Determine the value of the integral $I=\int_{0}^{1}\frac{\ln\left(1-a^2x^2\right)}{\sqrt{1-x^2}}dx$ Determine the value of the integral $$I(a)=\int_{0}^{1}\frac{\ln\left(1-a^2x^2\right)}{\sqrt{1-x^2}}dx, \: |a|\leq 1$$ 
My try:
$\to I'(a)=\int_{0}^{1}\frac{-2ax^2}{\left(1-a^2x^2\right)\sqrt{1-x^2}}dx$
Set $x=\cos t\to dx=-\sin tdt$
Hence $I'(a)=\int_{0}^{\frac{\pi}{2}}\frac{-2a\cos^2t}{1-a^2\cos^2t}dt=\pi\left(\frac{1}{a}-\frac{1}{\sqrt{1-a^2}}\right)\to I(a)=\pi\left(\ln|a|-\arcsin a\right)+C$
Question: Find C?

$$\begin{align}I(a) &= -\sum_{k=1}^{\infty} \frac{a^{2 k}}{k} \int_0^1 dx \frac{x^{2 k}}{\sqrt{1-x^2}}\\ &= -\frac{\pi}{2}\sum_{k=1}^{\infty} \frac1{k} \binom{2 k}{k} \left (\frac{a}{2} \right )^{2 k} \end{align} $$
$$I'(a) = -\frac{\pi}{2} \sum_{k=1}^{\infty} \binom{2 k}{k} \left (\frac{a}{2} \right )^{2 k-1} $$
$$a I'(a) = -\pi \sum_{k=1}^{\infty} \binom{2 k}{k} \left (\frac{a}{2} \right )^{2 k} = -\pi \left (\frac1{\sqrt{1-a^2}} -1\right )$$
$$\implies \begin{align}I(a) &= -\pi \int da \left (\frac1{a\sqrt{1-a^2}}-\frac1{a} \right )\\ &= \pi \log{\left (1+\sqrt{1-a^2}\right )} +C \end{align}$$
$$I(0)=0 \implies C=-\pi \log{2}$$
$$\therefore I(a) = \pi \log{\left (\frac{1+\sqrt{1-a^2}}{2} \right )}$$
To derive the result for the integral in the first line, see here.
