How can I solve this definite integral: $\int_{0}^{a}\frac{x^4dx}{\sqrt{a^2-x^2}}$ 
Evaluate $$\int_{0}^{a}\dfrac{x^4dx}{\sqrt{a^2-x^2}}$$

I tried taking $t$ as
$$t = \sqrt{a^2-x^2}$$
Thus my final integral became
$$\int_{0}^{a}(a^2-t^2)^{3/2}dt$$
but I couldn't go any further in solving this integral.
I also tried by taking $t$ as
$$t = a\sin^{-1}{x}$$
But I don't know how to solve the resulting integrand.
Also, can the king's rule be applied here? If yes then how?
 A: With $x=a\sin t$, which isn't quite what you said you tried, the integral is$$\begin{align}a^4\int_0^{\pi/2}\sin^4tdt&=\frac14a^4\int_0^{\pi/2}(1-\cos2t)^2dt\\&=\frac14a^4\int_0^{\pi/2}(1-2\cos2t+\cos^22t)dt\\&=\frac18a^4\int_0^{\pi/2}(3-4\cos2t+\cos4t)dt\\&=\frac18a^4[3t-2\sin2t+\tfrac14\sin4t]_0^{\pi/2}\\&=\frac{3\pi}{16}a^4.\end{align}$$
A: We assume ${a>0}$. As one of the comments suggested, taking ${x=a\sin(t)}$ would yield
$${\Rightarrow \int_{0}^{\frac{\pi}{2}}\frac{a^4\sin^4(t)}{a\cos(t)}a\cos(t)dt=a^4\int_{0}^{\frac{\pi}{2}}\sin^4(t)dt}$$
(notice we can cancel ${\sqrt{\cos^2(t)}}$ as ${\cos(t)}$ because of the positivity of ${\cos(t)}$ on the interval ${(0,\frac{\pi}{2})}$.
So in fact
$${\int_{0}^{a}\frac{x^4}{\sqrt{a^2-x^2}}dx=a^4\int_{0}^{\frac{\pi}{2}}\sin^4(t)dt=a^4\times\text{ const.}}$$
Can you take it from here and solve ${\int_{0}^{\frac{\pi}{2}}\sin^4(t)dt}$?
A: You can proceed as follows
Let $t=a\sin\theta\implies dt=a\cos\theta d\theta$
$$\int_{0}^{a}(a^2-t^2)^{3/2}dt=\int_{0}^{\pi/2}(a^2-a^2\sin^2\theta)^{3/2}a\cos\theta \ d\theta$$
$$=\int_{0}^{\pi/2}(a^{3}\cos^3\theta) a\cos\theta \ d\theta$$
$$=a^4\int_{0}^{\pi/2}\cos^4\theta d\theta$$
Using: $\color{blue}{\int_0^{\pi/2}\sin^m\theta\cos^n\theta\ d\theta=\dfrac{\Gamma(\frac{m+1}{2})\Gamma(\frac{n+1}{2})}{2\Gamma(\frac{m+n+2}{2})}}$ ,
$$=a^4\frac{\Gamma(\frac{4+1}{2})\Gamma(\frac{0+1}{2})}{2\Gamma(\frac{4+0+2}{2})}$$
$$=a^4\frac{\frac32\frac12\Gamma(\frac{1}{2})\Gamma(\frac{1}{2})}{2\cdot 2}$$
$$=a^4\frac{\frac34\sqrt{\pi}\cdot \sqrt{\pi}}{2\cdot 2}=\frac{3\pi}{16}a^4$$
A: Note $\left((t^3 + \frac32t)\sqrt{1-t^2}\right)’= 
\frac{\frac32-4t^4}{\sqrt{1-t^2} }$ and
$$\begin{align}
\int_{0}^{a}\frac{x^4dx}{\sqrt{a^2-x^2}} 
&= a^4\int_0^{1}\frac{t^4dt}{\sqrt{1-t^2}}\\
&=-\frac{a^4}4(t^3 + \frac32t)\sqrt{1-t^2}\bigg|_0^1
+\frac{3a^4}8\int_0^{1}\frac{dt }{\sqrt{1-t^2}}\\
&=0+\frac{3a^4}8\sin^{-1}(1) =\frac{3\pi}{16}a^4.\end{align}$$
A: There are already 4 solutions. instead,  I am going to evaluate the general integral $$I(k)=\int_{0}^{a} \frac{x^{k}}{\sqrt{a^{2}-x^{2}}} d x, \quad \textrm{ where }k=0,1,2, \cdots$$
by establishing a reduction formula on $k$.
$$
\displaystyle \begin{aligned}
I(k) &=\int_{0}^{a} \frac{x^{k}}{\sqrt{a^{2}-x^{2}}} d x \\
&=-\int_{0}^{a} x^{k-1} d \sqrt{a^{2}-x^{2}} \\
&=-\left[x^{k+1} \sqrt{a^{2}-x^{2}}\right]_{0}^{a}+(k-1) \int_{0}^{a} x^{k-2} \sqrt{a^{2}-x^{2}} \\
&=(k-1) \int_{0}^{a} \frac{x^{k-2}\left(a^{2}-x^{2}\right)}{\sqrt{a^{2}-x^{2}}} d x \\
&=(k-1) a^{2} I(k-2)-(k-1) I(k) \\
 \displaystyle I(k) &=\dfrac{(k-1) a^{2}}{k} I(k-2)\\ &= \vdots \\& = \displaystyle \left\{\begin{array}{l}
\dfrac{k-1}{k} a^{2} \cdot \dfrac{k-3}{k-2} a^{2} \cdots \dfrac{1}{2}a^2 I(0) \quad \text { if }k \textrm{ is even .} \\
\dfrac{k-1}{k} a^{2} \cdot \dfrac{k-3}{k-2} a^{2} \cdots \dfrac{2}{3} a^2I(1) \quad \text { if }k \textrm{ is odd. }
\end{array}\right.
\\ \\ \displaystyle  \displaystyle  &= \left\{\begin{array}{ll}
\dfrac{a^k(k-1) ! ! \pi}{2(k ! !)} & \text { if } k \text { is even. } \\
\dfrac{a^k(k-1) ! !}{2(k ! !)} & \text { if } k \text { is odd. }
\end{array}\right.\end{aligned}
$$
Now go back to the original integral
$$I(k)=\int_{0}^{a} \frac{x^{4}}{\sqrt{a^{2}-x^{2}}} dx =I(4)=\frac{3 \times 1 a^{4} \pi}{2(4 \times 2)}=\frac{3 a^{4} \pi}{16} \quad \blacksquare $$
Footnotes:

*

*$\displaystyle I(0)=\int_{0}^{a} \frac{d x}{\sqrt{a^{2}-x^{2}}}=\left[\sin ^{-1} \frac{x}{a}\right]_{0}^{a}=\frac{\pi}{2}$

*$ \displaystyle I(1) =\int_{0}^{a} \frac{x}{\sqrt{a^{2}-x^{2}}} d x =-\frac{1}{2}\left[\sqrt{a^{2}-x^{2}}\right]_{0}^{a} =\frac{a}{2}$
A: Bit of a roundabout method, but there's always everyone's favorite standard Euler substitution. Let
$$t=\frac{\sqrt{a^2-x^2}-a}x \implies x = -\frac{2at}{1+t^2} \implies dx = -\frac{2a(1-t^2)}{(1+t^2)^2} \, dt$$
to recover the integral of simpler rational function:
$$\begin{align*}
I &= \int_0^a \frac{x^4}{\sqrt{a^2-x^2}} \, dx \\[1ex]
&= \int_{-1}^\infty \frac{\frac{16a^4t^4}{(1+t^2)^4}}{\sqrt{a^2-\frac{4a^2t^2}{(1+t^2)^2}}} \, \frac{2a(1-t^2)}{(1+t^2)^2} \, dt \\[1ex]
&= 32a^4 \int_{-1}^\infty \frac{t^4}{(1+t^2)^5} \, dt
\end{align*}$$
Subsequently letting $t=\tan(u)$ will convert this back to Riemann'sPointyNose's integral.
