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

1 Answer
Feb 27, 2017

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)#