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
 A: This is quite similar to Random Variable's solution, just the starting integral is different to make the calculations a bit simpler. 
Consider
$$I(b)=\int_0^{\pi/2} \ln(b^2+\sin^2x)\,dx$$
$$\Rightarrow I'(b)=\int_0^{\pi/2} \frac{2b}{b^2+\sin^2x}\,dx=2b\int_0^{\pi/2} \frac{dx}{b^2+\cos^2x}$$
Factor out $\cos^2x$ from the denominator and rewrite $\sec^2x=1+\tan^2x$ to obtain:
$$I'(b)=2b\int_0^{\pi/2} \frac{\sec^2x\,dx}{b^2+1+b^2\tan^2x}\,dx$$
Use the substitution $\tan x=t$ and evaluating the resulting integral is easy so 
$$I'(b)=\frac{\pi}{\sqrt{1+b^2}} \Rightarrow I(b)=\pi\ln\left(b+\sqrt{1+b^2}\right)+C$$
For $b=0$, $I(0)=-\pi\ln 2$, hence $C=-\pi\ln2$
$$\Rightarrow \int_0^{\pi/2} \ln(b^2+\sin^2x)\,dx=\pi\ln\left(\frac{b+\sqrt{1+b^2}}{2}\right)$$
Replace $b$ with $1/\sqrt{\alpha}$ and you get:
$$\int_0^{\pi/2} \ln(1+\alpha \sin^2x)\,dx-\frac{\pi}{2}\ln \alpha=\pi\ln\left(\frac{1+\sqrt{1+\alpha}}{2\sqrt{\alpha}}\right)$$
$$\Rightarrow \int_0^{\pi/2} \ln(1+\alpha \sin^2x)\,dx=\pi\ln\left(\frac{1+\sqrt{1+\alpha}}{2}\right)$$
$\blacksquare$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{\pi/2}\ln\pars{1 + \alpha\sin^{2}\pars{x}}\,\dd x
     =\pi\,\ln\pars{1 + \root{1 + \alpha} \over 2}:\ {\large ?}}$

\begin{align}&
\partiald{}{\alpha}\color{#c00000}{\int_{0}^{\pi/2}\ln\pars{1 + \alpha\sin^{2}\pars{x}}\,\dd x}
=\partiald{}{\alpha}\int_{0}^{\pi/2}\ln\pars{1 + \alpha\cos^{2}\pars{x}}\,\dd x
\\[3mm]&=\int_{0}^{\pi/2}{\cos^{2}\pars{x} \over 1 + \alpha\cos^{2}\pars{x}}\,\dd x
={1 \over \alpha}\int_{0}^{\pi/2}
{\bracks{1 + \alpha\cos^{2}\pars{x}} - 1 \over 1 + \alpha\cos^{2}\pars{x}}\,\dd x
\\[3mm]&={\pi \over 2\alpha}-
{1 \over \alpha}\int_{0}^{\pi/2}{\dd x \over 1 + \alpha\cos^{2}\pars{x}}
={\pi \over 2\alpha}-
{1 \over \alpha}\int_{0}^{\pi/2}{\sec^{2}\pars{x}\,\dd x\over
\tan^{2}\pars{x} + 1 + \alpha}
\\[3mm]&={\pi \over 2\alpha} - {1 \over \alpha\root{1 + \alpha}}
\int_{0}^{\infty}{\dd t \over t^{2} + 1}
={\pi \over 2}\pars{{1 \over \alpha} - {1 \over \alpha\root{1 + \alpha}}}
\end{align}

\begin{align}&
\color{#c00000}{\int_{0}^{\pi/2}\ln\pars{1 + \alpha\sin^{2}\pars{x}}\,\dd x}
={\pi \over 2}\
\overbrace{\int_{0}^{\alpha}\pars{{1 \over t} - {1 \over t\root{1 + t}}}\,\dd t}
^{\ds{\mbox{Set}\ x \equiv 1 + \root{1 + t}\ \imp\ t = x^{2} - 2x}}
\\[3mm]&={\pi \over 2}\int_{2}^{1 + \root{1 + a}}{2\,\dd x \over x}
\end{align}

$$\color{#66f}{\large%
\int_{0}^{\pi/2}\ln\pars{1 + \alpha\sin^{2}\pars{x}}\,\dd x
=\pi\,\ln\pars{1 + \root{1 + \alpha} \over 2}}
$$

A: Differentiate $I_2$ (probably easier than $I_1$) with respect to $a$, then use the Weierstrass substitution to transform it into an integral that you can calculate with residues. I will look into it as well.
Edit: you can also integrate by parts to get rid of the log.
A: 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,
$$I(a) = \pi \ln \left(\frac{1 +\sqrt{1+a}}{2} \right) .$$
A: 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 $$ I(\alpha) =\pi \ln (1+\sqrt{1+\alpha})-\pi\ln 2 =\pi \ln \left(\frac{1+\sqrt{1+\alpha}}{2}\right) .$$
