How to Evaluate the Integral of y/sqrt(1-y^2) using Substitution

[yoleven]

Homework Statement

\int sin^-1y dy

The Attempt at a Solution

u=sin^-1y
du=\frac{1}{\sqrt{1-y^2}}
v=y
dv=dy

\int udv=uv- \int vdu

=ysin^-1y-\int \frac{y}{\sqrt{1-y^2}}

my main trouble is evaluating the integral at this point.
I would appreciate it if someone could show me how to evaluate the integral
\int \frac{y}{\sqrt{1-y^2}}

If it was \frac{1}{y} then I can see it would be ln x

 

[Mark44]

Assuming the work leading to your last integral is correct (I didn't check), you can use an ordinary substitution, u = 1 - y^2, du = -2u du. It's pretty straightforward.