Question

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

Answer 1

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

$\Rightarrow \sec x\tan xdx=dt$
$\therefore$ $I=\int t^{n-1}dt$
$=\frac{t^n}{n}+c$
$=\frac{{\sec }^{n}x}{n}+c$ .... $($ Resubstituting $t=\sec x)$