Answer
$-t \cot t + \ln |\sin t| + C$
Work Step by Step
Let $u = t, dv = \csc^2 t dt$ and $du = dt, v = - \cot t$
$\int t \csc^2 t dt$
$= -t \cot t - \int -\cot t dt$
$= -t \cot t + \int \frac{\cos t}{\sin t} dt$
$= -t \cot t + \int \frac{1}{z} dz$
$= -t \cot t + \ln |z| + C$
$= -t \cot t + \ln |\sin t| +C$