Question

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

• A. $\frac{{\sec }^{3}x}{3}+c$
• B. $-\frac{{\sec }^{3}x}{3}+c$
• C. $\frac{{\sec }^{4}x}{11}+c$
• D. $\frac{{\sec }^{4}x}{4}+c$

Answer 1

The correct option is A $\frac{{\sec }^{3}x}{3}+c$

$\int {\sec }^{2}x\left(\sec x\tan x\right)dx$

Let $\sec x=t\Rightarrow d\left(\sec x\right)=\sec x.\tan x.dx=dt$

$\int {\sec }^{2}x\left(\sec x\tan x\right)dx=\int t^{2}dt$

We know that, $\int x^n.dx=\frac{x^{n+1}}{n+1}$

$=\frac{t^{2+1}}{2+1}=\frac{t^3}{3}+1$

$=\frac{{\sec }^{3}x}{3}+c$ where $t=\sec x$

$\int {\sec }^{2}x(\sec x.\tan x)=\frac{{\sec }^{3}x}{3}+c$