We have
[tex]\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} \vec f(\vec r(t)) \cdot \dfrac{d\vec r}{dt} \, dt[/tex]
and
[tex]\vec f(\vec r(t)) = 5\sin(t) \, \vec\imath - \cos(t) \, \vec\jmath + t \, \vec k[/tex]
[tex]\vec r(t) = \sin(t)\,\vec\imath + \cos(t)\,\vec\jmath + t\,\vec k \implies \dfrac{d\vec r}{dt} = \cos(t) \, \vec\imath - \sin(t) \, \vec\jmath + \vec k[/tex]
so the line integral is equilvalent to
[tex]\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (5\sin(t) \cos(t) + \sin(t)\cos(t) + t) \, dt[/tex]
[tex]\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (6\sin(t) \cos(t) + t) \, dt[/tex]
[tex]\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (3\sin(2t) + t) \, dt[/tex]
[tex]\displaystyle \int_C \vec f \cdot d\vec r = \left(-\frac32 \cos(2t) + \frac12 t^2\right) \bigg_0^{\frac{3\pi}2}[/tex]
[tex]\displaystyle \int_C \vec f \cdot d\vec r = \left(\frac32 + \frac{9\pi^2}8\right) - \left(-\frac32\right) = \boxed{3 + \frac{9\pi^2}8}[/tex]