Prove $\int_0^a (a^2 - x^2)^n dx = a^{2n+1}\cdot\frac{(2n)!!}{(2n+1)!!}$ Prove 
$$\int_0^a (a^2 - x^2)^n dx = a^{2n+1}\cdot\frac{(2n)!!}{(2n+1)!!}$$  for $ n  \in N $ 
 My idea is to use binomial theorem and integrate partly, but I have not succeeded yet. Are there some hints? 

Note: @Heropup's approach seems to be the most adequate. But you can also succeed by applying the binomial theorem.

We obtain
  \begin{align*}
\int_{0}^a\left(a^2-x^2\right)^n\,dx&=\int_0^a\sum_{k=0}^n\binom{n}{k}\left(-x^2\right)^k\left(a^2\right)^{n-k}\,dx\\
&=\sum_{k=0}^n\binom{n}{k}(-1)^ka^{2n-2k}\int_0^ax^{2k}\,dx\\
&=\sum_{k=0}^n\binom{n}{k}(-1)^ka^{2n-2k}\frac{1}{2k+1}a^{2k+1}\\
&=a^{2n+1}\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{2k+1}\tag{1}
\end{align*}

There is a nice identity called Melzak's formula which can be used to solve (1). 

Melzak's formula: Let $f(x)$ be a polynomial in $x$ of degree $n$, i.e.
  \begin{align*}
f(x)=\sum_{k=0}^na_kx^k
\end{align*}
  Let $y$ be an arbitrary complex number. Melzak's formula states that for $y \ne 0,-1,-2,\ldots,-n$
  \begin{align*}
f(x+y)&=y\binom{y+n}{n}\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{f(x-k)}{y+k}\tag{2}\\
\end{align*}

The whole chapter 7 in H.W.Goulds Combinatorial Identities for Stirling Numbers is devoted to this formula. An easily accessible proof can be found in New proofs of Melzak's identity by U. Abel, H.W. Gould and J. Quaintance.

Setting $f(x)=1$ and $y=\frac{1}{2}$ we obtain from (2)
  \begin{align*}
\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{2k+1}=\frac{1}{\binom{\frac{1}{2}+n}{n}}
\end{align*}
Since
  \begin{align*}
\binom{\frac{1}{2}+n}{n}&=\frac{\left(n+\frac{1}{2}\right)\left(n-\frac{1}{2}\right)\cdots\left(n+\frac{1}{2}-(n-1)\right)}{n!}\\
&=\frac{1}{2^n}\cdot\frac{\left(2n+1\right)\left(2n-1\right)\cdots\left(3\right)}{n!}\\
&=\frac{(2n+1)!!}{2^nn!}\\
&=\frac{(2n+1)!!}{(2n)!!}
\end{align*}
  the claim follows.

