Question

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

Answer 1

$$\begin{gathered} \int x\left(\frac{\sec 2x-1}{\sec 2x+1}\right)dx \\ =\int x\left(\frac{{\displaystyle \frac{1}{\cos 2x}}-1}{{\displaystyle \frac{1}{\cos 2x}}+1}\right)dx \\ =\int x\left(\frac{1-\cos 2x}{1+\cos 2x}\right)dx \\ =\int x\left(\frac{2{\sin }^{2}x}{2{\cos }^{2}x}\right)dx\left[\because \left(1-\cos 2x\right)=2{\sin }^{2}x\mathrm{and}\left(1+\cos 2x\right)=2{\cos }^{2}x\right] \\ =\int x.{\tan }^{2}xdx \\ =\int x.\left({\sec }^{2}x-1\right)dx \\ =\int \underset{\mathrm{I}}{x}.\underset{\mathrm{II}}{{\sec }^{2}x}dx-\int xdx \\ =x\int {\sec }^{2}xdx-\int \left\{\frac{d}{dx}\left(x\right)\int \sec {}^{2}xdx\right\}dx-\frac{x^2}{2}+C_1 \\ =x\tan x-\int 1.\tan xdx-\frac{x^2}{2}+C_1 \\ =x\tan x-\log \left|\sec x\right|-\frac{x^2}{2}+C_{\mathit{2}}+C_1 \\ =x\tan x-\log \left|\sec \mathit{}x\right|-\frac{x^{\mathit{2}}}{2}+C\left(\mathrm{where}\mathit{}C=C_1+C_2\right) \end{gathered}$$