<s>[INST] 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.
 [/INST] 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 
$$\int^{\pi/2}_{-\pi/2}\frac{\ln(1+b\sin x )}{\sin x }\mathrm{d}x = \pi\sin^{-1} b $$
</s>