Solutions - 0050-01

(1) Substitute $x=\sin \theta$ and thus $dx=\cos \theta d\theta$. Adjust the bounds as follows:

$$\begin{array}{cc}x& =0\gg \gg \theta =0\\ x& =\frac{1}{2}\gg \gg \theta =\frac{\pi }{6}\end{array}$$

Rewrite the integral:

$$\begin{array}{cccc}& {\int }_0^{1/2}\frac{x^2}{\sqrt{1-x^2}}dx& \gg \gg {\int }_0^{\pi /6}\frac{{\sin }^2\theta \cos \theta }{\sqrt{1-{\sin }^2\theta }}d\theta & \\ & & & \\ & & \gg \gg {\int }_0^{\pi /6}{\sin }^2\theta d\theta & \text{(Note A)}\end{array}$$

---

(2) Use power-to-frequency conversion:

$$\begin{array}{cc}{\int }_0^{\pi /6}{\sin }^2\theta d\theta & \gg \gg \frac{1}{2}{\int }_0^{\pi /6}1-\cos 2\theta d\theta \\ & \\ & \gg \gg \frac{1}{2}{(\theta -\frac{1}{2}\sin 2\theta )|}_0^{\pi /6}\\ & \\ & \gg \gg \frac{\pi }{12}-\frac{\sin (\frac{\pi }{3})}{4}-\frac{1}{2}(0-0)\\ & \\ & \gg \gg \frac{\pi }{12}-\frac{\sqrt{3}}{8}\end{array}$$

---

Note A: Use $1-{\sin }^2\theta ={\cos }^2\theta$, then $\sqrt{{\cos }^2\theta }=|\cos \theta |$ and this equals $\cos \theta$ for $0\le \theta \le \pi /6$.