Evaluate the following integrals:
Question:

Evaluate the following integrals:

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

Solution:

Let $\mathrm{I}=\int \mathrm{x}\left(\frac{\sec 2 \mathrm{x}-1}{\sec 2 \mathrm{x}+1}\right) \mathrm{dx}$ it can be written $\mathrm{n}$ terms of $\cos \mathrm{x}$

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

$=\int x\left(\frac{\sec ^{2} x}{\cos ^{2} x}\right) d x$

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

$=\int x\left(\sec ^{2} x-1\right) d x$

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

Using integration by parts,

$=x \int \sec ^{2} x d x-\int \frac{d}{d x} x \int \sec ^{2} x d x-\frac{x^{2}}{2}$

$=x \tan x-\int \tan x d x-\frac{x^{2}}{2}$

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