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.
$\pi,\omega,\theta$ will render as $\pi,\omega,\theta$, which is better than Unicode characters
– FShrike
Jan 03 '22 at 19:44
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.
$\newcommand{\d}{\,\mathrm{d}}\newcommand{\b}{\mathcal{B}}$Cheating slightly, we can re-express the integral as:
$$\int_0^1x^{-1/2}(1-x)^{-1/2}\d x=\b\left(\frac{1}{2},\frac{1}{2}\right)$$
Where $\b$ is the Beta function, which satisfies for positive $x,y$ the relation:
$$\b(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
Where $\Gamma$ is the Gamma function.
Then your integral is:
$$\frac{(\Gamma(1/2))^2}{\Gamma(1)}=\pi$$
But maybe you are not familiar with this. I leave these for further reading.
Let's more elementarily evaluate this:
Insert $x=\sin^2u$ to find: $$\int_0^{\pi/2}\frac{2\sin x\cos x}{\sqrt{\sin^2x}\sqrt{1-\sin^2x}}\d x=2\int_0^{\pi/2}\frac{\sin x\cos x}{\sin x\cos x}\d x=2\int_0^{\pi/2}1\d x=\pi$$
These substitutions are ok for the interval $(0,\pi/2)$, and what's more, if you use the $\epsilon-\delta$ idea of evaluating this integral on $(\epsilon,1-\delta)$, you can use these substitutions to see that the integral is convergent, so the fact that division by $\sin 0,\cos\pi/2$ features doesn't matter:
Let $\epsilon,\delta\in(0,0.5)$ be fixed. $$\int_\epsilon^{1-\delta}x^{-1/2}(1-x)^{-1/2}\d x=\int_{\arcsin\sqrt{\epsilon}}^{\arcsin\sqrt{1-\delta}}2\d x=2(\arcsin\sqrt{1-\delta}-\arcsin\epsilon)$$
Independently of how $\epsilon,\delta\to0^+$, we see, by continuity of $\arcsin,\sqrt{}$ in these domains, that we get $2\arcsin 1=\pi$. Since this result is independent of the path/rate of approach, the integral is convergent and equal to $\pi$.
Alternatively, using $$x-x^2=\frac14-(x-\frac12)^2,$$ the integral is $$\int_0^1\frac{1}{\sqrt{\frac14-(x-\frac12)^2}}dx$$ $$=\left[\arcsin\left(\frac{x-\frac12}{\frac12}\right)\right]^1_0$$ $$=\pi$$
use $y=x-\frac12$, then: $$ I=\int_0^1\frac1{\sqrt {x(1-x)}}dx=\int^\frac12_{-\frac12}\frac1{\sqrt {\frac14-y^2} }dy $$ Subtituting $2y=\sin u$ we get: $$ I=\int^\frac\pi2_{-\frac\pi2}du=\pi $$
Notice that in the range $[0,1]$ we have
$$\int_0^1 \dfrac{\text{d}x}{\sqrt{x(1-x)}} \sim \int_0^1 \frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{2}+\frac{3 x^{3/2}}{8}+O\left(x^{5/2}\right)\ \text{d}x$$
Whence the convergence of the integral. For the exact result, you can check for any answer here around.
Observe that $$ \int_0^1\frac{dx}{\sqrt{x(1-x)}}{=2\int_0^\frac{1}{2}\frac{dx}{\sqrt{x(1-x)}} \\\le 2\int_0^\frac{1}{2}\frac{dx}{\sqrt{x\frac{1}{2}}} \\= 2\sqrt 2\int_0^\frac{1}{2}\frac{dx}{\sqrt{x}} \\=4 } $$
So let's just solve for the primitive function $\int_0^1\frac{dx}{\sqrt{x(1-x)}}=2\int_0^1\frac{d\sqrt{x}}{\sqrt{1-x}} =2\arcsin\sqrt{x}\Big|_0^1=2(\arcsin1-\arcsin0)=\pi$
