Question

Evaluate :$\int {\sec }^{4}xdx$

• A. $\tan x+\frac{1}{3}{\tan }^{3}x+C$
• B. $\tan x+\frac{1}{3}{\sec }^{3}x+C$
• C. $\tan x+{\tan }^{3}x+C$
• D. None of these

Answer 1

The correct option is A $\tan x+\frac{1}{3}{\tan }^{3}x+C$

$I=\int {\sec }^{4}xdx$

$I=\int {\sec }^{2}x{\sec }^{2}xdx$

$I=\int (1+{\tan }^{2}x){\sec }^{2}xdx$

Putting $\tan x=t$

${\sec }^{2}xdx=dt$

$I=\int (1+t^2)dt=t+\frac{t^3}{3}+C=\tan x+\frac{1}{3}{\tan }^{3}x+C$