Question

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

• A. $x\tan x+\log |\sec x|+c$
• B. $x\tan x-\log |\sec x|+c$
• C. $\frac{x}{2}\tan x-\log |\sec x|+c$
• D. $\frac{x}{2}\tan x+\log |\sec x|+c$

Answer 1

The correct option is B $x\tan x-\log |\sec x|+c$

We know, $\int u.vdx=u\int vdx$ $-\int (\frac{du}{dx}\int vdx)dx$

Now by integrating the functions by parts, taking $x$ as $u$ and $se{c}^{2}x$ as $v$.

$\int x.{\sec }^{2}xdx=x\int {\sec }^{2}xdx-\int 1(\int {\sec }^{2}xdx)dx$

$=x\tan x-\int \tan xdx+c$

$=x\tan x-\log |\sec x|+c$