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?
Best Answer
With $x=a\sin t$, which isn't quite what you said you tried, the integral is$$\begin{align}a^4\int_0^{\pi/2}\sin^4tdt&=\frac14a^4\int_0^{\pi/2}(1-\cos2t)^2dt\\&=\frac14a^4\int_0^{\pi/2}(1-2\cos2t+\cos^22t)dt\\&=\frac18a^4\int_0^{\pi/2}(3-4\cos2t+\cos4t)dt\\&=\frac18a^4[3t-2\sin2t+\tfrac14\sin4t]_0^{\pi/2}\\&=\frac{3\pi}{16}a^4.\end{align}$$