Question

∫x (sec2x-1)/ (sec2x+1) dx

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{I}=\int x\left[\frac{\sec 2x-1}{\sec 2x+1}\right]dx \\ =\int x\left(\frac{1-\cos 2x}{1+\cos 2x}\right)dx \\ =\int x\left(\frac{2{\sin }^{2}x}{2{\cos }^{2}x}\right)dx \\ =\int x{\tan }^{2}xdx \\ =\int x\left({\sec }^{2}x-1\right)dx \\ =\int x.{\sec }^{2}xdx-\int xdx \\ =x\int {\sec }^{2}xdx-\int \left[\frac{d}{dx}\left(x\right)\int {\sec }^{2}xdx\right]dx+\mathrm{C} \\ =x.\tan x-\int \tan xdx-\frac{x^2}{2}+\mathrm{C} \\ =x\tan x+\log \left|\cos x\right|-\frac{x^2}{2}+\mathrm{C} \end{gathered}$$