Answer
Please see proof in "step by step"
Work Step by Step
The expression on the RHS suggests that we bring
($\csc u+\cot u)$ "into play", that is, involve it in the integrand function...
$\displaystyle \csc u\cdot\frac{\csc u+\cot u}{\csc u+\cot u}=\frac{\csc^{2}u+\csc u\cot u}{\csc u+\cot u} \quad(*)$
Now, see page 135 (Summary of Differentiation Rules):
$\displaystyle \frac{d}{du}(\csc u+\cot u)=-\csc u\cot u-\csc^{2}u$
which equals $(-1)\times$(numerator) of (*).
So, if we substitute $\left[\begin{array}{l}
t=\csc u+\cot u\\
dt=-(\csc^{2}u+\csc u\cot u)du
\end{array}\right]$, we have
$\displaystyle \int \displaystyle \csc udu=\int\frac{\csc^{2}u+\csc u\cot u}{\csc u+\cot u}du$
... substitute $t=\csc u+\cot u$...
$=-\displaystyle \int\frac{1}{t}dt$ = Log Rule
$=-\ln|t|+C$
$=-\ln|\csc u+\cot u|+C$
