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My teacher wrote this on the blackboard: $$\int_0^z \frac{1}{\pi \sqrt{x(1-x)}}dx= \frac{2}{\pi} \arcsin(\sqrt{z})$$

But when I try to calculate the integral:

\begin{align} \int \frac{1}{\pi \sqrt{x(1-x)}}dx&=\int \frac{1}{\pi \sqrt{(\frac{1}{2})^2-(x-\frac{1}{2})^2}}dx\\ &= \int \frac{1}{\pi \sqrt{(\frac{1}{2})^2-k^2}}dk=2 \int \frac{1}{\pi \sqrt{1-(2k)^2}}dk\\ &= \int \frac{1}{\pi \sqrt{1-a^2}}da\\ &= \frac{1}{\pi} \arcsin(a)\\ &=\frac{1}{\pi} \arcsin\left(2(x-\frac{1}{2})\right) \end{align}

So

$$\int_0^z \frac{1}{\pi \sqrt{x(1-x)}}dx= \frac{1}{\pi} \arcsin(2z-1)-\frac{1}{\pi} \arcsin(-1)= \frac{1}{\pi} \arcsin(2z-1) - \frac{1}{2}$$

How I come to the solution of my teacher?

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2 Answers 2

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Your answer should be $$\frac {1}{\pi}\arcsin(2z-1)\color{red}{+}\frac 12$$

In fact both answers are equivalent. This is because if $$\phi=\frac{2}{\pi}\arcsin(\sqrt{z}),$$ then$$\sin(\pi\phi)=2\sqrt{z}\cos(\arcsin(\sqrt{z}))=2\sqrt{z}\sqrt{1-z}\Rightarrow\phi=\frac{1}{\pi}\arcsin(2\sqrt{z-z^2})$$

Meanwhile, if $$\theta=\frac {1}{\pi}\arcsin(2z-1)\color{red}{+}\frac 12,$$ then$$\sin(\pi\theta)=\sin(\arcsin(2z-1)+\frac{\pi}{2})$$ $$=\cos(\arcsin(2z-1))=\sqrt{1-(2z-1)^2}=2\sqrt{z-z^2}$$

Hence $$\phi=\theta$$

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if you make the substitution $x=u^2$ the integral becomes: $$ \int_0^{\sqrt{z}} \frac2{\pi \sqrt{1-u^2}}du $$

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  • $\begingroup$ thanks!, but under the assumption that $z>0$. $\endgroup$ Aug 14, 2016 at 16:14

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