Without using the Residue Theorem, evaluate $$\int _C\frac{ z\, \cos(z-1) \, dz}{z^4+z^2-20} $$
where $C$ is given by $x^2 + y^2 = 3$. Assume that C has positive orientation.
I know that $C$ is a simple, closed contour and that the integrand is continuous and continuously differentiable everywhere except at $z=2,$ $z=-2$, $z=i{\sqrt 5}$, and $z=-i{\sqrt 5}$.
I don't think I can apply the Cauchy-Goursat Theorem here since the integrand is not continuous or continuously differentiable everywhere. Perhaps one of Cauchy's other integral formulas would work, but I'm not sure which one to use or how to apply it to this problem.
Any help or suggestions would be greatly appreciated. Thank you!
EDIT: After thinking about this problem more carefully, I realized that $z=i\sqrt 5$ and $z=-i\sqrt5$ lie outside of the contour $C$, so I don't think I need to worry about those.
As for $z=2$ and $z=-2$, I think I can use the Cauchy-Goursat Theorem to break the integral up into two separate line integrals along contours that form small circles around $z=2$ and $z=-2$ and then find these integrals' sum.
For $z=2$ along contour $C_1$, I have $$\int _{C_1}\frac{ z\, \cos(z-1) \, dz}{(z^2+5)(z+2)} $$ $$= (2\pi i) \frac{z\, \cos(z-1) \, dz}{(z^2+5)(z+2)}$$
Evaluating the above expression at $z=2$, we have: $$= (2\pi i) \frac{(2)\,\cos((2)-1) \, dz}{(2^2+5)(2+2)}$$ $$=\frac{\pi i\, \cos(1)}{9}$$
For $z=-2$ along contour $C_2$, I have $$\int _{C_1}\frac{ z\, \cos(z-1) \, dz}{(z^2+5)(z-2)} $$ $$= (2\pi i) \frac{z\, \cos(z-1) \, dz}{(z^2+5)(z-2)}$$
Evaluating the above expression at $z=-2$, we have: $$= (2\pi i) \frac{(-2)\,\cos((-2)-1) \, dz}{((-2)^2+5)(-2+2)}$$ $$=\frac{\pi i\, \cos(-3)}{9}$$
So, when we sum the two together, we obtain:
$$\frac{(\pi i)[ \cos(-3) + \cos(1)]}{9}$$
Is this right, or am I way off?