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

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.

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.

Since the question is about convergence and not evaluating the integral, we can bound the integral directly. More precisely,

$$ \int_0^1 \frac{dx}{\sqrt{x(1-x)}}=\int_0^{1/2}\frac{dx}{\sqrt{x(1-x)}}+\int_{1/2}^{1}\frac{dx}{\sqrt{x(1-x)}}. $$ Using the substitution $u=1-x$ on the second integral, we get that this simplifies to $$ \int_0^{1/2}\frac{dx}{\sqrt{x(1-x)}}+\int_{1/2}^{0}\frac{-du}{\sqrt{(1-u)u}}=2\int_0^{1/2}\frac{dx}{\sqrt{x(1-x)}}. $$ Now, we break the square root into two factors: $\sqrt{x}\sqrt{1-x}$. Over the region of interest, $1-x$ varies between $1$ and $\frac{1}{2}$, so the smallest $\sqrt{1-x}$ could be is $\frac{1}{\sqrt{2}}$. Therefore, $$ 2\int_0^{1/2}\frac{dx}{\sqrt{x(1-x)}}\leq 2\sqrt{2}\int_0^{1/2}\frac{dx}{\sqrt{x}}=2\sqrt{2}\lim_{b\rightarrow 0}\int_b^{1/2}\frac{dx}{\sqrt{x}}. $$ This integral is straight-forward to integrate as $$ 2\sqrt{2}\lim_{b\rightarrow 0}\int_b^{1/2}\frac{dx}{\sqrt{x}}=2\sqrt{2}\lim_{b\rightarrow 0}\left.2\sqrt{x}\right|_{b}^{1/2}=2\sqrt{2}\lim_{b\rightarrow 0}\left(\frac{2}{\sqrt{2}}-2\sqrt{b}\right)=4. $$ Since we have shown that this integral is less than a constant and the integrand is positive everywhere, the integral converges.