Integrate polymonial of any degree

[hellbike]

how to integrate $$(1-x^2)^n$$ for $$n \in N$$ ? Limits of integral are from 0 to 1, but i don't think that matter.

(i tried to use latex for int, but it wasn't working).

 

[Cyosis]

Use the binomial expansion.

 

[hellbike]

$$\int_{0}^{1} (1-x^2)^n dx$$

using binomial expansion:

$$\int \sum_{k=0}^{n}( {n \choose k} (-x^2)^k) dx$$ = $$\sum_{k=0}^{n}( {n \choose k} \frac{x (-x^2)^k)}{2 k+1})$$

and going to definite integral:

$$\int_{0}^{1} (1-x^2)^n dx = \sum_{k=0}^{n}( {n \choose k} \frac{(-1)^k)}{2 k+1})$$

is this correct? Can anything else be done there?

And either "preview post" or latex is not working correctly on this forum.

How to solve this using integration by parts?

@edit
ok, nvm, done already.

 

[Gib Z]

That's correct. It can be written as $$\frac{\sqrt{\pi} \Gamma(1+n)}{2\Gamma(\frac{3+2n}{2})}$$ but that might be too advanced.

 

[l'Hôpital]

$$c_n = \int_{0}^{1} (1 - x^2)^{\frac{n-1}{n}} dx$$

Therefore,

$$c_{2n+1} = \frac{n-1}{2n} C_{2n-3}$$

I'm sure if you were to solve that explicitly, you would get what Gib Z got.