I want to evaluate this integral $\int_{0}^\frac{\pi}{2} \ln(1-a^2\sin^2x)dx$, where $|a|<1.$
First, I differentiate with respect to $a.$ Then, it would become a terrible integral. That is $$\int_{0}^\frac{\pi}{2} \frac{-2a\sin^2x}{1-a^2\sin^2x}dx.$$ After that, I think it may be down by cosine since it is symmetric with respect to $\frac{\pi}{4}$. However, it fails. I think that this integral may be done like $\sin^2x$.
Hope that teachers here could give me some hints to crack this done.
$$F(a)=\int_0^{\pi/2}\log\left[1-a^2\sin(x)^2\right]\mathrm dx\Rightarrow F(0)=0$$ $$F'(a)=\frac2a\int_0^{\pi/2}\frac{-a^2\sin(x)^2}{1-a^2\sin(x)^2}\mathrm dx$$ $$F'(a)=\frac2a\int_0^{\pi/2}\frac{1-a^2\sin(x)^2}{1-a^2\sin(x)^2}\mathrm dx-\frac2a\int_0^{\pi/2}\frac{\mathrm dx}{1-a^2\sin(x)^2}$$ $$F'(a)=\frac\pi a-\frac2a\int_0^{\pi/2}\frac{\mathrm dx}{1-a^2\sin(x)^2}$$ Then we consider $$J(a)=\int_0^{\pi/2}\frac{\mathrm dx}{1-a^2\sin(x)^2}$$ Let $x=t/2$ to get $$J(a)=\frac12\int_0^{\pi}\frac{\mathrm dt}{1-a^2\left(\frac{1-\cos t}2\right)}$$ $$J(a)=\int_0^{\pi}\frac{\mathrm dt}{a^2\cos t+2-a^2}$$ Then let $x=\tan(t/2)$ to get $$J(a)=2\int_0^\infty \frac1{a^2\frac{1-x^2}{1+x^2}+2-a^2}\frac{\mathrm dx}{1+x^2}$$ $$J(a)=2\int_0^\infty \frac{\mathrm dx}{a^2(1-x^2)+(2-a^2)(1+x^2)}$$ $$J(a)=\int_0^\infty \frac{\mathrm dx}{(1-a^2)x^2+1}$$ and since $$\int\frac{\mathrm dx}{ax^2+bx+c}=\frac2{\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}},\qquad \text{assumng}\ 4ac>b^2$$ We have $$J(a)=\frac\pi{2\sqrt{1-a^2}}$$ So we have $$F'(a)=\frac\pi a-\frac\pi{a\sqrt{1-a^2}}$$ Hence $$F(a)=\pi\int_0^a \frac{\sqrt{1-x^2}-1}{x\sqrt{1-x^2}}\mathrm dx$$ Setting $u=\sqrt{1-x^2}$, we have $$F(a)=\pi\int_1^{\sqrt{1-a^2}}\frac{u-1}{u\sqrt{1-u^2}}\frac{u\mathrm du}{\sqrt{1-u^2}}$$ $$F(a)=\pi\int_1^{\sqrt{1-a^2}}\frac{\mathrm du}{u+1}$$ $$F(a)=\pi\log\frac{1+\sqrt{1-a^2}}2$$
