how to integrate $\;I(a) = \int_0^1 \frac{\ln(1-a^2x^2)}{ x^2\sqrt{1-x^2}}dx$ How to solve this integral?
$$I(a) = \int_0^1 \frac{\ln(1-a^2x^2)}{x^2\sqrt{1-x^2}}\,dx$$
 A: EDIT 
There was a mistake in the original - I forgot a factor in the expansion of the log.  My apologies.
Expand the log in a Taylor series (assuming $|a| \lt 1$):
$$I(a) = -\sum_{k=0}^{\infty} \frac{a^{2 k+2}}{k+1} \int_0^1 dx \frac{x^{2 k}}{\sqrt{1-x^2}} = -\frac{\pi}{2} \sum_{k=0}^{\infty} \frac1{2^{2 k}}\binom{2 k}{k} \frac{a^{2 (k+1)}}{k+1}$$
(The derivation of the integral may be found here.)
The derivative of the sum is a well-known Taylor expansion in its own right.  Thus,
$$I'(a) = -\frac{\pi}{2} \frac{2 a}{\sqrt{1-a^2}} $$
$$I(a) = 0 \implies I(a) = -\pi \left (1-\sqrt{1-a^2} \right ) $$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
\,{\rm I}\pars{a}&\ =\
\overbrace{\color{#66f}{\large\int_{0}^{1}{\ln\pars{1 - a^{2}x^{2}} \over x^{2}\root{1 - x^{2}}}}
\,\dd x}^{\ds{\color{#c00000}{x\ \equiv\ \cos\pars{\theta}}}}\ =\
\int_{\pi/2}^{0}{\ln\pars{1 - a^{2}\cos^{2}\pars{\theta}} \over \cos^{2}\pars{\theta}\root{1 - \cos^{2}\pars{\theta}}}
\,\bracks{-\sin\pars{\theta}\dd\theta}
\\[5mm]&=\overbrace{\int_{0}^{\pi/2}
\ln\pars{1 - {a^{2} \over \tan^{2}\pars{\theta} + 1}}\sec^{2}\pars{\theta}
\,\dd\theta}^{\ds{\color{#c00000}{\tan\pars{\theta}\ \equiv t}}}\ =\
\int_{0}^{\infty}\ln\pars{1 - {a^{2} \over t^{2} + 1}}\,\dd t
\\[5mm]&=\int_{0}^{\infty}
\bracks{\ln\pars{t^{2} + 1 - {a^{2}}} - \ln\pars{t^{2} + 1}}\,\dd t
=2\int_{0}^{\infty}
\pars{{t^{2} \over t^{2} + 1} - {t^{2} \over t^{2} + 1 - a^{2}}}\,\dd t
\\[5mm]&=2\int_{0}^{\infty}
\pars{-\,{1 \over t^{2} + 1} + {1 - a^{2} \over t^{2} + 1 - a^{2}}}\,\dd t
=-2\int_{0}^{\infty}{\dd t \over t^{2} + 1}
+2\root{1 - a^{2}}\int_{0}^{\infty}{\dd t \over t^{2} + 1}
\\[5mm]&=-2\pars{1 - \root{1 - a^{2}}}\
\underbrace{\int_{0}^{\infty}{\dd t \over t^{2} + 1}}
_{\ds{\color{#c00000}{\pi \over 2}}}\ =\
\color{#66f}{\large -\pi\pars{1 - \root{1 - a^{2}}}}
\end{align}
A: Hint: Alternately, try differentiating under the integral sign with regard to a.
