Question

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

Answer 1

Given that

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

We know that

${\sec }^{2}x=1+{\tan }^{2}x$

Therefore,

$I=\int tan2x({\sec }^{2}2x-1)\sec 2xdx$

$I=\int tan2x\sec 2x({\sec }^{2}2x-1)dx$

Let $t=\sec 2x$

$\frac{dt}{dx}=2\sec 2x\tan 2x$

$\frac{dt}{2}=\sec 2x\tan 2xdx$

Therefore,

$I=\frac{1}{2}\int (t^2-1)dt$

$I=\frac{1}{2}(\frac{t^3}{3}-t)+C$

On putting the value of $t$, we get

$I=\frac{1}{2}(\frac{{\sec }^{3}2x}{3}-\sec 2x)+C$

$I=\frac{{\sec }^{3}2x}{6}-\frac{\sec 2x}{2}+C$

Hence, this is the answer.