Integral $\int_0^{\pi/2} \ln(1+\alpha\sin^2 x)\, dx=\pi \ln \frac{1+\sqrt{1+\alpha}}{2}$ $$
I_1:=\int_0^{\pi/2} \ln(1+\alpha\sin^2 x)\, dx=\pi \ln \frac{1+\sqrt{1+\alpha}}{2}, \qquad \alpha \geq -1.
$$
I am trying to prove this integral $I_1$.  We can write 
$$
\int_0^{\pi/2} \ln(\alpha(1/\alpha+\sin^2 x))dx=\int_0^{\pi/2} \left(\ln \alpha+\ln (\frac{1}{\alpha}+\sin^2 x)\right)dx=\frac{\pi}{2} \ln \alpha+I_2
$$
where
$$
I_2=\int_0^{\pi/2}\ln (\frac{1}{\alpha}+\sin^2 x) \,dx
$$
however I am not sure what that will do for us....  I also tried differentiating wrt $\alpha$ but didn't get placed.  How can we prove $I_1$ result?  Thanks

Using Feynman’s Technique Integration, we first differentiate the integral w.r.t. $\alpha$ and obtain
$$
\begin{aligned}
I^{\prime}(\alpha) &=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} x}{1+\alpha \sin ^{2} x} d x \\
&=\frac{1}{\alpha} \int_{0}^{\frac{\pi}{2}} \frac{1+\alpha \sin ^{2} x-1}{1+\alpha \sin ^{2} x} d x \\
&=\frac{\pi}{2 \alpha}-\frac{1}{\alpha} \int_{0}^{\frac{\pi}{2}} \frac{d x}{1+\alpha \sin ^{2} x} \\
&=\frac{\pi}{2 \alpha}-\frac{1}{\alpha} \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{\sec ^{2} x+\alpha \tan ^{2} x} d x \\
&=\frac{\pi}{2 \alpha}-\frac{1}{\alpha} \int_{0}^{\infty} \frac{d t}{1+(1+\alpha) t^{2}}, \text { where } t=\tan x \\
&=\frac{\pi}{2 \alpha}-\frac{1}{\alpha \sqrt{1+\alpha}}\left[\tan ^{-1}(\sqrt{1+\alpha }t)\right]_{0}^{\infty} \\
&=\frac{\pi}{2 \alpha}-\frac{\pi}{2 \alpha \sqrt{1+\alpha}}
\end{aligned}
$$
Integrating give back the integal
$$
\begin{aligned}
I(\alpha) &=\frac{\pi}{2} \ln |\alpha|-\frac{\pi}{2} \int \frac{d \alpha}{\alpha \sqrt{1+\alpha}} \\
&=\frac{\pi}{2} \ln |\alpha|-\pi\int \frac{d \sqrt{1+\alpha}}{(\sqrt{1+\alpha})^{2}-1}\\
&=\frac{\pi}{2}\left[\ln |\alpha|+\ln \left(\frac{\sqrt{1+\alpha}+1}{\sqrt{1+\alpha}-1}\right) \right ]+C\\
&=\frac{\pi}{2}\left[\ln |\alpha|+\ln \left(\frac{\sqrt{1+\alpha}+1)^{2}}{\mid \alpha \mid}\right) \right]+C\\
&=\pi \ln (\sqrt{1+\alpha}+1)+C
\end{aligned}
$$As $I(0)=0$ gives the value $C=-\pi\ln2 $ and hence we can conclude that $$ I(\alpha) =\pi \ln (1+\sqrt{1+\alpha})-\pi\ln 2 =\pi \ln \left(\frac{1+\sqrt{1+\alpha}}{2}\right) .$$
