I saw this problem:
$$\int\left(\frac{\sin(x)}{2\sin(x)- x(1+\cos(x))}\right)^2dx$$
I tried to solve this problem and I found a strange and unsatisfactory solution using differential equations.
Let $$I:=\int\left(\frac{\sin(x)}{2\sin(x)- x(1+\cos(x))}\right)^2dx$$ and let $x=2t$. Then $$I=\frac{1}{2}\int\left(\frac{\sin(t)\cos(t)}{\sin(t)\cos(t)- t\cos^2(t)}\right)^2dt=\frac{1}{2}\int\left(\frac{\sin(t)}{\sin(t)- t\cos(t)}\right)^2dt$$ Since $\frac{d}{dt}\frac{v}{u} = \frac{uv'-vu'}{u^2}$ and the integral is in the form $\frac{uv'-vu'}{u^2}$, such that $u =\sin(t)- t\cos(t)$ and $u' =t\sin(t)$, we need to find a function $v(t)$ such that $v'(t)(\sin(t)- t\cos(t)) -v(t)(t \sin(t)) =\sin^2(t)$ or $$-t( \sin(t)v(t) +\cos(t) v'(t) ) + v'(t) \sin(t) =\sin^2(t)$$ Since $\sin^2(t)$ is not a multiple of $t$, we can assume that $\sin(t)v(t) +\cos(t) v'(t)=0$, i.e., $v(t) =- \cos(t)$, and then $v'(t) \sin(t)=\sin^2(t)$ will solve this differential equation. So $$I=\frac{-\cos(t)}{2(\sin(t)- t\cos(t))}+C = \frac{-\cos(\frac{x}{2})}{2\sin(\frac{x}{2})- x\cos(\frac{x}{2})}+C$$
Although I found this strange solution to this integral, I want to see how to solve it using other methods.
Continue with $$I=\int\left(\frac{\sin t}{\sin t- t\cos t}\right)^2dt = \int \frac{1}{(t\cot t-1)^2}dt =\int \frac{dt}{y^2} $$ where $y= t\cot t-1$. Then $dy = (\cot t -t\csc^2 t)dt$ and \begin{align}\cot t\ dy = &\ (\cot^2t -t\cot t\csc^2 t)dt\\ =&\ (\csc^2t-1 -t\cot t\csc^2 t)dt=-y\csc^2t \ dt -dt \end{align} witch leads to $ dt= -y\csc^2t \ dt- \cot t\ dy$ and $$I= \int \frac{-y\csc^2t \ dt-\cot t\ dy}{y^2} = \int d\left(\frac {\cot t}{y} \right)= \frac {\cot t}{y}+C $$