Separable differential equation that results in a complicated integral

Question

I am having trouble solving this differential equation: $$y\left(3x+\frac{6x^2\sin^2(\frac x2)}{x-\sin x}\right)\,\mathrm dx=\frac{\sqrt x\,\mathrm dy}{(x-\sin x)^{\frac32}}.$$ I know it is separable but I cannot seem to be able to solve it. Is there even a way to do it analytically?

This is where I get stuck: $$\frac{\mathrm{d}y}{y} = (x^2 - x\sin x)^{\frac{1}{2}} \left( 3x - 3\sin x + 6x\sin^2\left( \frac{x}{2} \right) \right) \,\mathrm{d}x.$$ The left side is easy to integrate, but the right side is a problem.

Answer 1

Hint: With simplification $$2x^2y\ d\ln\left(x(x-\sin x)\right)=\dfrac{x^2\ dy}{(x(x-\sin x)^\frac32}$$

Answer 2

\begin{eqnarray} \frac{\mathrm{d}y}{y}&=&(x^2 - x\sin x)^{\frac{1}{2}} \left( 3x - 3\sin x + 6x\sin^2\left( \frac{x}{2} \right) \right) \,\mathrm{d}x\\ &=&3\sqrt{x}(x-\sin x)^{3/2}dx+3x^{3/2}(x-\sin x)^{1/2}(1-\cos x)\,dx \end{eqnarray} Integrating by parts, we get \begin{equation} \int3x^{3/2}(x-\sin x)^{1/2}(1-\cos x)\,dx=2x^{3/2}(x-\sin x)^{3/2}-3\int\sqrt{x}(x-\sin x)^{3/2}dx\tag{1} \end{equation} which subtracts out the other term, giving

\begin{equation} \ln y=2x^{3/2}(x-\sin x)^{3/2}+c \end{equation}

ADDENDUM: The OP seems to have vanished. In case my answer was too terse, I am including the integration by parts steps.

$$\int3x^{3/2}(x-\sin x)^{1/2}(1-\cos x)\,dx$$

Tableau Form

\begin{align} &+&3x^{3/2}&&(x-\sin x)^{1/2}(1-\cos x)\\ &-&\frac{9}{2}x^{1/2}&&\frac{2}{3}(x-\sin x)^{3/2} \end{align}

giving the result in equation (1).