Answer
$$\frac{{4\sqrt {{u^2} - 4} }}{{{u^2}}}$$
Work Step by Step
$$\eqalign{
& \sin \left( {2{{\sec }^{ - 1}}\frac{u}{2}} \right) \cr
& {\text{From the triangle we have that}} \cr
& \sec \theta = \frac{u}{2} \cr
& \theta = {\sec ^{ - 1}}\frac{u}{2} \cr
& \sin \left( {2{{\sec }^{ - 1}}\frac{u}{2}} \right) = \sin 2\theta \cr
& \sin \left( {2{{\sec }^{ - 1}}\frac{u}{2}} \right) = 2\sin \theta \cos \theta \cr
& \sin \left( {2{{\sec }^{ - 1}}\frac{u}{2}} \right) = 2\left( {\frac{{{\text{opposite side}}}}{{{\text{hypotenuse}}}}} \right)\left( {\frac{{{\text{adjacent side}}}}{{{\text{hypotenuse}}}}} \right) \cr
& \sin \left( {2{{\sec }^{ - 1}}\frac{u}{2}} \right) = 2\left( {\frac{{\sqrt {{u^2} - 4} }}{u}} \right)\left( {\frac{{\text{2}}}{u}} \right) \cr
& \sin \left( {2{{\sec }^{ - 1}}\frac{u}{2}} \right) = \frac{{4\sqrt {{u^2} - 4} }}{{{u^2}}} \cr} $$
