Definite integral of inverse sine using a trigonometric substitution.

We use a trigonometric substitution to compute the definite integral of inverse sine on the interval [0,1].

Given the integral from [0,1] of arcsin(x), we let x=sin(theta) and dx=cos(theta)d(theta) and transform the limits of integration to theta-space. This gives us the classic integration by parts integral theta*cos(theta)d(theta), so we let u=theta and du=cos(theta)d(theta) and apply the integration by parts formula to finish the integral.