What is the integral of inverse sine?

Question

This is a difficult integration problem that will use BOTH integration by parts AND u-substitution.

Answer 1

∫Sin^-1xdx

Let u = Sin^-1x. Then du = 1 / √(1 - x²) dx.

Let dv = dx. Then v = x.

So, using integration by parts, ∫Sin^-1xdx = xSin^-1x - ∫x / √(1 - x²)dx

Now, let w = 1 - x². Then dw = -2xdx. So, xdx = -(1/2)dw.

∫Sin^-1xdx = xSin^-1x + (1/2)∫w^-1/2dw = xSin^-1x + √w + C = xSin^-1x + √(1 - x²) + C