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How to integrate

$$\int_a^1 \frac{\arcsin x}{\sqrt{x^2-a^2}} dx?$$

This integral appeared in

Demkov Yu. N., Ostrovsky V. N., Berezina (Avdonina) N. B., "Uniqueness of the Firsov Inversion Method and Focusing Potentials", Sov. Phys. JETP 33, 867-70, (1971).

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    This is an interesting question, so please add some context in order to prevent downvotes. – 2017-01-25
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    By differentiation under the integral sign, for any $a\in(0,1)$ the answer is given by $$\frac{\pi}{2}\log\left(1+\sqrt{1-a^2}\right).$$ – 2017-01-25
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    You may also consider that $\arcsin(x)$ has a simple Fourier-Chebyshev series. – 2017-01-25
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    Thanks Jack. I added the source of the integral in the question. – 2017-01-25
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    Can you be more specific Jack, about the differentiate under the integral? Also does $\arcsin(x)/\sqrt{x^2-a^2}$ have an anti derivative? – 2017-01-25
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    @Yuanyuan: not an elementary one. My suggestion is simply to consider $$\frac{\partial}{\partial a}\int_{a}^{1}\frac{\arcsin x}{\sqrt{x^2-a^2}}\,dx.$$ – 2017-01-25

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Assuming $a\in(0,1)$, the first temptation may be to use Feynman's trick. But we have to be careful since the $a$ parameter appears both in the integration range and in the integrand function. So we may start wondering about different approaches$\ldots$ $$\begin{eqnarray*} I(a) &=& \int_{a}^{1}\frac{\arcsin x}{\sqrt{x^2-a^2}}\,dx \\ &=& \int_{a}^{1}\int_{0}^{x}\frac{1}{\sqrt{(1-z^2)(x^2-a^2)}}\,dz\,dx\\ (z\mapsto xz)\quad&=& \int_{a}^{1}\int_{0}^{1}\frac{x}{\sqrt{(1-x^2 z^2)(x^2-a^2)}}\,dz\,dx\\(x\mapsto\sqrt{x})\quad &=& \frac{1}{2}\int_{a^2}^{1}\int_{0}^{1}\frac{1}{\sqrt{(1-x z^2)(x-a^2)}}\,dz\,dx\end{eqnarray*}$$ Now it is a good moment for switching the integrals. If we integrate with respect to $x$ for first: $$ I(a) = \int_{0}^{1}\left.\arcsin\sqrt{\frac{1-x z^2}{1-a^2 z^2}}\right|_{x=a^2}^{x=1}\,\frac{dz}{z}$$ and it is easier to compute $I'(a)$, since now $a$ has disappeared from the integration range.
After that, it is just tedious to check that: $$ I'(a) = -\frac{\pi a}{2\sqrt{1-a^2}(1+\sqrt{1-a^2})} = \frac{d}{da}\left[\frac{\pi}{2}\log\left(1+\sqrt{1-a^2}\right)\right]$$ and since $I(0)$ is a famous integral, equal to $\frac{\pi}{2}\log 2$, we get: $$\boxed{ I(a) = \color{red}{\frac{\pi}{2}\,\log\left(1+\sqrt{1-a^2}\right)} }$$ as wanted.

I was trying to exploit the hidden symmetry given by $$ \frac{1}{\sqrt{x^2-a^2}}=\frac{d}{da}\,\arcsin\left(\frac{a}{x}\right)$$ but I was not able to conclude from there. I leave such observation to more skilled integrators than me, it might be useful. The Poisson kernel or the generating function for Legendre/Chebyshev polynomials might be useful too, and it is remarkable that the final outcome is just the very first contribute we get from integration by parts, namely $\arcsin(x)\log\left(x+\sqrt{x^2-a^2}\right)$ evaluated at $x=1$.