$dr = b(\cos \theta \, dr + r \sin \theta \, d\theta)$
$dr = b\cos \theta \, dr + br \sin \theta \, d\theta$
$dr - b\cos \theta \, dr = br \sin \theta \, d\theta$
$(1 - b\cos \theta) \, dr = br \sin \theta \, d\theta$
$\dfrac{(1 - b\cos \theta) \, dr}{r(1 - b\cos \theta)} = \dfrac{br \sin \theta \, d\theta}{r(1 - b\cos \theta)}$
$\dfrac{dr}{r} = \dfrac{b \sin \theta \, d\theta}{1 - b\cos \theta}$
$\displaystyle \int \dfrac{dr}{r} = \int \dfrac{b \sin \theta \, d\theta}{1 - b\cos \theta}$
$\ln r = \ln (1 - b\cos \theta) + \ln c$
$\ln r = \ln c(1 - b\cos \theta)$
$r = c(1 - b\cos \theta)$ answer