Question on evaluating $\int_{C}\frac{e^{iz}}{z(z-\pi)}dz$ without the residue theorem

Using partial fraction expansion we have, for every $R>\pi$,

$$\oint_{|z|=R} \frac{e^{iz}}{z(z-\pi)}\,dz=\frac1\pi\oint_{|z|=R} \frac{e^{iz}}{z}\,dz-\frac1\pi\oint_{|z|=R}\frac{e^{iz}}{z-\pi}\,dz$$

Now finish by using Cauchy's Integral Formula (or the residue theorem).