Applying Cauchy's Integral Formula to Higher Power Denominators

[Dissonance in E]

If i have a closed pathintegral of the form:
(sin(z)+3cos(z) + 3e^z)/((z-(pi/2)^2)
How is cauchys integral formula applicable? If I split the integral into partial fractions won't i still get A/(z-pi/2) + B/(z-pi/2)^2, the B part of which won't be applicable to cauchys formula since the denominator is still squared.

In short, how do i apply cauchys formula to integrals with a denominator with a power higher than 1.

 

[HallsofIvy]

Apparently, by "Cauchy's formula" you are referring to
f(a)= \frac{1}{2\pi i}\cint_\gamma \frac{f(z)}{z- a} dz

However, it can be easily generalized to
f^{(n)}(a)= \frac{n!}{2\pi i}\cint_\gamma \frac{f(z)}{(z- a)^{n+1}}dz
where the left sided is the nth derivative of f evaluated at z= a.

In your case, that integral is equal to the derivative of f(z)= sin(z)+ 3cos(z)+ 3e^z evaluated at z= \pi/2, multiplied by 2\pi i.

 

[snipez90]

...multiplied by 2\pi i.

 

[HallsofIvy]

Right, thanks. I will now edit my post so I can pretend I didn't make that foolish mistake!