Question

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

Answer 1

$I=\int \frac{{\tan }^4\sqrt{x}.{\sec }^2\sqrt{x}}{\sqrt{x}}dx\sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}}=dt\Rightarrow dx=2\sqrt{x}dtI=\int {\tan }^{4}t{\sec }^{2}tdt$

Let $\tan t=u,{\sec }^{2}tdt=du$

$I=\int u^{4}du=\frac{u^5}{5}+C=\frac{{\tan }^{5}t}{5}+C=\frac{{\tan }^5\sqrt{x}}{5}+C$