Question

Evaluate: $\int {\tan }^{3}2x\sec 2xdx$

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

Answer 1

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

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

putting $2x=t$

$\Rightarrow dx=\frac{dt}{2}$

$=\frac{1}{2}{\int }^{}{\tan }^{3}t\sec tdt$

$=\frac{1}{2}{\int }^{}{\tan }^{2}t\tan t\sec tdt$

$=\frac{1}{2}{\int }^{}\left({\sec }^{2}t-1\right)\tan t\sec tdt$

putting $sect=y$ $\Rightarrow \sec t\tan t=dy$

$=\frac{1}{2}{\int }^{}\left(y^2-1\right)dy$

$=\frac{1}{2}\left(\frac{y^3}{3}-y\right)+c$

$=\frac{1}{6}\left({\sec }^{3}t-3\sec t\right)+c$

$=\frac{1}{6}\left({\sec }^{3}2x-3\sec 2x\right)+c$