Show that $$\int_{0}^\frac{\pi}{2} \ln(1+a\sin(x))\csc(x)dx = \frac{\arcsin(a)}{2}(\pi-\arcsin(a))$$ for $|a|\leq1$
I have been able to use Leibniz's rule to find a closed form for $a\geq1$, but I'm really struggling to find the above close form for $|a|\leq1$. I've tried series, but that didn't seem to get anywhere. And it seems that Leibniz's rule only gets the result for $a\geq1$.
Another form for the function is
$$\int_{0}^1 \frac{\ln\left(m^2+2mx+1\right)}{m}dm- \frac{\pi^2}{24}$$
I don't really want a direct answer (while that would be appreciated). Just a nudge in the right direction.
Edit: Also avoiding polylogarithms would be appreciated.
Best Answer
For simplicity we will use $\alpha =\sin a$ so let's consider:
$$I(a):=\int_0^\frac{\pi}{2}\frac{\ln(1+\sin a\sin x)}{\sin x}dx$$
Note that $\sin a$ is always inside $[-1,1]$ so it's equivalent to $|\alpha|\le 1$. Also put $x \to \pi-x$, then average the two integrals to see that: $$2\int_{0}^\frac{\pi}{2} \frac{\ln(1+\sin a\sin x)}{\sin x}dx=\int_{0}^{\pi} \frac{\ln(1+\sin a\sin x)}{\sin x}dx$$
$$\Rightarrow I(a)=\frac12 \int_{0}^\pi \frac{\ln(1+\sin a\sin x)}{\sin x}dx\Rightarrow I'(a)=\frac12 \int_0^{\pi}\frac{\cos a}{1+\sin a\sin x}dx$$ $$\overset{\tan \frac{x}{2}=t}=\int_0^\infty \frac{\cos a}{1+\sin a\frac{2t}{1+t^2}}\frac{1}{1+t^2}dt=\int_0^\infty \frac{\cos a}{(t+\sin a)^2+\cos^2 a}dt$$ $$=\arctan\left(\frac{t+\sin a}{\cos a}\right)\bigg|_0^\infty =\frac{\pi}{2}-a$$ Now we integrate to get back: $$I(a)=\int\left(\frac{\pi}{2}-a\right)da=\frac{\pi a}{2}-\frac{a^2}{2}+C$$ $$I(0)=0\Rightarrow C=0 \Rightarrow I(a)=\frac{a}{2}(\pi-a)$$ If you really want to have $\alpha$ instead of $\sin a$ then just put $a=\arcsin \alpha$ in the above result.