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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?

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  • $\begingroup$ Hint: Use the Euler substitution $\endgroup$ – Dr. Sonnhard Graubner May 2 '19 at 16:52
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    $\begingroup$ $\arcsin\sqrt{x}$ and $\frac{1}{2}\arcsin(2x-1)$ seems differ by a constant only. $\endgroup$ – Yuta May 2 '19 at 16:54
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    $\begingroup$ I think they're both correct results. Make sure they are differentiated. $\endgroup$ – georg May 2 '19 at 16:56
  • $\begingroup$ @Yuta is correct. Indeed, $\arcsin(2x-1)=\frac{\pi}{2}-\arccos(2x-1)=\frac{\pi}{2}-2\arccos\sqrt{x}=2\arcsin\sqrt{x}-\frac{\pi}{2}$. $\endgroup$ – J.G. May 2 '19 at 19:36
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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 $$ -\frac{\pi}{4}=c $$

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

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  • $\begingroup$ Thank you for your answer :) $\endgroup$ – DaniVaja May 2 '19 at 17:31
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Hint:

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

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