Can somebody find an elementary proof of the following identity:
$$ \arcsin ( 1 - 2x) + 2 \arcsin(\sqrt{x}) = \frac\pi2 $$
I noticed it while solving the following integral:
$$ I = \int \frac{\sqrt{x}}{\sqrt{1 - x}} = -2 \sqrt{1 - x}\sqrt{x} + \int \frac{\sqrt{1 - x}}{\sqrt{x}} $$
where the first equality follows after applying an integration by parts with $f = \sqrt{x}$ and $\mathrm{d} g = 1/\sqrt{1 - x}$. For the sake of simplicity we omit $\mathrm{d}x$ in each integral. Adding the integral to itself:
\begin{align} 2I = \int \frac{\sqrt{x}}{\sqrt{1 - x}} &= -2 \sqrt{1 - x}\sqrt{x} + \int \left(\frac{\sqrt{1 - x}}{\sqrt{x}} + \frac{\sqrt{x}}{\sqrt{1- x}}\right) \\&= -2 \sqrt{1-x}\sqrt{x} + \int\frac{1}{\sqrt{x}\sqrt{1-x}}\end{align}
The last integral on the RHS evaluates to $\arcsin(1 - 2x)$, so $$I = - \sqrt{1-x}\sqrt{x} + \frac{1}{2}\arcsin(1 - 2x) + C.$$
On the other hand, the integral can also be evaluated by applying a $u$-sub with $u = \sqrt{x}$. We find that:
$$ I = 2 \int\frac{u^2}{\sqrt{1 - u^2}} =2 \left(-u \sqrt{1 - u^2} + \int\sqrt{1 - u^2}\right) = - u\sqrt{1 - u^2} + \arcsin u + C_2 $$
So then it follows that $I$ is also equal to $-\sqrt{x}\sqrt{1-x} + \arcsin{\sqrt{x}} + C_2$. Equate the results of the two methods and plug in a random point to find $C - C_2$ and the subsequent identity.
from Hot Weekly Questions - Mathematics Stack Exchange
najkim
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