Integration of $\int\frac{1}{\sqrt{x-x^2}}$dx

Question

I have recently been trying out some questions on improper integrals and in an intermediate step, I was supposed to find the value of I where

$$I=\int\frac{1}{\sqrt{x-x^2}}dx$$

I tried using $x = \sin^2\theta$ then the answer is $\arcsin$$\sqrt{x}$

But on trying to evaluate the integral using the method of completing the squares, Answer is coming out to be different.

Please help!

Answer 1

Completing the square you can write $$\int {dx \over \sqrt{{1 \over 4}-(x-{1 \over 2})^2}}=2\int {dx \over \sqrt{1-(2x-1)^2}}$$ substitute $u=2x-1, du=2dx$ and then it's trivial to see that your integral is $\arcsin(u)+C=arcsin(2x-1)+C$. Substituiting $x=\sin(\theta)$ in the original integral left me with a $\sin^{{-1 \over 2}}(\theta)$ term which doesn't seem easy to handle.

Answer 2

Maybe try and compare with my completion of the square.
$$\int\frac{1}{\frac{1}{4}-(x-\frac{1}{2})^2}dx$$

From here, simply substitute $u=x-\frac{1}{2}$ and then $u=\frac{\sin(\alpha)}{2}$ to end up with $$\arcsin(2x-1)$$