Question

Integrate:$\int \left({\tan }^{3}2x\sec 2x\right)dx$

• A. $I=\frac{1}{2}[\frac{{\sec }^{3}2x}{5}-\sec 2x]+C$
• B. $I=\frac{1}{2}[\frac{{\sec }^{3}2x}{3}-\sec 2x]+C$
• C. $I=\frac{1}{2}[\frac{{\sec }^{3}2x}{3}+\sec 2x]+C$
• D. $Noneofthese$

Answer 1

The correct option is B $I=\frac{1}{2}[\frac{{\sec }^{3}2x}{3}-\sec 2x]+C$

Consider the given integral.

$I=\int \left({\tan }^{3}2x\sec 2x\right)dx$

$I=\int \left({\tan }^{2}2x\tan 2x\sec 2x\right)dx$

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

Let $t=\sec 2x$

$\frac{dt}{dx}=\sec 2x\tan 2x\left(2\right)$

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

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

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

Hence, this is the answer.