Question

Find $\int \frac{1}{1+tanx}dx$.

Answer 1

$\begin{array}{c}\int \frac{dx}{1+\tan x}\\ =\int \frac{\cos xdx}{\cos x+\sin x}\\ =\frac{1}{2}\int \frac{\left(\sin x+\cos x\right)+\left(\cos x-\sin x\right)}{\left(\sin x+\cos x\right)}dx\\ =\frac{x}{2}+\frac{1}{2}\int \frac{\left(\cos x-\sin x\right)}{\sin x+\cos x}dx\\ \sin x+\cos x=t\\ =\frac{x}{2}+\frac{1}{2}\int \frac{dt}{t}\\ I=\frac{x}{2}+\frac{1}{2}\log \left(\sin x+\cos x\right)+C\end{array}$