How to Integrate this function $\int(1-x^2)^ndx$ The actual question was to find the area bounded by the curve $y=\int_{-1}^{1}(1-x^2)^ndx$ and coordinate axes. But I haven't came across these type of problems with power $n$.
 A: Using the binomial theorem, $(1-x^2)^n=\sum_{k=0}^n\binom{n}{k}(-1)^kx^{2k}$ and therefore
$$\int (1-x^2)^n\,dx=\sum_{k=0}^n\binom{n}{k}(-1)^k\frac{x^{2k+1}}{2k+1}+C$$
and
$$\int_{-1}^1 (1-x^2)^n\,dx=\sum_{k=0}^n\binom{n}{k}(-1)^k\frac{2}{2k+1}$$

ALTERNATIVE APPROACH TO EVALUATE THE DEFINITE INTEGRAL:
First, note that $\int_{-1}^1 (1-x^2)^n\,dx=2\int_0^1 (1-x^2)^n\,dx$.  Next, enforce the substitution $x\to \sqrt{x}$ to obtain
$$\begin{align}
\int_{-1}^1 (1-x^2)^n\,dx&=2\int_0^1 (1-x)^n x^{-1/2}\,dx\\\\
&=2\,B(n+1,1/2)\\\\
&=2\frac{\Gamma(n+1)\Gamma(1/2)}{\Gamma(n+3/2)}\\\\
&=2\frac{n!\Gamma(1/2)}{(n+1/2)!\Gamma(1/2)}\\\\
&=2\frac{n!}{\frac32\,\frac52\,\frac72\,\cdots\,\frac{2n+1}{2}}\\\\
&=2\frac{2^n\,n!}{(2n+1)!!}\\\\
&=2\frac{(2n)!!}{(2n+1)!!}
\end{align}$$
as expected!
A: By parts,
$$I_n=\int(1-x^2)^ndx=x(1-x^2)^n+2n\int x^2(1-x^2)^{n-1}dx.$$
But $x^2=1-(1-x^2)$, so that
$$I_n=x(1-x^2)^n+2nI_{n-1}-2nI_n,$$ and
$$I_n=\frac{x(1-x^2)^n}{2n+1}+\frac{2n}{2n+1}I_{n-1}.$$
With $I_0=x$, you can compute for any $n$.

When evaluating the definite integral, the first term vanishes and
$$I_n=\frac{2(2n)!!}{(2n+1)!!}.$$
