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.
Best Answer
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!