## Calculus (3rd Edition)

$C=\dfrac{1}{\pi}$ and $P(-\dfrac{1}{2} \leq X \leq \dfrac{1}{2})=\dfrac{1}{3}$
We are given that $p(x)=\dfrac{C}{\sqrt {1-x^2}}$ Suppose that $x =\sin a \implies dx=\cos a da$ Consider $I=\int_{-1}^1 \dfrac{C}{\sqrt {1-x^2}}dx \\=\int_{-1}^1 \dfrac{C}{\sqrt {1-\sin^2 a}}dx\\=[C \sin^{-1} x]_{-1}^1 \\=C[\sin^{-1} (1)-\sin^{-1} (-1)]\\=\pi C$ The condition will be satisfied when $\int_{-1}^1 \dfrac{C}{\sqrt {1-x^2}}dx=C \pi=1$ or, $C=\dfrac{1}{\pi}$ We will find $P(-\dfrac{1}{2} \leq X \leq \dfrac{1}{2})=\int_{-\frac{1}{2}}^{\frac{1}{2}} \dfrac{1}{\pi \sqrt {1-x^2}} dx \\=\dfrac{1}{\pi}[ \sin^{-1} x]_{-\frac{1}{2}}^{\frac{1}{2}} \\=C[\sin^{-1} (1/2)-\sin^{-1} (-1/2)]\\=\dfrac{1}{3}$