Integrals — Class 12 Maths Solution

Let $I = \int {x{{\sec }^2}x} dx$
$= x\int {{{\sec }^2}x} dx - \int {\left( {\cfrac{{d\left( x \right)}}{{dx}} \cdot \int {{{\sec }^2}x} dx} \right)} dx$

$= x\left( {\tan x} \right) - \int {\left( 1 \right)\tan xdx} = x\tan x + \log \left( {\cos x} \right) + C$