# Cauchy Goursat problem

Let k be the rectangle with corners $-2-2i,2-2i,2+i,-2+i$. Evaluate the integral: $$\int_k \frac{\cos(z)}{z^4}dz$$ Would the best way to do this problem be to integrate along each contour line using the Cauchy-Goursat theorem: $$\int_k f(z) \, dz = \int_b^a f(z(t))\frac{dz(t)}{dt} \, dt$$ Is there a way to do one integral and include all of the contour lines on the rectangle. Thanks in advance for the help.

By Cauchy-Goursat, $$\int_k\frac{f(z)}{(z-z_0)^{n+1}}dz=\frac{2\pi i}{n!}f^{(n)}(z_0)$$ and so $$\int_k \frac{\cos(z)}{z^4}dz=\frac{2\pi i}{3!}\sin(0)=0$$