TITLE: A definite integral­
QUESTION [5 upvotes]: $$\int_{-\frac\pi2}^{\frac\pi2}\frac{\ln(1+b\sin x)}{\sin x} dx \\|b|<1$$
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.

REPLY [5 votes]: $\newcommand{\+}{^{\dagger}}
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$\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
$$\color{#00f}{\large%
\int_{-\pi/2}^{\pi/2}{\ln\pars{1 + b\sin\pars{x}} \over \sin\pars{x}}\,\dd x
=\pi\,\arcsin\pars{b}}
$$