Answer
$\frac{1}{\sqrt {u^2+1}}$
Work Step by Step
Step 1. Letting $cot^{-1}u=t$, we have $cot(t)=u$ and $-\frac{\pi}{2}\le t \le \frac{\pi}{2}$
Step 2. Form a right triangle with sides $x=|u|,y=1,r=\sqrt {u^2+1}$ and angle $|t|$ facing $y$.
Step 3. As $sin(t)$ and $cot(t)$ have the same sign in the above interval, we have $sin(cot^{-1}u)=sin(t)=\frac{1}{\sqrt {u^2+1}}$
