<s>[INST] 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
 [/INST] Let $ \displaystyle I(a) =  \int_{0}^{\pi /2} \ln(1+ a \sin^{2}x) \, dx$.
Then differentiating under the integral sign, $$I'(a) = \int_{0}^{\pi /2} \frac{\sin^{2} x}{1+a \sin^{2} x} \, dx = \int_{0}^{\pi /2} \frac{1}{a+ \csc^{2} x} \, dx .$$
Now let $u = \cot x$.
Then
$$ \begin{align} I'(a) &= \int_{0}^{\infty} \frac{1}{a+1+u^{2}} \frac{1}{1+u^{2}} \, du \\ &= \frac{1}{a} \int_{0}^{\infty} \left(\frac{1}{1+u^{2}} - \frac{1}{1+a+u^{2}} \right) \, du \\ &= \frac{1}{a} \left(\frac{\pi}{2} - \frac{1}{1+a} \int_{0}^{\infty} \frac{1}{1+\frac{u^{2}}{1+a}} \, du \right)  \\ &=\frac{1}{a} \left(\frac{\pi}{2} - \frac{1}{\sqrt{1+a}} \int_{0}^{\infty} \frac{1}{1+v^{2}} \, dv \right)  \\  &= \frac{\pi}{2a} \left(1 - \frac{1}{\sqrt{1+a}} \right). \end{align}$$
Then integrating back,
$$ \begin{align} I(a) &= \frac{\pi}{2} \int \frac{1}{a} \left(1 - \frac{1}{\sqrt{1+a}} \right) \, da \\ &= \frac{\pi}{2} \int \frac{1}{u^{2}-1} \left(1 - \frac{1}{u} \right) 2u \, du \\ &= \pi \int \frac{1}{1+u} \, du \\ &= \pi \ln \left(1+ \sqrt{1+a} \right) + C. \end{align}$$
And since $I(0) = 0$, $C = -\pi \ln 2$.
Therefore,
$$I(a) = \pi \ln \left(\frac{1 +\sqrt{1+a}}{2} \right) .$$
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