primitives of $f(x)=\frac{1}{2\sqrt{x-x^{2}}}$ I have this function $f:(0,1)\rightarrow \mathbb{R},f(x)=\frac{1}{2\sqrt{x-x^{2}}}$   and I need to find the primitives of $f(x)$
So I calculated $\int \frac{1}{2\sqrt{x-x^{2}}}\,dx$ and I got $\frac{1}{2}\arcsin(2x-1)+C$ but the right answer is $\arcsin\sqrt{x}+C$ .Where's my mistake?How to start?
 A: If $\theta=\arcsin \sqrt{x}$, then $\sin\theta=\sqrt{x}$, so $\sin^2\theta=x$. 
Since $\sin^2\theta=\frac{1-\cos(2\theta)}{2}$, we get
$$\frac{1-\cos(2\theta)}{2}=x$$
We deduce $-\cos(2\theta)=2x-1$. 
Then, $\sin(2\theta-\pi/2)=2x-1$. 
So, (maybe up to a constant) we have $\arcsin(2x-1)=2\theta-\pi/2$.
Which means that $\theta=\frac{1}{2}\arcsin(2x-1)+\pi/4$. So you have that $\arcsin \sqrt{x}$ and $\frac{1}{2}\arcsin(2x-1)$ differ by a constant. So, both results are okay.
A: The derivative of
$$
f(x)=\frac{1}{2}\arcsin(2x-1)
$$
is
$$
f'(x)=\frac{1}{2}\frac{1}{\sqrt{1-(2x-1)^2}}\cdot2
$$
The square root is
$$
\sqrt{1-4x^2+4x-1}=2\sqrt{x-x^2}
$$
so your result is correct.
Quite likely, the book used the substitution $t=\sqrt{x}$, so the integral becomes
$$
\int\frac{1}{2\sqrt{t^2-t^4}}\cdot2t\,dt=\int\frac{1}{1-t^2}\,dt=\arcsin t+C
$$
There is no problem if you instead complete the square observing that
$$
2\sqrt{x-x^2}=\sqrt{4x-4x^2}=\sqrt{1-(2x-1)^2}
$$
as you probably did.
Just by completeness, we can conclude that, over $(0,1)$,
$$
\frac{1}{2}\arcsin(2x-1)=\arcsin\sqrt{x}+c
$$
for a constant $c$. Evaluating at $x=0$ (it is possible, because the two functions can be extended by continuity to $[0,1]$) we find
$$
-\frac{\pi}{4}=c
$$
A: Hint:
$$\int\frac{dx}{2\sqrt{x-x^{2}}}=\int\frac{dx}{2\sqrt x\sqrt{1-x}}=\int\frac{d\sqrt x}{\sqrt{1-(\sqrt x)^2}}.$$
