A definite integral­ $$\int_{-\frac\pi2}^{\frac\pi2}\frac{\ln(1+b\sin x)}{\sin x} dx \\|b|<1$$
I tried putting $-x$ using properties of definite integral, but that doesn't really help.
$$I=\int_{-\frac\pi2}^{\frac\pi2}\frac{-\ln(1-b\sin x)}{\sin x}dx$$
I don't think adding these 2 equations would yield anything. And I just cannot think of a good substitution.
Edit : Plugging b=1, it gives 4.9xxx . as $\pi ^2 \approx 10$, the value is $\pi \arcsin b$. Other values for b confirm this.
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$\ds{\mbox{Let's}\ {\cal I}\pars{b} \equiv
    \int_{-\pi/2}^{\pi/2}{\ln\pars{1 + b\sin\pars{x}} \over \sin\pars{x}}
     \,\dd x:\ {\large ?}}$ with $\ds{b \in {\mathbb R}\,,\ \verts{b} < 1.\quad}$
$\ds{{\cal I}\pars{0} = 0}$

With
  Weierstrass Substitution $\ds{t \equiv \tan\pars{x \over 2}}$:
  \begin{align}
{\cal I}'\pars{b}&=\int_{-\pi/2}^{\pi/2}{\dd x \over 1 + b\sin\pars{x}}
=\int_{-1}^{1}{2\,\dd t/\pars{1 + t^{2}} \over 1 + b\bracks{2t/\pars{1 + t^{2}}}}
=2\int_{-1}^{1}{\dd t \over t^{2} + 2bt + 1}
\\[3mm]&=2\int_{-1}^{1}{\dd t \over \pars{t + b}^{2} + 1 - b^{2}}
=2\int_{-1 + b}^{1 + b}{\dd t \over t^{2} + 1 - b^{2}}
\\[3mm]&={2 \over \root{1 - b^{2}}}
\int_{\pars{-1 + b}/\root{1 - b^{2}}}^{\pars{1 + b}/\root{1 - b^{2}}}
{\dd t \over t^{2} + 1}
\\[3mm]&={2 \over \root{1 - b^{2}}}\bracks{%
\arctan\pars{1 + b \over \root{1 - b^{2}}}
-\arctan\pars{-1 + b \over \root{1 - b^{2}}}}
\\[3mm]&={2 \over \root{1 - b^{2}}}\bracks{%
{\pi \over 2} - \arctan\pars{\root{1 - b^{2}} \over 1 + b}
-\arctan\pars{-1 + b \over \root{1 - b^{2}}}}
\\[3mm]&={2 \over \root{1 - b^{2}}}\braces{%
{\pi \over 2} - \overbrace{%
\bracks{\arctan\pars{1 - b \over \root{1 - b^{2}}}
+\arctan\pars{-1 + b \over \root{1 - b^{2}}}}}^{\ds{=\ 0}}}
\\[3mm]&={\pi \over \root{1 -b^{2}}}
\end{align}

Then
$$\color{#00f}{\large%
\int_{-\pi/2}^{\pi/2}{\ln\pars{1 + b\sin\pars{x}} \over \sin\pars{x}}\,\dd x
=\pi\,\arcsin\pars{b}}
$$
A: I have been trying to compute the antiderivative; I tried several change of variable with absolutely nos success; as already mentioned, integration by parts makes the problem still more complex.
Using a CAS (brut force), the antiderivative which came out is just a nightmare for me $$-\text{Li}_2\left(\frac{\tan
   \left(\frac{x}{2}\right)}{\sqrt{b^2-1}-b}\right)-\text{Li}_2\left(-\frac{\tan
   \left(\frac{x}{2}\right)}{b+\sqrt{b^2-1}}\right)+\log \left(\tan
   \left(\frac{x}{2}\right)\right) \left(-\log \left(\frac{\tan
   \left(\frac{x}{2}\right)}{\sqrt{b^2-1}+b}+1\right)-\log
   \left(\left(\sqrt{b^2-1}+b\right) \tan \left(\frac{x}{2}\right)+1\right)+\log (b
   \sin (x)+1)+\log \left(1-i \tan \left(\frac{x}{2}\right)\right)+\log \left(1+i
   \tan \left(\frac{x}{2}\right)\right)\right)+\frac{1}{2} \text{Li}_2\left(-\tan
   ^2\left(\frac{x}{2}\right)\right)$$  Computing the integral leads, after simplifications, to $$-\text{Li}_2\left(\frac{1}{\sqrt{b^2-1}-b}\right)-\frac{1}{2} \text{Li}_2\left(\frac{1}{\left(b+\sqrt{b^2-1}\right)^2}\right)+2
   \text{Li}_2\left(\frac{1}{b+\sqrt{b^2-1}}\right)+\text{Li}_2\left(b+\sqrt{b^2-1}\right) $$ which is a real valued function of $b$. Trying to use the definition and properties of the dilogarithm took me to a dead end.
Thanks to your edit and clever remark, I verified that the result is exactly what you said, that is to say that $$\int_{-\frac\pi2}^{\frac\pi2}\frac{\ln(1+b\sin x)}{\sin x} dx = \pi \sin ^{-1}(b)$$
I checked that the Taylor series built at $b=0$ are identical.
Congratulations for having found the result. I am really sorry to have not been able to simplify these monsters.
Added later after Lucian's suggestion
After Lucian's so good suggestion, let us consider $$I'(b)=\int_{-\frac\pi2}^{\frac\pi2}\frac{dx}{1+b\sin x}$$ First of all, the antiderivative is given by $$\frac{2 \tan ^{-1}\left(\frac{b+\tan
   \left(\frac{x}{2}\right)}{\sqrt{1-b^2}}\right)}{\sqrt{1-b^2}}$$ Integrated between the given bounds $$I'(b)=\frac{2 \left(\tan ^{-1}\left(\sqrt{\frac{1-b}{1+b}}\right)+\tan
   ^{-1}\left(\sqrt{\frac{1+b}{1-b}}\right)\right)}{\sqrt{1-b^2}}=\frac{\pi }{\sqrt{1-b^2}}$$   and the final result for $I(b)$.
A: From the Taylor series expansion of $\ln(1+x)$ we know that
$$\ln(1+b\sin x) = b\sin x - \frac{b^2\sin^2x}{2} +\frac{b^3\sin^3x}{3} - \cdots$$
Therefore
$$\frac{\ln(1+b\sin x)}{\sin x} = b - \frac{b^2\sin x }{2} +\frac{b^3\sin^2 x }{3}-\cdots$$
Integrating both sides of the equation
$$\int^{\pi/2}_{-\pi/2}\frac{\ln(1+b\sin x )}{\sin x }\mathrm{d}x = \int^{\pi/2}_{-\pi/2} b - \frac{b^2\sin x }{2} +\frac{b^3\sin^2 x }{3}-\cdots \mathrm{d}x$$
Observe that $\int^{\pi/2}_{-\pi/2}\sin^nx\mathrm{d}x = 0$ if n is an odd number. The above equation is then equivalent to
$$\int^{\pi/2}_{-\pi/2}\frac{\ln(1+b\sin x )}{\sin x }\mathrm{d}x = \int^{\pi/2}_{-\pi/2} b  +\frac{b^3\sin^2 x }{3}+\frac{b^5\sin^4 x }{5}\cdots \mathrm{d}x$$
As $\sin^{2n}x$ is an even function therefore
$$\int^{\pi/2}_{-\pi/2}\frac{\ln(1+b\sin x )}{\sin x }\mathrm{d}x =2 \int^{\pi/2}_0 b  +\frac{b^3\sin^2 x }{3}+\frac{b^5\sin^4 x }{5}\cdots \mathrm{d}x$$
Using the fact that $\int_0^{\pi/2}\sin^nx\mathrm{d}x = \frac{(n-1)\cdot(n-3)\cdots3\cdot1}{n\cdot(n-2)\cdots4\cdot2}$ if n is even (See this page for the proof)
$$\int^{\pi/2}_{-\pi/2}\frac{\ln(1+b\sin x )}{\sin x }\mathrm{d}x =2\cdot\frac{\pi}{2}\Bigg[ b+\frac{b^3}{3}\frac{1}{2}+\frac{b^5}{5}\frac{3\cdot1}{4\cdot2}\cdots\Bigg]$$
$$=\pi\Bigg[ b+\frac{b^3}{3}\frac{1}{2}+\frac{b^5}{5}\frac{3\cdot1}{4\cdot2}\cdots\Bigg]$$
The term in the parenthesis is precisely the Taylor series expansion (Thanks to your edit. I was completely clueless at this point) for $\sin^{-1} b $. Therefore the integral is 
$$\int^{\pi/2}_{-\pi/2}\frac{\ln(1+b\sin x )}{\sin x }\mathrm{d}x = \pi\sin^{-1} b $$
