Show the $\arcsin$ identity: $ \arcsin(1 - 2x) + 2\arcsin(\sqrt{x}) = \pi / 2$ 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}} &amp;= -2 \sqrt{1 - x}\sqrt{x} + \int \left(\frac{\sqrt{1 - x}}{\sqrt{x}} + \frac{\sqrt{x}}{\sqrt{1- x}}\right) \\&amp;= -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.
Here’s an elementary proof. Let $x=\sin^2\phi$. Then, since $\cos(2\phi)=1-2\sin^2\phi$, we have
$$\arcsin(1-2x)=\arcsin(1-2\sin^2\phi)=\arcsin(\cos(2\phi))=\frac{\pi}{2}-2\phi$$
and
$$2\arcsin(\sqrt{x})=2\arcsin(\sin\phi)=2\phi$$
From which the result follows elementarily (without calc).
Let $x=u^2$ with $0\le u\le1$ and rewrite the equation $\arcsin(1-2x)+2\arcsin(\sqrt x)=\pi/2$ as
$$\arcsin(1-2u^2)={\pi\over2}-2\arcsin u$$
Think of each side as an angle, $\theta_L$ and $\theta_R$. Clearly $-\pi/2\le\theta_L\lt\pi/2$, that being the range of the arcsine function.  Because $u$ is nonnegative, we have $0\le2\arcsin u\le\pi$, and hence $-\pi/2\le\theta_R\le\pi/2$ as well.  Thus to prove that $\theta_L=\theta_R$, it suffices to show that $\sin\theta_L=\sin\theta_R$. Obviously
$$\sin\theta_L=\sin(\arcsin(1-2u^2))=1-2u^2$$
while standard trig identities take care of $\theta_R$:
$$\sin\theta_R=\sin\left({\pi\over2}-2\arcsin u \right)=\cos(2\arcsin u)=1-2\sin^2(\arcsin u)=1-2u^2$$