Chapter 9: First-Order Differential Equations - Section 9.2 - First-Order Linear Equations - Exercises 9.2 - Page 537: 13

$r=-\csc \theta \ln (\cos \theta)+c \csc \theta$

Here, we have $r'+\cot (\theta) r=\sec \theta$ The integrating factor is: $I=e^{\int \cot \theta d\theta}=\sin \theta$ Now, $\sin \theta [r'+\cot \theta r]=\sin \theta \sec \theta$ This gives: $\int [\sin \theta (r)]' dt=\tan \theta d\theta$ $\implies \sin (\theta) (r)=-\ln |\cos \theta|+c$ so, the general solution is: $r=-\csc \theta \ln (\cos \theta)+c \csc \theta$