How do we prove that $\int \frac{1}{\sqrt{x-x^2}}\mathrm{d}x$ equals $\arcsin(2x - 1) + C$?

Question

My friend and I were looking at MIT Integration Bee problems, and we both tried to solve $$\int\frac{1}{\sqrt{x-x^2}}dx$$ However, we both got differing answers and he won't tell me how he got his.

I did the following:

$$\int\frac{1}{\sqrt{x-x^2}}dx=\int\frac{1}{x\sqrt{\frac{1}{x}-1}}dx$$ $$Let\space u=\sqrt{\frac{1}{x}-1}\Longleftrightarrow du=\frac{-1}{2x^2\sqrt{\frac{1}{x}-1}}dx$$ $$x=\frac{1}{u^2+1}$$

$$-2\int\frac{1}{u^2+1}du=-2\arctan{u}+C=-2\arctan({\sqrt{\frac{1}{x}-1}})+C$$

My friend got the answer as

$$\arcsin({2x-1})+C$$

I haven't been able to figure out how he arrived at his answer. Can they show me how he might have gotten his answer? Thank you.

Answer 1

Here it is how he obtained the proposed result: \begin{align*} \int\frac{\mathrm{d}x}{\sqrt{x - x^{2}}} & = \int\frac{\mathrm{d}x}{\sqrt{1/4 - (x - 1/2)^{2}}}\\\\ & = \int\frac{2\mathrm{d}x}{\sqrt{1 - (2x - 1)^{2}}}\\\\ & = \int\frac{\mathrm{d}(2x - 1)}{\sqrt{1 - (2x - 1)^{2}}}\\\\ & = \int\frac{\mathrm{d}u}{\sqrt{1 - u^{2}}}\\\\ & = \arcsin(u) + c\\\\ & = \arcsin(2x - 1) + c \end{align*}

Hopefully this helps!

Answer 2

The quickest solution would be to recognize the hidden chain rule:

$$\int\frac{dx}{\sqrt{x-x^2}} = \int \frac{1}{\sqrt{1-x}}\cdot\frac{dx}{\sqrt{x}} = \int\frac{1}{\sqrt{1-(\sqrt{x})^2}}\cdot 2d(\sqrt{x}) = 2\arcsin\sqrt{x}$$

Plotting the difference of each of these solutions, we can see that your solution and your friend's solution both give constants on the domain of the original integrand, $(0,1)$, meaning all solutions are correct.

Answer 3

Letting $x=\sin^2 \theta$ yields $$ \begin{aligned} I=\int\frac{d x}{\sqrt{x(1-x)}} &=\int\frac{2 \sin \theta \cos \theta d \theta}{\sin \theta \cos \theta} =2 \theta=2\arcsin \sqrt x +C \end{aligned} $$