# 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?

• Hint: Use the Euler substitution – Dr. Sonnhard Graubner May 2 '19 at 16:52
• $\arcsin\sqrt{x}$ and $\frac{1}{2}\arcsin(2x-1)$ seems differ by a constant only. – Yuta May 2 '19 at 16:54
• I think they're both correct results. Make sure they are differentiated. – georg May 2 '19 at 16:56
• @Yuta is correct. Indeed, $\arcsin(2x-1)=\frac{\pi}{2}-\arccos(2x-1)=\frac{\pi}{2}-2\arccos\sqrt{x}=2\arcsin\sqrt{x}-\frac{\pi}{2}$. – J.G. May 2 '19 at 19:36

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 $$-\frac{\pi}{4}=c$$

• Thank you very much for your help:)I understood. – DaniVaja May 2 '19 at 17:31

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.

• Thank you for your answer :) – DaniVaja May 2 '19 at 17:31

Hint:

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