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!

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.
