Question

How do you integrate `int 1/(xsqrt(4x^2-1) )dx` using trigonometric substitution?

Answer 1

`\color{red}{\int \frac{1}{x\sqrt{4x^2-1}}\ dx}=\color{blue}{\sec^{-1}(2x)+C`

Explanation:

Given integral

`\int \frac{1}{x\sqrt{4x^2-1}}\ dx`

`=2\int \frac{1}{2x\sqrt{(2x)^2-1}}\ dx`

Let `2x=\sec t\implies 2dx=\sec t\tan t\ dt\ \ \or \ \ dx=\sect \tan t\ dt/2`

`=2\int \frac{1}{\sec t\sqrt{sec^2t-1}}\ sec t\tan t\ dt/2`

`=\int \frac{\sec t\tan t}{\sec t\tan t}dt`

`=\int dt`

`=t+C`

`=\sec^{-1}(2x)+C`