You can parameterize the curve [tex]C[/tex] and compute the integral along each component curve, or you can use the fact that [tex]f(x,y)=(\sin x,\cos y)[/tex] is the continuous gradient of a function [tex]F(x,y)[/tex] and observe that the line integral is path independent.
In other words, there is a function [tex]F(x,y)[/tex] such that [tex]\nabla F(x,y)=f(x,y)[/tex], so the integral along any curve [tex]C[/tex] from the points [tex]\mathbf a[/tex] to [tex]\mathbf b[/tex] is simply
[tex]\displaystyle\int_Cf(x,y)\,\mathrm dS=\int_C(P(x,y)\,\mathrm dx+Q(x,y)\,\mathrm dy)=F(\mathbf b)-F(\mathbf a)[/tex]
You have [tex]\nabla F(x,y)=(\sin x,\cos y)[/tex], and so
[tex]\dfrac{\partial F(x,y)}{\partial x}=\sin x\implies F(x,y)=-\cos x+g(y)[/tex]
while
[tex]\dfrac{\partial F(x,y)}{\partial y}=\cos y=g'(y)\implies g(y)=\sin y+K[/tex]
where [tex]K[/tex] is an arbitrary constant. So we've found that
[tex]F(x,y)=\sin y-\cos x+K[/tex]
which means the line integral has a value of
[tex]F(-4,4)-F(3,0)=(\sin4-\cos 4)-(\sin0-\cos3)=\sin4-\cos4+\cos3[/tex]
