definite integral $\int_{-1}^1x^2\sqrt{4-x^2}dx$ I have solved the problem and obtained the value of the integral $2\pi/3-\sqrt{3}/2$.
In the actual solution of this problem I have seen that $x$ was substituted by $2\sin z$ and upper limit $1$ was substituted by $\pi/6$ and lower limit $-1$ by $-\pi/6$.
Now the question is why was the upper limit substituted by $\pi/6$ instead of $5\pi/6$ or anything equivalent to $\arcsin\tfrac12$?
Actually,$$\sin(\pi-y)=\sin y\implies\sin\tfrac{5\pi}{6}=\sin\tfrac{\pi}{6}\ne-\sin\tfrac{\pi}{6}.$$We need a $z$ satisfying $\sin z=-\sin\tfrac{\pi}{6}$. Since the sine function is odd, $z=-\tfrac{\pi}{6}$ will do.
Alternative argument: since the original integrand is even,$$\begin{align}\int_{-1}^1x^2\sqrt{4-x^2}dx&amp;=2\int_0^1x^2\sqrt{4-x^2}dx\\&amp;=2\int_0^{\pi/6}16\sin^2z\cos^2zdz\\&amp;=\int_{-\pi/6}^{\pi/6}16\sin^2z\cos^2zdz,\end{align}$$as the new integrand is even too.
It must be $2 \sin{z} = \pm1 \Rightarrow z = \arcsin{\pm\frac{1}{2}} = \pm\frac{\pi}{6}$
That's all.