Solve the differential equation \(dr+(2r\cot\theta+\sin2\theta)\,d\theta=0\)

Answer: \(r\sin^2\theta+\dfrac{\sin^4\theta}{2}=c\)

\(\dfrac{dr}{d\theta}+(2\cot\theta)r=-\sin2\theta\), linear in \(r\) with \(P=2\cot\theta\), \(Q=-\sin2\theta\).

I.F. \(=e^{\int2\cot\theta\,d\theta}=e^{2\log\sin\theta}=\sin^2\theta\).

Solution: \(r\sin^2\theta=\displaystyle\int-\sin2\theta\sin^2\theta\,d\theta+c=-\int2\sin^3\theta\cos\theta\,d\theta+c\).

Put \(\sin\theta=t\), \(\cos\theta\,d\theta=dt\): \(-\displaystyle\int2t^3dt=-\dfrac{t^4}{2}=-\dfrac{\sin^4\theta}{2}\).

\(\therefore r\sin^2\theta=-\dfrac{\sin^4\theta}{2}+c\), i.e. \(r\sin^2\theta+\dfrac{\sin^4\theta}{2}=c\) (or \(2r\sin^2\theta+\sin^4\theta=c_1\)).