# Different answer from using definition of contour integral to using Cauchy's Integral Formula

The contour integral is:
$$\int_{\Gamma}\frac{\cos z + i \sin z}{(z^2 + 36)(z+2)} \mathrm{d} z$$
where $\Gamma$ is the circle centred at the origin, with radius 3, traversed once positively.

I computed this contour integral to be $\frac{\pi i e^{-2}}{20}$, however, WolframAlpha is claiming that it's equal to zero when I try to do it by parametrisation with $z:= e^{it}$ for $0 \leq t \leq 2\pi$.

Why does W|A give a different answer? Am I wrong then with my original answer?

You gave Wolfram Alpha the wrong integral to do. Your parametrization was $z = e^{it}$, but this gives a circle of radius $1$, not a circle of radius $3$. With the correct parametrization $z = 3 e^{it}$, Wolfram Alpha gives the correct answer.
Let $f(z)=\frac{e^{iz}}{z^2+36}$. Then, by the integral formula:
$$\int_\Gamma\frac{e^{iz}}{(z+2)(z^2+36)}\,dz=\int_\Gamma\frac{f(z)}{z+2}\,dz= 2 \pi i f(-2)=\frac{\pi i e^{-2i}}{20}.$$
This is $$\int_\Gamma\frac{e^{iz}}{(z+2)(z^2+36)}\,dz.$$ The only pole inside $\Gamma$ is at $z=-2$ and is simple. The residue is $$\frac{e^{-2i}}{40}$$ so the integral is $$\frac{\pi i e^{-2i}}{20}.$$