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? 
 A: If you do $x=a\sin\theta$ then since $x$ goes from $0$ to $a$, $\theta$ goes from $0$ to $\frac\pi2$. So,\begin{align}\int_0^a(a^2-x^2)^{\frac32}\,\mathrm dx&=\int_0^{\frac\pi2}(a^2-a^2\sin^2\theta)^{\frac32}a\cos\theta\,\mathrm d\theta\\&=a^4\int_0^{\frac\pi2}\cos^4\theta\,\mathrm d\theta\\&=\frac{3a^4\pi}{16}.\end{align}The last equality comes from the fact that$$\cos^2\theta=\frac{\cos(2\theta)+1}2$$and therefore\begin{align}\cos^4\theta&=\frac{\cos^2(2\theta)+2\cos(2\theta)+1}4\\&=\frac{\frac{\cos(4\theta)+1}2+2\cos(2\theta)+1}4\\&=\frac{\cos(4\theta)+4\cos(2\theta)+3}8.\end{align}
A: Use Principal values
When $0=a\sin\theta\implies\theta=\arcsin0=?$
When $a=a\sin\theta\implies\theta=\arcsin1=\dfrac\pi2$
A: $$\dfrac{d(x(a^2-x^2)^n)}{dx}=(a^2-x^2)^n+nx(a^2-x^2)^{n-1}(-2x)$$
$$=(a^2-x^2)^n+2n(a^2-x^2-a^2)(a^2-x^2)^{n-1}$$
$$=(1+2n)(a^2-x^2)^n-2na^2(a^2-x^2)^{n-1}$$
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 $n=3/2,1/2$
$$J_{-1/2}=\int_0^a\dfrac{dx}{\sqrt{a^2-x^2}}=\dfrac1{|a|}\arcsin\dfrac xa\big|_0^a=\dfrac\pi{2|a|}$$
