Question

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

Answer 1

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

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

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

Putting $\tan x=t$

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

$\therefore I=\int (1+t^2)dt$

$=\int dt+\int t^{2}dt$

$=t+\frac{t^3}{3}+C$

$=\tan x+\frac{1}{3}{\tan }^{2}x+C[\because t=\tan x]$