Finding the antiderivative of x² √(9-x²)

[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 a lot, 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.