Need help in evaluating $\int_{-1}^1 (1-x^2)^k, k \in \mathbb{N}$ Can someone tell me how to evaluate this integral please?
$$\int_{-1}^1 (1-x^2)^k, k \in \mathbb{N}$$
I tried using the substitution x = sin(t), which would allow me to express this as:
$$\int_{-1}^1 cos^{2k+1}(t) dt$$
but this doesn't really help. Any other tricks?
 A: Here is a standard route using integration by parts,
$$\begin{align}
I_k&=\int_{-1}^{1}(1-x^2)^kdx
\\\\&=\left[x(1-x^2)^k\right]_{-1}^{1}+2k\int_{-1}^{1}x^2(1-x^2)^{k-1}dx
\\\\&=0+2k\int_{-1}^{1}\left[(1-(1-x^2))(1-x^2)^{k-1}\right]dx
\\\\&=2kI_{k-1}-2kI_{k}
\end{align}
$$ then, with $I_0=2,\,I_1=\frac43,$ one gets
$$
I_{k}=\frac{2k}{2k+1}\cdot I_{k-1}, \quad k\ge1,
$$ finally

$$
I_k=\int_{-1}^{1} (1-x^2)^k dx = \frac{2^{2k+1}(k!)^2}{(2k+1)!}, \quad k\ge1.
$$

A: Enforcing the substitution $x\to x^{1/2}$ reveals
$$\begin{align}
\int_0^1 (1-x^2)^k\,dx&=2\int_0^1(1-x^2)^k\,dx\\\\
&=\int_0^1 x^{-1/2}(1-x)^k\,dx\\\\
&=B(1/2,k+1)\\\\
&=\frac{\Gamma(1/2)\Gamma(k+1)}{\Gamma(k+3/2)}\\\\
&=\frac{\sqrt{\pi}\,k!}{(k+1/2)\Gamma(k+1/2)}\\\\
&=\frac{\sqrt{\pi}\,k!\Gamma(k)}{(k+1/2)2^{1-2k}\sqrt{\pi}\Gamma(2k)}\\\\
&=2\frac{4^k(k!)^2}{(2k+1)!}
\end{align}$$
A: When we expand
$$(1-x^2)^k$$
into a binomial, we get
$$_kC_0-_kC_1x^2+_kC_2x^4-...+_kC_kx^{2k}(-1)^k$$
and when we integrate termwise, we get
$$_kC_0x-\frac{1}{3}{_k}C_1x^3+\frac{1}{5}{_k}C_2x^5-...+\frac{1}{2k}{_k}C_kx^{2k}(-1)^k$$
Now, since we are integrating from $-1$ to $1$, and since the function is symmetric about the $y$-axis, it is the same as twice the integral from $0$ to $1$, which is easily evaluated:
$$2\bigg({_k}C_0-\frac{1}{3}{_k}C_1+\frac{1}{5}{_k}C_2-...+\frac{1}{2k+1}{_k}C_k(-1)^{k}\bigg)$$
And with the help of Wolfram Alpha (thanks, Wolfram!) we find this to be
$$\frac{2(2k)!!}{(2k+1)!!}$$
A: This is simply a Beta function. Using the substitution $t=x^2$, this reduces to
$$2\int_0^1 \frac{1}{2\sqrt{t}} (1-t)^k\ dt$$
$$\int_0^1 t^{-\frac{1}{2}} (1-t)^k$$
$$B\left(\frac{1}{2},k+1\right)$$
which can be expressed in a lot of different ways (see the linked Wikipedia page). 
