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?

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}$
