Did I Solve the Integral x^2√(1-x^2) dx Correctly?

[Fabio010]

People, today i had a exam in math analysis and there was a integral to solve:

∫ x^2√(1-x^2) dx

ok, i started to think about the trigonometric substitution. x= sint

but, with that substitution now i have a ∫sin^2tcos^2t dt

so i have to do something like ∫(1-cos^2t)(cos^2t) dt ok and i thought (no thanks...)
i never learned how to solve a integral with the trigonometric formula, so solve something like
∫cos^4t dt takes a lot of time.


So i tried t = √(1-x^2)

dt/dx = -x/(√(1-x^2) )

So now i have a integral

-∫(x^2*t*√(1-x^2))/(x) dt
-∫(x^2*t*t)/(x) dt
-∫(x*t^2) dt

as we know t = √(1-x^2) so x= 1-t^2

-∫((1-t^2)t^2) dt = -∫t^2 - t^4 dt

ok now it is easy...

Please tell me that i did it in the correct way!

 

[CompuChip]

You can easily check by differentiating, however I think that you'll find you're off by a factor of x then.

The problem seems to be

as we know t = √(1-x^2) so x= 1-t^2

if t² = 1 - x² then x² = 1 - t².
I don't know how much that helps you though.

 

[Fabio010]

omfg...

what a stupid error.

Damn. ok i should do it by trigonometric substitution.

 

[CompuChip]

I just made the observation that
x^2 \sqrt{1 - x^2} \propto x \cdot \frac{d}{dx} (1 - x^2)^{3/2}
so maybe you can try partial integration.