Answer
$\sin(2t)+t+C$
Work Step by Step
Recall that $\sin 3x=3\sin x-4\sin^{3}x$ and $\csc x=\frac{1}{\sin x}$. Therefore,
$\int \csc t \sin 3t\,dt=\int \frac{1}{\sin t} (3\sin t-4\sin^{3}t)dt$
$=\int (3-4\sin^{2}t)\,dt=3t-4\int \sin^{2}t\,dt$
We know that $\sin^{2}x=\frac{1}{2}(1-\cos 2x)$. Therefore
$3t-4\int \sin^{2}t\,dt=3t-4\times\frac{1}{2}\int(1-\cos 2t)\,dt$
$=3t-2(t-\frac{\sin 2t}{2})+C=t+\sin2t+C$
