Evaluate the following integrals:
$\int x\left(\frac{\sec 2 x-1}{\sec 2 x+1}\right) d x$
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,
$=x \int \sec ^{2} x d x-\int \frac{d}{d x} x \int \sec ^{2} x d x-\frac{x^{2}}{2}$
$=x \tan x-\int \tan x d x-\frac{x^{2}}{2}$
$=x \tan x-\log |\sec x|-\frac{x^{2}}{2}+c$
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