Good afternoon, I'm doing Exercise 2.19.3 about p.d.f function in a introductory probability course:
Could you please verify if my proof looks fine or contains logical gaps/errors? Thank you so much for your help!
My attempt:
By Euler's substitution, let $\sqrt{x(1-x)} = xt$. Then $t \to +\infty$ as $x \to 0^+$ and $t \to 0$ as $x \to 1^-$.
$$\int_{0^+}^{1^-} (x(1-x))^{-1/2} \, \mathrm{d}x = \int_{0^+}^{1^-} \dfrac{1}{\sqrt{x(1-x)}} \, \mathrm{d}x = \int_{0}^{+\infty} \dfrac{2}{1+t^2} \, \mathrm{d}t = \arctan (t) \Big|_0^{+\infty} = \frac{\pi}{2}$$
In order for $f$ to be p.d.f, $$\int_{0^+}^{1^-} C(x(1-x))^{-1/2} \, \mathrm{d}x =1$$
This implies $$C\frac{\pi}{2} =1$$
Hence $C = 2/\pi$.