Answer
\[\ln|\cos\theta+\sin\theta|+C\]
Work Step by Step
Let \[I=\int\frac{1-\tan\theta}{1+\tan\theta}d\theta\]
\[I=\int\frac{1-\left(\frac{\sin\theta}{\cos\theta}\right)}{1+\left(\frac{\sin\theta}{\cos\theta}\right)}d\theta\]
\[I=\int\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}\;d\theta\;\;\;\ldots (1)\]
Substitute $\; \cos\theta+\sin\theta=t\;\;\;\ldots (2)$
\[\Rightarrow (\cos\theta-\sin\theta)d\theta=dt\]
(1) becomes
\[I=\int\frac{dt}{t}=\ln|t|+C\]
Where $C$ ia constant of integration
From (2)
\[I=\ln|\cos\theta+\sin\theta|+C\]
Hence,
\[I=\ln|\cos\theta+\sin\theta|+C\;\;.\]
