$\int_{0}^{\pi/2}\ln\left(1+4\sin^4 x\right)\mathrm{d}x$ and the golden ratio We already know that, for any real number $t$ such that $t\geq-1$,

$$
\int_{0}^{\pi/2} \ln \left(1+t \sin^2 x\right) \mathrm{d}x = \pi \ln \left( \frac{1+\sqrt{1+t}}{2} \right).
$$

Prove that

$$
\int_{0}^{\pi/2} \ln \left(1+4\sin^4 x\right) \mathrm{d}x = \pi \ln \left( \frac{\varphi+\sqrt{\varphi}}{2} \right)
$$ 
  where $\displaystyle \varphi = \frac{1+\sqrt{5}}{2}$ is the golden ratio.

 A: Hint: $\quad1+t^2\sin^4x~=~(1-ti\sin^2x)\cdot(1+ti\sin^2x),\qquad\log(ab)=\log a+\log b,\quad$ and 
$\Re(\pm~ti)>-1$.
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{\pi/2}\ln\pars{1 + 4\sin^{4}\pars{x}}\,\dd x
     =\pi\ln\pars{\varphi + \root{\varphi} \over 2}:\ {\large ?}}$

\begin{align}&
\partiald{}{\mu}\int_{0}^{\pi/2}\ln\pars{1 + \mu\sin^{4}\pars{x}}\,\dd x
=\int_{0}^{\pi/2}{\sin^{4}\pars{x} \over 1 + \mu\sin^{4}\pars{x}}\,\dd x
\\[3mm]&={\pi \over 2\mu} -{1 \over \mu}\color{#c00000}{\int_{0}^{\pi/2}
{\dd x \over 1 + \mu\sin^{4}\pars{x}}}
\end{align}

\begin{align}&\color{#c00000}{\int_{0}^{\pi/2}%
{\dd x \over 1 + \mu\sin^{4}\pars{x}}}
=\int_{0}^{\pi/2}{\csc^{4}\pars{x} \over \csc^{4}\pars{x} + \mu}\,\dd x
=\ \overbrace{\int_{0}^{\pi/2}%
{\cot^{2}\pars{x} + 1  \over \bracks{\cot^{2}\pars{x} + 1}^{2}+ \mu}
\,\csc^{2}\pars{x}\,\dd x}^{\ds{\cot\pars{x} \equiv t}}
\\[3mm]&=-\int_{\infty}^{0}{t^{2} + 1 \over \pars{t^{2} + 1}^{2} + \mu}\,\dd t
=\Re\int_{0}^{\infty}{\dd t \over t^{2} + 1 + \root{\mu}\ic}
\\[3mm]&=\Re\bracks{{1 \over \root{1 + \root{\mu}{\ic}}}
\int_{0}^{\infty/\root{1 + \root{\mu}{\ic}}}{\dd t \over t^{2} + 1}}
={\pi \over 2}\,\Re\pars{{1 \over \root{1 + \root{\mu}{\ic}}}}
\end{align}

\begin{align}&\color{#66f}{%
\large\int_{0}^{\pi/2}\ln\pars{1 + 4\sin^{4}\pars{x}}\,\dd x}
={\pi \over 2}\,\Re\int_{0}^{4}\overbrace{%
{1 \over \mu}\pars{1 - {1 \over \root{1 + \root{\mu}\ic}}}\,\dd\mu}
^{\ds{t \equiv \root{1 + \root{\mu}\ic}}}
\\[3mm]&=2\pi\,\Re\int_{1}^{\root{1 + 2\,\ic}}{\dd t \over 1 + t}
=\color{#66f}{\large 2\pi\,\Re\ln\pars{1 + \root{1 + 2\ic} \over 2}}
\approx 1.1565078476153109133
\end{align}

It agrees with the OP proposed answer
$\color{#000}{\large\quad\ds{\pi\,\ln\pars{\varphi + \root{\varphi} \over 2}}}$.
A: Let's put it with real numbers.
We have

$$
\int_{0}^{\pi/2} \ln \left(1+t\sin^4 x\right) \mathrm{d}x 
= \pi \ln \left( \frac{1}{4} \sqrt{ 1 + \sqrt{1+t} } 
\left( \sqrt{1 + \sqrt{1+t} } + \sqrt{2} \right) \right), \quad t \geq -1.
$$ 

