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 $\pi \arcsin b$.
Best Answer
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
$$I(b)=\pi\arcsin(b)$$