Apply integration by parts to establish some integral formulas

Using integration by parts, prove the validity of the following formula:
$\int (a^2 - x^2)^n \, dx = \frac{x(a^2 - x^2)^n}{2n+1} + \frac{2a^2n}{2n+1} \int (a^2 - x^2)^{n-1} \, dx + C.$
Using part (a) evaluate the definite integral:
$\int_0^a (a^2 - x^2)^{\frac{5}{2}} \, dx.$

Proof. Let
$\begin{align*} u &= (a^2 - x^2)^n & \implies && du &= (-2nx)(a^2 - x^2)^{n-1} \, dx \\ dv &= dx & \implies && v &= x. \end{align*}$

Then,

$\begin{align*} \int (a^2-x^2)^n \, dx &= \int u \, dv \\ &= uv - \int v \, du \\ &= x(a^2-x^2)^n + 2n \int x^2(a^2-x^2)^{n-1} \, dx + C. \end{align*}$

Using the identity $x^2 = x^2 + a^2 - a^2 = -(a^2-x^2) + a^2$ , we have

$\begin{align*} &&\int (a^2-x^2)^n \, dx &= x (a^2 - x^2)^n \\[9pt] &&& \qquad + 2n \left( - \int (a^2-x^2)^n \, dx + a^2 \int (a^2-x^2)^{n-1} \, dx \right) \\[9pt] \implies && (2n+1) \int (a^2-x^2)^n \, dx &= x(a^2-x^2)^n + 2na^2 \int (a^2-x^2)^{n-1} \, dx \\[9pt] \implies && \int (a^2-x^2)^n \, dx &= \frac{x(a^2-x^2)^n}{2n+1} + \frac{2na^2}{2n+1} \int (a^2-x^2)^{n-1} \, dx. \qquad \blacksquare \end{align*}$
Now, using part (a) we can evaluate
$\begin{align*} \int (a^2-x^2)^{\frac{5}{2}} \, dx &= \left. \left( \frac{x(a^2-x^2)^{\frac{5}{2}}}{6} + \frac{5a^2}{6} \int (a^2-x^2)^{\frac{3}{2}} \, dx \right) \right|_0^a \\[9pt] &= \frac{5a^2}{6} \int_0^a (a^2-x^2)^{\frac{3}{2}} \, dx \\[9pt] &= \frac{5a^2}{6} \left( \frac{x(a^2-x^2)^{\frac{3}{2}}}{4} + \frac{3a^2}{4} \int (a^2-x^2)^{\frac{1}{2}} \, dx \right) \Bigg \rvert_0^a \\[9pt] &= \frac{15a^4}{24} \int_0^a (a^2 - x^2)^{\frac{1}{2}} \, dx \\[9pt] &= \left( \frac{5a^4}{8} \right) \left( \frac{1}{2} \right) \int_{-a}^a (a^2-x^2)^{\frac{1}{2}} \, dx \\[9pt] &= \left( \frac{5a^4}{16} \right) \left( \frac{1}{2} \right) 2 \int_{-a}^a (a^2-x^2)^{\frac{1}{2}} \, dx \\[9pt] &= \left( \frac{5a^6}{32} \right) \pi \\[9pt] &= \frac{5\pi}{32} a^6. \end{align*}$