Finding $ \int_{-\pi /2}^{\pi /2} \frac{\log (1 + b \sin x)}{\sin x}\,\mathrm dx$ given $|b|<1$ 
Find $$\int_{-\pi/2}^{\pi/2}\frac{\log(1+b\sin x)}{\sin x}\,\mathrm dx$$given that $|b|<1$.  

I split the integral into$$I=\int_0^{\pi/2}f(x)\,\mathrm dx+\int_{-\pi/2}^0f(x)\,\mathrm dx$$
For the second term made the substitution $x =-t$ and further solved $I$ to get$$I=\int_{0} ^{\pi / 2} \frac{\log \frac{1 + b \sin x}{1- b \sin x}}{\sin x}\,\mathrm dx$$ 
I do not know how to proceed further. The answer is $\pi \arcsin b$.
 A: Let $I(b)=\int_{-\pi/2}^{\pi/2}\frac{\log(1+b\sin(x))}{\sin(x)}\,dx$.  Differentiating reveals
$$\begin{align}
I'(b)&=\int_{-\pi/2}^{\pi/2} \frac1{1+b\sin(x)}\,dx\\\\
&=2\left(\frac{\arctan\left(\sqrt{\frac{1+b}{1-b}}\right)+\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)}{\sqrt{1-b^2}}\right)\\\\
&=\frac{\pi}{\sqrt{1-b^2}}\tag1
\end{align}$$
Using $I(0)=0$ and integrating $(1)$ yields
$$I(b)=\pi\arcsin(b)$$
A: This is not an answer but a curiosity
Using the tangent half-angle substitution $x=2\tan^{-1}(t)$, we have
$$I=\int_0^{\frac \pi 2}\csc (x) \log \left(\frac{1+b \sin (x)}{1-b \sin (x)}\right)=\int_0^1\frac{1}{t}\log \left(1+\frac{4 b t}{t^2-2 b t+1}\right)\,dt$$ Using Taylor series
$$\frac{1}{t}\log \left(1+\frac{4 b t}{t^2-2 b t+1}\right)=2\sum_{n=0}^\infty \frac {a_+^{2 n+1}+a_-^{2 n+1}}{2n+1} t^{2n}\quad \text{where}\quad \color{blue}{a_\pm=b\pm\sqrt{b^2-1}}$$
$$I=2\sum_{n=0}^\infty  \frac {a_+^{2 n+1}+a_-^{2 n+1}}{(2n+1)^2}$$
$$I=\color{blue}{\frac 12\Bigg[a_+ \Phi
   \left(a_+^2,2,\frac{1}{2}\right)+a_- \Phi
   \left(a_-^2,2,\frac{1}{2}\right)\Bigg]}$$ where appears  the Lerch transcendent function.
I have not be able to simplify this expression
