Answer
$$
\int \frac{1-\tan \theta}{1+\tan \theta} d \theta =\ln |\cos \theta+\sin \theta|+C
$$
Work Step by Step
$$
\begin{aligned}
\int \frac{1-\tan \theta}{1+\tan \theta} d \theta &=\int \frac{\frac{\cos \theta}{\cos \theta}-\frac{\sin \theta}{\cos \theta}}{\frac{\cos \theta}{\cos \theta}+\frac{\sin \theta}{\cos \theta}} d \theta\\
& =\int \frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta} d \theta \\
&=\ln |\cos \theta+\sin \theta|+C
\end{aligned}
$$
where $C$ is a constant