evaluation of some integral I'm looking for a direct proof of the following identity:
$$\frac{1}{\pi}\int_1^x\frac{dt}{t\sqrt{t-1}}\arcsin\left(\sqrt{\frac{x-t}{y-t}}\right)=\arcsin\sqrt{\frac{x}{y}}-\arcsin\sqrt{\frac{1}{y}},$$
for $1<x<y$.
I know this holds because the integral above is a related to some probability (concerning brownian motion) that can be evaluated by some other means.
Any idea will be welcome!
 A: One way out is probably to differentiate both sides, and try to show they are equal. 
Indeed, we think of fixed $y$ and note that the equality is true for $x=1$. Next, the derivative of the left-hand side is (one term disappear if one uses the general formula)
$$
\frac{1}{2\pi\sqrt{y-x\vphantom{)}}}\int_1^x \frac{1}{t\sqrt{(t-1)(x-t)}}\,dt
$$
and the derivative of the right-hand side is
$$
\frac{1}{2\sqrt{x(y-x)}}.
$$
It thus remains to prove that
$$
\int_1^x \frac{1}{t\sqrt{(t-1)(x-t)}}\,dt = \frac{\pi}{\sqrt{x}}.
$$
Can you proceed from here?
Update
Since the integral in the left-hand side is a bit difficult to handle, I give some further steps:
Do the change of variable $t=(x-1)u+1$ (this sends $t=1$ to $u=0$ and $t=x$ to $u=1$). You will (hopefully) get the integral
$$
\int_0^1 \frac{1}{\sqrt{u(1-u)}(1+(x-1)u)}\,du
$$
Next, do the change of variables $u=\frac{s^2}{1+s^2}$. This will send $u=0$ to $s=0$ and $u=1$ to $s=+\infty$, and the integral becomes
$$
\int_0^{+\infty}\frac{2}{1+s^2 x}\,ds.
$$
I'm sure you can continue from here.
