Question

Evaluate $\int x\frac{(\sec 2x-1)}{(\sec 2x+1)}dx$

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

Answer 1

The correct option is B $x\tan x-\log |\sec x|-\frac{x^2}{2}+c$

$\int x\left(\frac{\sec 2x-1}{\sec 2x+1}\right)dx=\int x\left(\frac{1-\cos 2x}{1+\cos 2x}\right)dx$

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

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

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

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

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

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

$=x\tan x-\int \tan xdx-\frac{x^2}{2}+c$

$=x\tan x-\log |\sec x|-\frac{x^2}{2}+c$

$\int x\left(\frac{\sec 2x-1}{\sec 2x+1}\right)dx=x\tan x-\log |\sec x|-\frac{x^2}{2}+c$