Integral with root calculation

Question

The integral is this:
$\int_0^1$ $x^{2}$$\sqrt{1-x^{2}} dx$ I've tried several things, but i couldn't come to a conclusion.

Answer 1

If we substitute $x = \sin \varphi$, then we get

$$\int_0^1 x^2\sqrt{1-x^2}\,dx = \int_0^{\pi/2} \sin^2\varphi \sqrt{\cos^2\varphi}\cos\varphi\,d\varphi = \int_0^{\pi/2}\sin^2\varphi\cos^2\varphi\,d\varphi.$$

Now using $\sin\varphi \cos\varphi = \frac12\sin (2\varphi)$, we rewrite that as

$$\begin{align} \int_0^{\pi/2} \sin^2\varphi\cos^2\varphi\,d\varphi &= \frac14 \int_0^{\pi/2} \sin^2 (2\varphi)\,d\varphi\\ &= \frac18 \int_0^\pi \sin^2 \psi\,d\psi. \end{align}$$

Now you can write $\sin^2\psi = \frac12\left(1-\cos (2\psi)\right)$ to evaluate the integral, or you might know that the average of $\sin^2\psi$ (and of $\cos^2\psi$) over a period is $\frac12$, and hence see that the end result is

$$\int_0^1x^2\sqrt{1-x^2}\,dx = \frac{\pi}{16}.$$