Integration using reduction formulas issue. Show $(2n+1)I_n=2na^2I_{n-1}$ where $I_n=\int_{0}^{a}(a^2-t^2)^ndt$. I have a question where I am given $I_n=\int_{0}^{a}(a^2-t^2)^ndt$ and I am asked to show

$$(2n+1)I_n=2na^2I_{n-1}$$

I then decided to let $u=(a^2-t^2)^n$ and $\dfrac{dv}{dt}=1$. I proceeded to use integration by parts giving:
$$[_{a}^{0}t(a^2-t^2)^n]-(-2n)\int_{0}^{a}t(a^2-t^2)^{n-1}dt$$
I can spot that I have $I_{n-1}$ in the second half of the equation however I have a $t$ which I have no idea how to get rid of,any ideas?
 A: Enforcing the substitution $t=a\sin(x)$, we find that 
$$\begin{align}
I_n&=\int_0^a (a^2-t^2)^n\,dt\\\\
&=a^{2n+1}\int_0^{\pi/2} \cos^{2n+1}(x)\,dx\\\\
&=a^{2n+1}\int_0^{\pi/2}\cos^{2n-1}(x)\,dx-a^{2n+1}\int_0^{\pi/2}\cos^{2n-1}(x)\sin^2(x)\,dx\\\\
&=a^2I_{2n-1}-a^{2n+1}\int_0^{\pi/2}\cos^{2n-1}(x)\sin^2(x)\,dx\\\\
\end{align}$$
Now, integrate by parts with $u=\sin(x)$ and $v=-\frac1{2n} \cos^{2n}(x)$
A: If you will allow for the use of the beta function $\text{B}(m,n)$, where the beta function is defined by
$$\text{B}(m,n) = \int^1_0 x^{m - 1} (1 - x)^{n - 1} \, dx,$$
we can also solve your problem.
Starting with your integral
$$I_n = \int^a_0 (a^2 - t^2)^n \, dt,$$
where I will assume $a > 0$ and $n > - 1$, setting $x = t^2/a^2, dt = a/(2 \sqrt{x}) dx$ while for the limits of integration $(0,a) \mapsto (0,1)$ so that the integral becomes
$$I_n = \frac{a^{2n + 1}}{2} \int^1_0 x^{-1/2} (1 - x)^n \, dx = \frac{a^{2n + 1}}{2} \int^1_0 x^{\frac{1}{2} - 1} (1 - x)^{(n + 1) - 1} \, dx.$$
As the integral $I_n$ is now exactly in the form for the beta function, it can be written as
$$I_n = \frac{a^{2n + 1}}{2} \text{B} \left (\frac{1}{2}, n + 1 \right ).$$
Shifting the index $n \mapsto n - 1$ yields
$$I_{n - 1} = \frac{a^{2n - 1}}{2} \text{B} \left (\frac{1}{2}, n \right ).$$
Making use of the following property for the beta function, namely
$$\text{B}(m,n) = \frac{\Gamma (m) \Gamma (n)}{\Gamma (m + n)},$$
where $\Gamma (x)$ is the gamma function, we have
\begin{align*}
\frac{I_n}{I_{n - 1}} &= \frac{\frac{a^{2n + 1}}{2} \text{B} \left (\frac{1}{2}, n + 1 \right )}{\frac{a^{2n - 1}}{2} \text{B} \left (\frac{1}{2}, n \right )}\\
&= a^2 \cdot \frac{\Gamma \left (\frac{1}{2} \right ) \Gamma (n + 1)}{\Gamma \left (n + \frac{3}{2} \right )} \cdot \frac{\Gamma \left (n + \frac{1}{2} \right )}{\Gamma \left (\frac{1}{2} \right ) \Gamma (n)}.
\end{align*}
And since $\Gamma (x + 1) = x \Gamma (x)$, we can write this as
\begin{align*}
\frac{I_n}{I_{n - 1}} &= a^2 \cdot \frac{n \Gamma (n)}{\left (n + \frac{1}{2} \right ) \Gamma \left (n + \frac{1}{2} \right )} \cdot \frac{\Gamma \left (n + \frac{1}{2} \right )}{\Gamma (n)} = \frac{2n a^2}{(2n + 1)},
\end{align*}
or
$$(2n + 1) I_n = 2n a^2 I_{n - 1},$$
as required.
