Using differentiation under integral sign evaluate $\int_{0}^{\frac {\pi}{2}} \ln (\frac {a+b\sin x}{a-b\sin x}) \cdot \frac {dx}{\sin x} $

Question

Using differentiation under integral sign evaluate $\int_{0}^{\frac {\pi}{2}} \ln (\frac {a+b\sin x}{a-b\sin x}) \cdot \frac {dx}{\sin x} $

My Attempt: Given $$\int_{0}^{\frac {\pi}{2}} \ln (\frac {a+b\sin x}{a-b\sin x}) \cdot \frac {dx}{\sin x}$$ Let, $$F(a)=\int_{0}^{\frac {\pi}{2}} \ln (\frac {a+b\sin x}{a-b\sin x}) \cdot \frac {dx}{\sin x}$$ Differentiating both sides w.r.t $a$ $$\frac {dF(a)}{da}=\int_{0}^{\frac {\pi}{2}} \frac {a-b\sin x}{a+b\sin x} \times \frac {(a-b\sin x)-(a+b\sin x)}{(a-b\sin x)^2} \times \frac {dx}{\sin x}$$ $$\frac {dF(a)}{da} = \int_{0}^{\frac {\pi}{2}} \frac {-2b\sin x}{a+b\sin x} \times \frac {dx}{\sin x}$$ $$\frac {dF(a)}{da} = -2b \int_{0}^{\frac {\pi}{2}} \frac {dx}{a+b\cdot \frac {2\tan (\frac {x}{2})}{1+\tan^2 (\frac {x}{2})}}$$ $$\frac {dF(a)}{da} = -\frac {2b}{a} \int_{0}^{\frac {\pi}{2}} \frac {\sec^2 (\frac {x}{2})}{\tan^2 (\frac {x}{2}) + \frac {2b}{a} \tan (\frac {x}{2})+1} dx$$ $$\textrm {put} \tan (\frac {x}{2}) = t$$ $$\textrm {so}, \sec^2 (\frac {x}{2}) dx = 2dt$$ $$\textrm {Also, the limit of integration becomes 0 t0 1}$$ Now, $$\frac {dF(a)}{da} = -\frac {4b}{a} \int_{0}^{1} \frac {dt}{t^2 + \frac {2b}{a} t +1 } $$ $$\frac {dF(a)}{da} = -\frac {4b}{a} \times \frac {a}{\sqrt {a^2 - b^2}} [\tan^{-1} \frac {1+\frac {b}{a}}{\sqrt {\frac {a^2 -b^2}{a^2}}} - \tan^{-1} \frac {\frac {b}{a}}{\sqrt {\frac {a^2 - b^2}{a^2}}}]$$ How do I evaluate further?

Answer 1

If I am not mistaken $$F(a)=\int\log \left(\frac {a+b\sin x}{a-b\sin x}\right) \, \frac {dx}{\sin x} \implies \frac {dF(a)}{da} =2b\int \frac{dx}{b^2 \sin ^2(x)-a^2}$$ Now, use $\tan(x)=t$ to get $$\frac {dF(a)}{da}= 2b \int \frac{dt} { \left(b^2-a^2\right)t^2-a^2}$$ which is not bad (take care about the bounds).