Question

Evaluate:$\int {\tan }^{3}2x\sec 2xdx$

Answer 1

$I=\int {\tan }^{3}2x\sec 2xdx$

$I=\int {\tan }^{2}2x\tan 2x\sec 2x.\frac{1}{2}d\left(2x\right)$

$I=\frac{1}{2}\int ({\sec }^{2}2x-1)\tan 2x\sec 2xd\left(2x\right)$

$u=\sec 2x$

$du=\sec 2x\tan 2xd\left(2x\right)$

$=\frac{1}{2}\int (u^2-1)du$

$=\frac{1}{2}(\frac{u^3}{3}-u)=\frac{1}{2}(\frac{{\sec }^{3}2x}{3}-\sec 2x)$