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?

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: $\color{blue}{\int_0^{\pi/2}\sin^m\theta\cos^n\theta\ d\theta=\dfrac{\Gamma(\frac{m+1}{2})\Gamma(\frac{n+1}{2})}{2\Gamma(\frac{m+n+2}{2})}}$ ,
$$=a^4\frac{\Gamma(\frac{4+1}{2})\Gamma(\frac{0+1}{2})}{2\Gamma(\frac{4+0+2}{2})}$$
$$=a^4\frac{\frac32\frac12\Gamma(\frac{1}{2})\Gamma(\frac{1}{2})}{2\cdot 2}$$
$$=a^4\frac{\frac34\sqrt{\pi}\cdot \sqrt{\pi}}{2\cdot 2}=\frac{3\pi}{16}a^4$$
