Question

Integrate $\int x{\sec }^{2}xdx$

Answer 1

Consider given the given intigration,

Let,

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

$\because \int u.vdx=u\int vdx-\int (\frac{du}{dx}\int vdx)dx$

$\therefore \int x.{\sec }^{2}xdx=x\int {\sec }^{2}xdx-\int (\frac{dx}{dx}\int {\sec }^{2}x)dx$

$=x\tan x-\int 1.\tan xdx$

$=x\tan x-\log \sec x+C$

Hence, this is the answer.