{x}.\frac{1}{2\sqrt{x}}dx=dt$

$\Rightarrow \frac{{\sec }^2\sqrt{x}}{\sqrt{x}}dx=edt$

$\therefore I=\int t^4.2dt=2\int t^{4}dt$

$=\frac{2}{5}{t}^5$, substituting $t=\tan \sqrt{x}$

We get $I=\frac{2}{5}{\tan }^5\sqrt{x}+C$