Question

Prove that $\int x\frac{(\sec 2x-1)}{(\sec 2x+1)}dx=$

Answer 1

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

$\Rightarrow \int \frac{\frac{1}{\cos 2x}-1}{\frac{1}{\cos 2x}+1}dx$

$\Rightarrow \int x\frac{1-\cos 2x}{1+\cos 2x}dx$

$\Rightarrow \int x\frac{1-1+2{\sin }^{2}x}{1+2{\cos }^{2}x-1}dx$

$\Rightarrow \int x\frac{{\sin }^{2}x}{{\cos }^{2}x}dx$

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

$\therefore {\tan }^{2}x={\sec }^{2}x-1$

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

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

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

$\int IIIdx=I\int IIdx-\int (\frac{d}{dx}I\cdot \int IIdx)dx$

$\Rightarrow x\int {\sec }^{2}xdx-\int \left(\frac{dx}{dx}\int {\sec }^{2}xdx\right)dx-\int xdx$

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

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

Hence, this is the answer.