Question

The value of $\int {\tan }^{3}2x\sec 2xdx$ is equal to :

• A. $\frac{1}{3}{\sec }^{3}2x-\frac{1}{2}\sec 2x+C$
• B. $-\frac{1}{6}{\sec }^{3}2x-\frac{1}{2}\sec 2x+C$
• C. $\frac{1}{6}{\sec }^{3}2x-\frac{1}{2}\sec 2x+C$
• D. $\frac{1}{3}{\sec }^{3}2x+\frac{1}{2}\sec 2x+C$

Answer 1

The correct option is C $\frac{1}{6}{\sec }^{3}2x-\frac{1}{2}\sec 2x+C$

We know that

$\text{sec}x=\int \tan x\text{sec}xdx,{\tan }^{2}x+1={\text{sec}}^{2}x$

Using these

$\int ({\text{sec}}^{2}2x-1)\tan 2x\text{sec}2xdx$

$\frac{1}{6}{\text{sec}}^{3}2x-\frac{1}{2}\text{sec}2x+C$