Question

$\int {\sec }^{n}x\tan xdx$ is equal to

• A. $\frac{{\sec }^{n}x}{n}+c$
• B. $\frac{{\sec }^{2}x}{n}+c$
• C. $\frac{\tan x}{n}+c$
• D. $\frac{\left({\sec }^{n}x\right)\tan x}{n}+c$

Answer 1

The correct option is A $\frac{{\sec }^{n}x}{n}+c$

$I=\int {\sec }^{n}x\tan xdx$

So we can write as,

$I=\int {\sec }^{n-1}x\sec x\tan xdx$

Let $\sec n=y$

$\sec n\tan ndn=dy$ (diff. wrt n)

$\therefore I=\int y^{n-1}dyI=\frac{y^n}{n}+cI=\frac{{\sec }^{n}x}{n}+c\because y=\sec x$