Integrating by parts 3 times,
$$ \begin{align} \int_{0}^{\pi /2} \frac{x^{4}}{\sin^{4} x} \ dx &= - \frac{x^{4}}{3} \cot(x) \left(\csc^{2} (x) +2 \right) \Bigg|^{\pi/2}_{0} + \frac{4}{3} \int_{0}^{\pi /2} x^{3} \cot (x) \left(\csc^{2} (x) +2 \right) \ dx \\ &= \frac{4}{3} \int_{0}^{\pi /2} x^{3} \cot (x) \left(\csc^{2} (x) +2 \right) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx + \frac{4}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \csc^{2}(x) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx - \frac{2}{3}x^{3} \cot^{2}(x) \Bigg|^{\pi/2}_{0} + 2 \int_{0}^{\pi /2} x^{2} \cot^{2}(x) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx + 2 \int_{0}^{\pi /2} x^{2} \cot^{2} (x) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx -2x^{2} \Big( x + \cot(x) \Big) \Bigg|^{\pi/2}_{0} +4 \int_{0}^{\pi /2} x\Big(x+ \cot(x) \Big) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx - \frac{\pi^{3}}{4} + 4 \int_{0}^{\pi /2} x^{2} \ dx + 4 \int_{0}^{\pi /2} x \cot(x) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx - \frac{\pi^{3}}{12} + 4 \int_{0}^{\pi /2} x \cot(x) \ dx . \end{align}$$
In general, $$ \int_{a}^{b} f(x) \cot(x) \ dx = 2 \sum_{n=1}^{\infty} \int_{a}^{b} f(x) \sin (2nx) \ dx .$$
See here.
So $$ \begin{align} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx &= 2 \sum_{n=1}^{\infty} \int_{0}^{\pi /2} x^{3} \sin (2nx) \ dx \\ &= 2 \sum_{n=1}^{\infty} \left(\frac{(-1)^{n-1} \pi^{3}}{16n} - \frac{(-1)^{n-1} 3\pi}{8n^{3}} \right) \\ &= \frac{\pi^{3}}{8} \ln (2) - \frac{3 \pi}{4} \eta(3) \\ &= \frac{\pi^{3}}{8} \ln (2) - \frac{9 \pi }{16} \zeta(3). \end{align}$$
And
$$ \begin{align} \int^{\pi /2}_{0} x \cot(x) \ dx &= 2 \sum_{n=1}^{\infty} \int_{0}^{\pi /2} x \sin(2nx) \ dx \\ &= -\frac{\pi}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \\ &= \frac{\pi \ln 2}{2} . \end{align}$$
Therefore,
$$ \begin{align} \int_{0}^{\pi /2} \frac{x^{4}}{\sin^{4} x} \ dx &= \frac{8}{3} \left( \frac{\pi^{3}}{8} \ln (2) - \frac{9 \pi }{16} \zeta(3) \right) - \frac{\pi^{3}}{12} + 4 \left(\frac{\pi \ln 2}{2} \right) \\ &= - \frac{\pi^{3}}{12} + 2 \pi \ln(2) + \frac{\pi^{3}}{3} \ln (2) - \frac{3 \pi}{2} \zeta(3) . \end{align}$$
Let's consider the integral
\begin{align}I(\alpha)&=\int_0^{\Large\frac{\pi}{2}}\frac{\ln\,(1+\cos\alpha\,\sin\,\phi)}{\sin\,\phi}\;d\phi\quad\Rightarrow\quad\phi\mapsto \frac{\pi}{2}-\phi\\
&=\int_0^{\Large\frac{\pi}{2}}\frac{\ln\,(1+\cos\alpha\,\cos\,\phi)}{\cos\,\phi}\;d\phi, \qquad 0 < \alpha < \pi.\end{align}
Differentiating $I(\alpha)$ with respect to $\alpha$, we have
\begin{align}
{I}'(\alpha) &= \int_0^{\Large\frac{\pi}{2}} \frac{\partial}{\partial\alpha} \left(\frac{\ln(1 + \cos\alpha \cos \phi)}{\cos \phi}\right)\,d\phi \\
&=-\int_0^{\Large\frac{\pi}{2}}\frac{\sin \alpha}{1+\cos \alpha \cos \phi}\,d\phi \\
&=-\int_0^{\Large\frac{\pi}{2}}\frac{\sin \alpha}{\left(\cos^2 \frac{\phi}{2}+\sin^2 \frac{\phi}{2}\right)+\cos \alpha\,\left(\cos^2\,\frac{\phi}{2}-\sin^2 \frac{\phi}{2}\right)}\,d\phi \\
&=-\frac{\sin\alpha}{1-\cos\alpha} \int_0^{\Large\frac{\pi}{2}} \frac{1}{\cos^2\frac{\phi}{2}}\frac{1}{\left[\left(\frac{1+\cos \alpha}{1-\cos \alpha}\right) +\tan^2 \frac{\phi}{2} \right]}\,d\phi \\
&=-\frac{2\,\sin\alpha}{1-\cos\alpha} \int_0^{\Large\frac{\pi}{2}}\,\frac{\frac{1}{2}\,\sec^2\,\frac{\phi}{2}}{\left[\,\left(\dfrac{2\,\cos^2\,\frac{\alpha}{2}}{2\,\sin^2\,\frac{\alpha}{2}}\right) + \tan^2\,\frac{\phi}{2} \right]} \,d\phi \\
&=-\frac{2\left(2\,\sin\,\frac{\alpha}{2}\,\cos\,\frac{\alpha}{2}\right)}{2\,\sin^2\,\frac{\alpha}{2}}\,\int_0^{\Large\frac{\pi}{2}}\,\frac{1}{\left[\left(\dfrac{\cos \frac{\alpha}{2}}{\sin\,\frac{\alpha}{2}}\right)^2\,+\,\tan^2\,\frac{\phi}{2}\,\right]}\,d\left(\tan\,\frac{\phi}{2}\right)\\
&=-2\cot \frac{\alpha}{2}\,\int_0^{\Large\frac{\pi}{2}}\,\frac{1}{\left[\,\cot^2\,\frac{\alpha}{2} + \tan^2\,\frac{\phi}{2}\,\right]}\,d\left(\tan \frac{\phi}{2}\right)\,\\
&=-2\,\left.\tan^{-1} \left(\tan \frac{\alpha}{2} \tan \frac{\phi}{2} \right) \right|_0^{\Large\frac{\pi}{2}}\\
&=-\alpha
\end{align}
Therefore:
$$I(\alpha) = C - \frac{\alpha^2}{2}$$
However by definition, $I\left(\frac{\pi}{2}\right) = 0$, hence $C = \dfrac{\pi^2}{8}$ and
$$I(\alpha) = \frac{\pi^2}{8}-\frac{\alpha^2}{2}.$$
The integral we want to evaluate is
$$I(0) = \int_0^{\Large\frac{\pi}{2}}\frac{\ln\,(1+\sin\,\phi)}{\sin\,\phi}\;d\phi=\frac{\pi^2}{8}.$$
Best Answer
The integral can be generalized and solved in terms of dilogarithms: $$ (1) \quad \int_0^1 \log{\big( (1+\sqrt{1+y\ x})/2\big)} \frac{dx}{x} = -\frac{1}{2}\Big\{ \zeta(2) + \text{Li}_2\big( (1-\sqrt{1+y})/2\big) + $$ $$ + \text{Li}_2\big( (1+\sqrt{1+y})/2\big) - \log{(-y/4)} \log{\big( (1+\sqrt{1+y})/2\big)} \big) \Big\}$$
Note that for $y=4,$ the analytic continuation of the dilogarithm must be used because the last two terms will have an imaginary part. In the dilog term on the top line of (1), the argument is $\rho=1/2(\sqrt{5} - 1).$ The dilog at this argument and its square has been known since 1780 (long before Ramanujan) and is due to Landen: $$ \text{Li}_2(\rho) = \pi^2/10 - \log^2\rho \ , \ \text{Li}_2(\rho^2) = \pi^2/15 - \log^2\rho $$
My proof of (1) ultimately depends on finding a close-form expression for $$ (2) \quad \sum_{k=1}^\infty \frac{(-y/4)^k}{k^2}\binom{2k}{2} = \frac{1}{\pi}\int_0^1 \text{Li}_2(-x\,y)\frac{dx}{\sqrt{x(1-x)}}.$$ (The answer to both is within the curly braces of (1).) I doubt my contribution is original. Mathematica knows about Landen's formulas and plugging $y=4$ and letting it do the simplification gives $I=\pi^2/15.$