Integrate \(\int_{0}^{a} (a^{2}-x^{2})^{n-1/2} \mathrm{d} x\)

From Victor Moll’s attempt to solve all of the integrals in Gradshteyn and Ryzhik, we have

\begin{equation}
\int\limits_{0}^{a} (a^{2}-x^{2})^{n-1/2} \mathrm{d} x = a^{2n} \frac{(2n-1)!!}{(2n)!!} \frac{\pi}{2}
\label{eq:1608041}
\tag{1}
\end{equation}

Moll’s solution is at the link, so let us solve this another way. We first use the substitution used by Moll, let \(x=av\) to obtain

\begin{equation}
\int\limits_{0}^{a} (a^{2}-x^{2})^{n-1/2} \mathrm{d} x = a^{2n} \int\limits_{0}^{1} (1-v^{2})^{n-1/2} \mathrm{d} v
\label{eq:1608042}
\tag{2}
\end{equation}

We designate the integral on the right hand side as I and make the substitution \(z = v^{2}\)

\begin{equation}
\mathrm{I} = \int\limits_{0}^{1} (1-v^{2})^{n-1/2} \mathrm{d} v = \frac{1}{2} \int\limits_{0}^{1} z^{-1/2} (1-z)^{n-1/2} \mathrm{d} z
\label{eq:1608043}
\tag{3}
\end{equation}

From the beginning this integral looked like a gamma or beta function, now, in this form it is obviously a beta function. We then have

\begin{equation}
\mathrm{I} = \frac{1}{2} \mathrm{B}\left(\frac{1}{2}, n + \frac{1}{2} \right)
\label{eq:1608044}
\tag{4}
\end{equation}

and thus

\begin{equation}
\int\limits_{0}^{a} (a^{2}-x^{2})^{n-1/2} \mathrm{d} x = a^{2n} \frac{1}{2} \mathrm{B}\left(\frac{1}{2}, n + \frac{1}{2} \right)
\label{eq:1608045}
\tag{5}
\end{equation}

Now we must show that the right hand sides of equations \eqref{eq:1608041} and \eqref{eq:1608045} are equal or

\begin{equation}
\mathrm{B}\left(\frac{1}{2}, n + \frac{1}{2} \right) \overset{\underset{\mathrm{?}}{}}{=} \pi \frac{(2n-1)!!}{(2n)!!}
\label{eq:1608046}
\tag{6}
\end{equation}

We begin by expressing the beta function in terms of gamma functions

\begin{equation}
\mathrm{B}\left(\frac{1}{2}, n + \frac{1}{2} \right) = \frac{\Gamma(\frac{1}{2}) \Gamma(n + \frac{1}{2})}{\Gamma(n+1)} = \sqrt{\pi}
\frac{\Gamma(n + \frac{1}{2})}{\Gamma(n+1)}
\label{eq:1608047}
\tag{7}
\end{equation}

Now we invoke the following two relationships between the gamma function and the double factorial, found at Wikipedia and Wolfram Math World respectively

\begin{equation}
(2n-1)!! = \frac{2^{n} \Gamma(n + \frac{1}{2})}{\sqrt{\pi}} \quad \mathrm{and} \quad (2n)!! = 2^{n}n! = 2^{n} \Gamma(n+1)
\end{equation}

Substituting these expressions into \eqref{eq:1608046} shows that it is indeed an equality.

Putting everything together, we have
\begin{equation}
\int\limits_{0}^{a} (a^{2}-x^{2})^{n-1/2} \mathrm{d} x = a^{2n} \frac{(2n-1)!!}{(2n)!!} \frac{\pi}{2} = a^{2n} \frac{1}{2} \mathrm{B}\left(\frac{1}{2}, n + \frac{1}{2} \right)
\end{equation}