antiderivative help

[SeReNiTy]

can anyone help me find an antiderivative for (x^2)((9-(x^2))^(1/2))

[HallsofIvy]

When you see \sqrt{1- x^2} or anything like that, you should think cos(x)=\sqrt{1- sin^2(x)}- and use a trig substitution.

In this problem, factor a "9" out of the squareroot to get 3x^2\sqrt{1- \frac{x^2}{9}}. Now make the substitution x= 3sin(θ).

dx= 3cos(θ)dθ and \sqrt{1- \frac{x^2}{9}} becomes \sqrt{1- sin^2(\theta)}= cos(\theta). The entire integrand becomes sin2(θ)cos2(θ)dθ. You will need to use trig substitutions to integrate that.

[phoenixthoth]

I think you may also be able to do it by parts if you let u=x and dv=the rest.

[SeReNiTy]

When you see \sqrt{1- x^2} or anything like that, you should think cos(x)=\sqrt{1- sin^2(x)}- and use a trig substitution.

In this problem, factor a "9" out of the squareroot to get 3x^2\sqrt{1- \frac{x^2}{9}}. Now make the substitution x= 3sin(θ).

dx= 3cos(θ)dθ and \sqrt{1- \frac{x^2}{9}} becomes \sqrt{1- sin^2(\theta)}= cos(\theta). The entire integrand becomes sin2(θ)cos2(θ)dθ. You will need to use trig substitutions to integrate that.

Thanks alot, i was really just wondering if it was possible to do with the scope of my year 12 specialist maths course, and it seems as i can't. As that substitution i am not familiar with and intergration by parts is not on the course, is there another way with only linear substitution, substitution by "u" or partial fractions?

[Zurtex]

There are a few different ways, you could use the + (a - a) method, by-parts and substitution. But at some point in all of them you are either going to have to make a trigonometric substitution or put it into standard form (which is also basically using a trig substitution but without the effort of workings). Quite simply because the anti derivative has inverse sine in it.