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?

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 $v=-\frac1{2n} \cos^{2n}(x)$
