Answer
Diverges
Work Step by Step
Suppose $f(t) =\dfrac{1}{t-\sin t} ; g(t) = \dfrac{1}{t^3}$
Then $\lim\limits_{t \to 0} \dfrac{f(t)}{g(t)}=\lim\limits_{t \to 0} \dfrac{t^{3}}{t-\sin t}\\=\lim\limits_{t \to 0} \dfrac{3 t^{2}}{1-\cos t} \\=\lim\limits_{t \to 0} \sin t\\=\lim\limits_{t \to 0} \dfrac{6}{\cos t}\\=6$
Now, $\int_{0}^{1} \dfrac{d t}{t^{3}} =\lim\limits_{t \to 0^{+}} [-\dfrac{1}{2 t^{2}}]_{b}^{1} =\lim\limits_{t \to 0^{+}} [-\dfrac{1}{2}-(-\dfrac{1}{2 b^{2}})]=+\infty$
Thus the integral diverges.