# For this trig substitution

1. Dec 15, 2007

### frasifrasi

For the integral $$\int frac{x^3}{sqrt{1-x^2}} dx}$$
==> okay...

what I meant was:

int of x^3 over sqrt(1-x^2)

--I trig substitute to get sin^(3)(x)cosxdx over cos x

and end up with sin^3(x)...this is obviously wrong, can anyone point out what i did wrong?

2. Dec 15, 2007

### Avodyne

It seems right to me. Now you can try u=cos(x).

3. Dec 15, 2007

### rock.freak667

Why is that wrong for? Although you should use another variable and not use x again

4. Dec 15, 2007

### frasifrasi

Ok, i did the substitution and got
-sqrt(1-x^2) + 1/3*[sqrt(1-x^2)]^3

But the answer key has -sqrt(1-x^2) + 1/3*(1-x^2)^3
--> does anyone know why?

Thank you so much.

5. Dec 16, 2007

### frasifrasi

anyone?

6. Dec 16, 2007

### cristo

Staff Emeritus
Perhaps you should show your steps, instead of basically saying "this is my answer, can someone do it themselves and see if they get the same."

If you spend a little time to show your working, it is more likely that someone will be willing to spend their time helping you.