Definite Integral of $(a^2-x^2)^\frac{3}{2}$ 
Prove the following:
  $$\frac{4c}3\int\limits_0^a\left(a^2 - x^2\right)^\frac32\,\mathrm dx = \frac{\pi a^4c}4.$$

Taking $x = a\sin\theta$, how will the limit change? 

$$\dfrac{d(x(a^2-x^2)^n)}{dx}=(a^2-x^2)^n+nx(a^2-x^2)^{n-1}(-2x)$$
$$=(a^2-x^2)^n+2n(a^2-x^2-a^2)(a^2-x^2)^{n-1}$$
$$=(1+2n)(a^2-x^2)^n-2na^2(a^2-x^2)^{n-1}$$
Integrate both sides with respect to $x$ and writing $\displaystyle I_m=\int(a^2-x^2)^mdx$
$$(1+2n)I_n-2na^2I_{n-1}=x(a^2-x^2)^n$$
Now if $\displaystyle J_m=\int_0^a(a^2-x^2)^mdx$
$$(1+2n)J_n-2na^2J_{n-1}=0$$
Set $n=3/2,1/2$
$$J_{-1/2}=\int_0^a\dfrac{dx}{\sqrt{a^2-x^2}}=\dfrac1{|a|}\arcsin\dfrac xa\big|_0^a=\dfrac\pi{2|a|}$$
