Integrate then differentiate?

$\frac{dI}{da}=\int_0^{\pi/2} \frac{\partial}{\partial a} \ln{\frac{1+a\sin{x}}{1-a\sin{x}}}\frac{dx}{\sin{x}}=\int_0^{\pi/2} \frac{1-a\sin{x}}{1+a\sin{x}}\frac{\sin{x}-(-\sin{x})}{(1-a\sin{x})^2}\frac{dx}{\sin{x}}=\int_0^{\pi/2} \frac{2dx}{(1+a\sin{x})(1-a\sin{x})}=\int_0^{\pi/2} \Big(\frac{1}{1+a\sin{x}}+\frac{1}{1-a\sin{x}}\Big)dx$ Then I perform the tangent half-angle substitution $t=\tan{\frac{u}{2}}$ after the substitution $u=\frac{\pi}{2}-x$ . $=\int_0^{\pi/2}\Big(\frac{1}{1+a\cos{u}}+\frac{1}{1-a\cos{u}}\Big)du=\int_0^1 \Big(\frac{1}{1+a\frac{1-t^2}{1+t^2}}+\frac{1}{1-a\frac{1-t^2}{1+t^2}}\Big)\frac{2}{1+t^2}dt=2\int_0^1\Big(\frac{1}{1+a+(1-a)t^2}+\frac{1}{1-a+(1+a)t^2}\Big)dt$ $=\frac{2}{\sqrt{1-a^2}}\Big[\arctan{\sqrt{\frac{1+a}{1-a}}t}+\arctan{\sqrt{\frac{1-a}{1+a}}t}\Big]_0^1=\frac{2}{\sqrt{1-a^2}}\Big(\arctan{\sqrt{\frac{1+a}{1-a}}}+\arctan{\sqrt{\frac{1-a}{1+a}}}\Big)=\frac{\pi}{\sqrt{1-a^2}}$