Parameterize the line segment [tex]C[/tex] by
[tex]\mathbf r(t)=\langle x(t),y(t)\rangle=(1-t)\langle0,1\rangle+t\langle3,5\rangle=\langle3t,1+4t\rangle[/tex]
where [tex]t\in[0,1][/tex]. Then
[tex]\mathrm ds=\|\mathbf r'(t)\|\,\mathrm dt[/tex]
[tex]\mathrm ds=\|\langle3,4\rangle\|\,\mathrm dt[/tex]
[tex]\mathrm ds=5\,\mathrm dt[/tex]
So the integral is
[tex]\displaystyle\int_Cx\sin y\,\mathrm ds=5\int_0^1x(t)\sin y(t)\,\mathrm dt[/tex]
[tex]=\displaystyle5\int_0^13t\sin(1+4t)\,\mathrm dt[/tex]
[tex]=\dfrac{15}{16}(\sin5-\sin-4\cos5)\approx-2.7516[/tex]
