Answer
$r=-\csc \theta \ln \cos \theta+c \csc \theta$
Work Step by Step
The standard form of the given equation is:
$r'+\cot \theta r=\sec \theta$ ....(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.
$\int [\sin \theta (r)]' dt=\tan \theta d\theta$
or, $\sin \theta (r)=-\ln |\cos \theta|+c$
Thus, the general solution is as follows:
$r=-\csc \theta \ln \cos \theta+c \csc \theta$
