Don't love me, no more

Let $\displaystyle I(a) = \int_{0}^{\frac{\pi}{2}} \frac{\ln(1+a\sin^{2}(x))}{\sin^{2}(x)}dx$

Differentiating with respect to $a$

So $\displaystyle I'(a) = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1+a\sin^{2}(x)} =\int_{0}^{\frac{\pi}{2}} \frac{dx}{1+a\cos^{2}(x)}$

Multiplying and dividing by $\sec^{2}(x)$

$\displaystyle I'(a)=\int_{0}^{\frac{\pi}{2}} \frac{\sec^{2}(x)}{1+\tan^{2}(x)+a}dx$

Substituting $\displaystyle \tan(x) = z$

$\displaystyle I'(a)=\int_{0}^{\infty} \frac{1}{1+z^{2}+a}dz = \frac{1}{\sqrt{1+a}}\frac{\pi}{2}$

( The integral $\displaystyle \int\frac{1}{a^{2}+z^{2}}dz$ is well known and equals $\displaystyle \frac{1}{a}\tan^{-1}(\frac{z}{a})$ without the integration constant )

So $\displaystyle I(a) = \pi\sqrt{a+1} + C$

Now we have $\displaystyle I(0) = 0$

So $\displaystyle C = -\pi$

So $I(1) = (\sqrt{2}-1)\pi$ which is our answer.

Relevant wiki Differentiation under the integral sign/Leibnitz Rule