# Evaluate this integral

1. Jan 20, 2008

### Vash

1. The problem statement, all variables and given/known data

elvaluate this integral (arcsinx)^2dx

2. Relevant equations

(arcsinx)^2dx

3. The attempt at a solution

integration by parts, let u= arcsinx and make y=arcsinx for easier integration. Once i plug it into the parts equation it turns into a mess. Any help would be superb.

2. Jan 20, 2008

### G01

This should be solvable by using substitution and integration by parts one after the other.

HINT: Let sin u = x.

3. Jan 20, 2008

### Vash

That's what I've been doing and it still doesn't work out for me....oh well, I'll keep trying.

4. Jan 20, 2008

### rocomath

$$\int(\sin^{-1}x)^2dx$$

$$t=\sin^{-1}x$$
$$dt=\frac{dx}{\sqrt{1-x^2}}$$

$$x=\sin t$$
$$dt=\frac{dx}{\sqrt{1-\sin^2 t}}$$

Continue simplifying and see what you can get.

Last edited: Jan 20, 2008
5. Jan 20, 2008

### Vash

im getting xarcsinx+sqrt(1-x^2)

6. Jan 20, 2008

### rocomath

Final answer? Take the derivative and see if you get your Integral.

7. Jan 20, 2008

### Vash

Im not sure how to do it your way so I just set it up by integration by parts. u=arcsinx, du=1/(sqrt(1-x^2)) dv=dx v=x. When I do it i get my answer above.

8. Jan 20, 2008

### rocomath

After simplifying, the Integral becomes ...

$$\int t^2 \cos t dt$$

Last edited: Jan 20, 2008