Answer
$r=\dfrac{\sin^2 \theta}{3}+\dfrac{c}{\sin \theta}=\dfrac{\sin^2 \theta}{3}+c (\csc \theta)$
Work Step by Step
The standard form of the given equation is:
$r'+\cot \theta r=\sin \theta\cos \theta$ ....(1)
The integrating factor is: $e^{\int \cot \theta d\theta}=\sin \theta$
In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides.
$\int [\sin \theta (r)]' dt=\sin^2 \theta\cos \theta d\theta$
or, $\sin \theta (r)=\dfrac{\sin^3 \theta}{3}+c$
Thus, the general solution is as follows:
$r=\dfrac{\sin^2 \theta}{3}+\dfrac{c}{\sin \theta}=\dfrac{\sin^2 \theta}{3}+c (cosec \theta)$
