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

Answer 1

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

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

Put $t=\sec 2x\Rightarrow t^2={\sec }^{2}2x\Rightarrow t^2-1={\sec }^{2}2x-1$

$dt=2.\sec 2x\tan 2xdx$

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

Now,

$y=\int {\tan }^{3}2x\sec 2x.\frac{dt}{2\sec 2x\tan 2x}$

$=\frac{1}{2}\int {\tan }^{2}2xdx$

$=\frac{1}{2}\int ({\sec }^{2}2x-1)dt$

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

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

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

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