Answer
$(\dfrac{20}{9})[\sin (6)-3 \cos 6-\sin 3]$
Work Step by Step
$I=\int_{3}^{6}(4/3) \times(t-3) \sin (t) \sqrt {{(\dfrac{dx}{dt})^{2}}+{(\dfrac{dy}{dt})^{2}}}dt=\int_{3}^{6}(4/3) \times(t-3) \sin (t) \sqrt {{(\dfrac{4}{3})^{2}}+{(1)^{2}}}dt$
$=\int_{3}^{6}(4/3) \times(t-3) \sin (t) (\dfrac{5}{3})dt$
$=(20/9)[ \int (t-3) \sin t dt]_{3}^{6}$
$= (\dfrac{20}{9})[-(t-3) \cos t +\int \cos t dt ]_{3}^{6}$
$= (\dfrac{20}{9})[-(t-3) \cos t + \sin t dt ]_{3}^{6}$
$= (\dfrac{20}{9})[\sin (6)-3 \cos 6-\sin 3]$
