Question

Find the integrals of the function ${\tan }^{3}2x\sec 2x$

Answer 1

${\tan }^{3}2x\sec 2x={\tan }^{2}2x\tan 2x\sec 2x$
$=({\sec }^{2}2x-1)\tan 2x\sec 2x$
$={\sec }^{2}2x\cdot \tan 2x\sec 2x-\tan 2x\sec 2x$
$\therefore \int {\tan }^{3}2x\sec 2xdx=\int {\sec }^{2}2x\tan 2x\sec 2xdx-\int \tan 2x\sec 2xdx$
$=\int {\sec }^{2}2x\tan 2x\sec 2xdx-\frac{\sec 2x}{2}+C$
Let $\sec 2x=t$
$\therefore 2\sec 2x\tan 2xdx=dt$
$\therefore \int {\tan }^{3}2x\sec 2xdx=\frac{1}{2}\int t^{2}dt-\frac{\sec 2x}{2}+C$
$=\frac{t^3}{6}-\frac{\sec 2x}{2}+C$
$=\frac{(\sec 2x{)}^3}{6}-\frac{\sec 2x}{2}+C$