Answer
\begin{aligned} I = \int(\csc x-\sec x)(\sin x+\cos x) d x = \ln \cos x +\ln \sin x +C \\
\end{aligned}
Work Step by Step
Given $$ \int(\csc x-\sec x)(\sin x+\cos x) d x $$
So, we get
\begin{aligned} I&= \int(\csc x-\sec x)(\sin x+\cos x) d x \\
&= \int(\csc x \sin x-\sec x\sin x+\csc x \cos x-\sec x\cos x) d x \\
&= \int(1-\tan x+\cot x-1) d x \\
&= \int( -\tan x+\cot x ) d x \\
&= \int( -\frac{\sin x}{\cos x} +\frac{\cos x}{\sin x} ) d x \\
&= \ln \cos x +\ln \sin x +C \\
\end{aligned}
