Show that $$\int(a^2-x^2)^n dx=\dfrac{x(a^2-x^2)^n}{2n+1}+\dfrac{2na^2}{2n+1}\int(a^2-x^2)^{n-1}dx$$
My try (integrating by parts): $$\int(a^2-x^2)^n dx=x(a^2-x^2)^n-\int xn(a^2-x^2)^{n-1}(-2x)dx\\ =x(a^2-x^2)^n-\dfrac{n}{n}\int nx(a^2-x^2)^{n-1}d(a^2-x^2)\\ =x(a^2-x^2)^n-\int xd(a^2-x^2)^n=x(a^2-x^2)^n-x(a^2-x^2)^n+\int (a^2-x^2)^n dx,$$ so we basically arrive at $0=0$, which is true, but not very useful. Any ideas?
Some thoughts that may (or may not) be useful. In order to get the denominator in the RHS which is $2n+1$, we actually have to find an expression for $I_n=\int (a^2-x^2)^n dx$ in which we must have $I_n$, namely $(-2nI_n)$ (as $I_n+2nI_n=(1+2n)I_n$), and $I_{n-1}$ as well.
P.S. I am not yet familiar with the approach that uses substitution for solving integrals, so I'd rather integrate by parts on this occasion (as I think it would probably work).
$\begin{align}I_n=\int(a^2-x^2)^n dx &=x(a^2-x^2)^n-\int xn(a^2-x^2)^{n-1}(-2x)dx\\ &=x(a^2-x^2)^n-2n\int (a^2-x^2)^{n-1}(a^2-x^2-a^2)\\ &=x(a^2-x^2)^n-2n I_n+2na^2 I_{n-1} \end{align}$
Collect $I_n$'s together and finish.