# Determine whether $\int_{0}^{1}\frac{dx}{\sqrt{x(1-x)}}$ converges or diverges.

calculus

I'm having trouble determining whether $$\int_{0}^{1}\frac{dx}{\sqrt{x(1-x)}}$$ converges or diverges.

My first thought was to do the substitution $$u=\sqrt{x(1-x)}$$ which algebraically could gives me something meaningful. But the problem I face with this substitution is that my integration boundaries become $$u:0\rightarrow 0$$ and such an integral is always 0, right? I'm not sure on that one. Would love some guidance.

#### Best Answer

As Ninad Munshi have said : try using $$x=\sin ^{2} \theta$$ as substitution to solve the integral, or allow me to solve it for you using another way also :

Take the integral: $$\int \frac{1}{\sqrt{(1-x) x}} d x$$ Factor powers: $$=\int \frac{1}{\sqrt{1-x} \sqrt{x}} d x$$ For the integrand $$\frac{1}{\sqrt{1-x} \sqrt{x}}$$,

substitute $$u=\sqrt{x}$$ and $$d u=\frac{1}{2 \sqrt{x}} d x$$ : $$=2 \int \frac{1}{\sqrt{1-u^{2}}} d u$$ The integral of $$\frac{1}{\sqrt{1-u^{2}}}$$ is $$\sin ^{-1}(u)$$ : $$=2 \sin ^{-1}(u)+$$ constant

Substitute back for $$u=\sqrt{x}$$ : $$=2 \sin ^{-1}(\sqrt{x})+\text { constant }$$ Which is equivalent for restricted $$x$$ values to: Answer: $$=-2 \sin ^{-1}(\sqrt{1-x})+\text { constant }$$ Then evaluate it from 0 to 1 to get : $$\pi$$, the least-known constant in mathematics.

As a plus, I want to add why your substitution didn't work, a remark that no one did comment on it here or gave an answer why it didn't work:

It's because your substitution is not increasing only(or decreasing only) between 0 and 1, and that's an important condition for doing u-sub.