Question

${\tan }^{3}2x\sec 2x$

Answer 1

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

$2x=t$

$2.dx=dt$

$\Rightarrow \frac{1}{2}\int {\tan }^{3}t\sec t.dt$

$\sec t=y$

$\sec t\tan t.dt=dy$

${\tan }^{2}t=1-{\sec }^{2}t$

$=1-y^2$

$\Rightarrow \frac{1}{2}\int (1-y^2).dy$

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