Trig substitution for x³/√(1-x²) gives sin³(x)

[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?

 

[Avodyne]

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

 

[rock.freak667]

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

 

[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.

 

[frasifrasi]

anyone?

 

[cristo]

anyone?

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.