Question

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

Answer 1

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

Let $u=\sec x$

then $du=\sec x\tan xdx$

So,

$\int {\sec }^{2}x\tan xdx=\int udu$

$=\frac{u^2}{2}$

Substituting value of u we get

$=\frac{{\sec }^{2}x}{2}+C$