Explicit form of the function $F\left(p\right)=\int_{0}^{\, \pi/2}\ln\left(1+p\sin^2\left(t\right)\right)\text{d}t$ Let $x \in \mathbb{R}^{+}$, I wonder how to prove that
$$
F\left(x\right)=\int_{0}^{\, \pi/2}\ln\left(1+x\sin^2\left(t\right)\right)\text{d}t=\pi \ln\left(\frac{1}{2}\left(1+\sqrt{1+x}\right)\right)
$$
It would beautifully show that
$$
\int_{0}^{\, \pi/2}\ln\left(1+4\sin^2\left(t\right)\right)\text{d}t=\pi \ln\left(\varphi\right)
$$
I've no idea how to proceed, except maybe that
$$
F\left(x\right)=\pi\ln\left(\sqrt{1+x}\right)-x\int_{0}^{\, \pi/2}\frac{\sin\left(2t\right)}{1+x\sin^2\left(t\right)}\text{d}t
$$
Then I guess the second will be the missing part, but how to compute it easily ?
 A: Following Jack D'Aurizio's idea: differentiate under the integral sign to get:
\begin{align}
F'(x)=\int^{\pi/2}_0 \frac{\sin^2(t)}{1+x\sin^2(t)}\,dt
\end{align}
This can be easily done by setting $t=\arctan(z)$:
\begin{align}
F'(x)=\int^\infty_0 \frac{z^2}{(1+(1+x)z^2)(1+z^2)}\,dz
\end{align}
Which can be easily solved by partial fraction decomposition or Residue Theorem to get:
\begin{align}
F'(x)=\frac{\pi}{2+2\sqrt[]{1+x}+2x}
\end{align}
Integrating this back for example with substitution $1+x=u^2$ you finally get:
\begin{align}
F(x)=\pi\ln\left(1+\sqrt[]{x+1}\right)+C
\end{align}
By setting $x=0$ and noticing that $\ln(1)=1$ we get: 
\begin{align}
\pi\ln(2)+C=0
\end{align}
Hence $C=\pi\ln(1/2)$. So:

\begin{align}
F(x)=\int^{\pi/2}_0 \ln\left(1+x\sin^2(t)\right)\,dt=\pi\ln\left( \frac{1+\sqrt[]{1+x}}{2}\right)
\end{align}

A: Consider
$$ F'(x) =\int_0^{\pi/2} \frac{\sin^2 t}{1+x\sin^2t}dt = \int_0^{\pi/2} \frac{1}{\csc^2t + x}dt $$
Let $u = \cot t$
$$ \begin{align} 
F'(x) &= \int_0^\infty \frac{1}{(u^2+1)(u^2+1+x)}du \\
&= \frac{1}{x}\int_0^\infty  \left(\frac{1}{u^2+1} - \frac{1}{u^2+1+x} \right) du \\
&= \frac{1}{x}\left(\frac{\pi}{2} - \frac{\pi}{2\sqrt{1+x}}\right) \\
&= \frac{\pi}{2\sqrt{1+x}(\sqrt{1+x}+1)}
\end{align} $$
Since $F(0)=0$, we have
$$ \begin{align} 
F(x) &= \frac{\pi}{2}\int_0^x \frac{1}{\sqrt{1+t}(\sqrt{1+t}+1)}dt \\
&= \pi \int_1^{\sqrt{1+x}} \frac{1}{1+u}du \\
&= \pi\ln \left(\frac{1 + \sqrt{1+x}}{2}\right) 
\end{align} $$
