Question

How do you evaluate `int arcsin(x) / x^2 dx` from 1/2 to 1?

Answer 1

Substitution

Explanation:

`I = int_{1/2}^1 (\sin^(-1)(x))/(x^2) dx`
`y = \sin^(-1)(x)`
`x = \sin y`
`dx = \cosy dy`
`I = int_{pi/6}^{pi/2} (y \cosy )/(\sin^2y) dy`
`= int_{pi/6}^{pi/2} (y\coty\cscy dy)`
`= int_{pi/6}^{pi/2} y.d(-csc y)`
Use the integration by parts.

`I= [-y csc y] + \int_{pi/6}^{pi/2} csc y dy`
`= [-(pi/2)+(pi/3)]-ln|csc y + cot y|`
`= ln|2+sqrt(3)| -(pi/6)`