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.