A Trig Integral

Question

Does the integral \begin{align} \int_{0}^{\pi/2} \cos(x) \, \ln\left( \frac{1 + a^{2} \sin(x)}{1 - a^{2} \sin(x)} \right) \, dx \end{align} have a closed form and what is changed if the limits are changed to $(- \frac{\pi}{2}, \frac{\pi}{2})$ ?

Answer 1

First step I'd try is to use the properties of logarithms.

$$\int_0^\frac{\pi}2\cos x[\ln(1+a^2\sin x)-\ln(1-a^2\sin x)]dx$$ $$u=a^2\sin x,du=a^2\cos xdx$$ $$\frac1{a^2}\int_0^{a^2}\ln(1+u)-\ln(1-u)du=$$ $$\frac1{a^2}\left[(1+u)\ln(1+u)-(1+u)+(1-u)\ln(1-u)-(1-u)\right]|_0^{a^2}=$$ $$\frac1{a^2}\left[(1+a^2)\ln(1+a^2)+(1-a^2)\ln(1-a^2)-2+2\right]=$$ $$\frac{(1+a^2)\ln(1+a^2)+(1-a^2)\ln(1-a^2)}{a^2}$$

If the lower limit is changed to $-\frac{\pi}2$, the result is $0$. You are integrating an odd function.