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

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Answer 1 · 2018-07-23T13:45:07

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

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#