A direct proof for $\int_0^x \frac{- x \ln(1-u^2)}{u \sqrt{x^2-u^2}} \, \mathrm{d} u = \arcsin^2(x)$ I have been trying to evaluate
$$ f(x) \equiv \int \limits_0^\infty - \ln\left(1 - \frac{x^2}{\cosh^2 (t)}\right) \, \mathrm{d} t $$
for $x \in [0,1]$ and similar integrals recently. I know that
$$ \int \limits_0^\infty \frac{\mathrm{d} t}{\cosh^z (t)} = \frac{2^{z-2} \Gamma^2 (\frac{z}{2})}{\Gamma(z)} $$
holds for $\operatorname{Re} (z) > 0$, so by expanding the logarithm I found that
$$ f(x) = \frac{1}{2} \sum \limits_{n=1}^\infty \frac{(2n)!!}{n^2 (2n-1)!!} x^{2n} \, .$$
But the right-hand side is the power series of the arcsine squared, so $f(x) = \arcsin^2 (x)$.
On the other hand, the substitution $u = \frac{x}{\cosh(t)}$ in the original integral leads to the representation
$$ f(x) = \int \limits_0^x \frac{- x \ln(1-u^2)}{u \sqrt{x^2-u^2}} \, \mathrm{d} u \, ,$$
for which Mathematica (or WolframAlpha if you're lucky) gives the correct result. 
I would like to compute this integral without resorting to the above power series and thereby find an alternative proof for the expansion. I have tried to transform the integral into the usual form
$$ \arcsin^2 (x) = \int \limits_0^x \frac{2 \arcsin(y)}{\sqrt{1-y^2}} \, \mathrm{d} u $$
and thought about using the relations
$$ \arcsin(x) = \arctan\left(\frac{x}{\sqrt{1-x^2}}\right) = 2 \arctan\left(\frac{x}{1+\sqrt{1-x^2}}\right) \, , $$
but to no avail. Maybe the solution is trivial and I just cannot see it at the moment, maybe it is not. Anyway, I would be grateful for any ideas or hints.
 A: Let $u=x \sin (\theta)$
\begin{eqnarray*}
-\int_0^{\pi/2} \frac{\ln(1-x^2 \sin^2(\theta))}{\sin(\theta)} d \theta
\end{eqnarray*}
Now expand the logarithms
\begin{eqnarray*}
\sum_{n=1}^{\infty} \int_0^{\pi/2} \frac{1}{n} x^{2n} \sin^{2n-1}(\theta) d \theta
\end{eqnarray*}
Now use 
\begin{eqnarray*}
 \int_0^{\pi/2}   \sin^{2n-1}(\theta) d \theta= \frac{(2n-2)!!}{(2n-1)!!}.
\end{eqnarray*}
Finally use the result you state in the question
\begin{eqnarray*}
 \frac{1}{2} \sum \limits_{n=1}^\infty \frac{(2n)!!}{n^2 (2n-1)!!} x^{2n}=(\sin^{-1}(x))^2
\end{eqnarray*}
and we are done.
A: I have finally managed to put all the pieces together, so here's a solution that does not use the power series:
Let $u = x v$ to obtain
$$ f(x) = \int \limits_0^1 \frac{- \ln(1 - x^2 v^2)}{v \sqrt{1-v^2}} \, \mathrm{d} v \, . $$
Now we can differentiate under the integral sign (justified by the dominated convergence theorem) and use the substitution $v = \sqrt{1 - w^2}\, .$ Then the derivative is given by
\begin{align}
f'(x) &= 2 x \int \limits_0^1 \frac{v}{(1-x^2 v^2) \sqrt{1-v^2}} \, \mathrm{d} v = 2 x \int \limits_0^1 \frac{\mathrm{d} w }{1-x^2 + x^2 w^2} \\
&= \frac{2}{\sqrt{1-x^2}} \arctan \left(\frac{x}{\sqrt{1-x^2}}\right) = \frac{2 \arcsin (x)}{\sqrt{1-x^2}}
\end{align}
for $x \in (0,1)$. Since $f(0)=0 \, ,$ integration yields
$$ f(x) = f(0) + \int \limits_0^x \frac{2 \arcsin (y)}{\sqrt{1-y^2}} \, \mathrm{d} y = \arcsin^2 (x)$$
for $x \in [0,1]$ as claimed. 
