What will be $\oint \frac{1}{\sin(πz)} dz$ over a circle of radius 1.5?
If we use Cauchy's residue theorem to calculate $\oint \ \frac 1 {\sin(π z)}dz$ we will get $6\pi i$ but upon calculating $\int_0^{2π}\ \frac {{1.5i}e^{ix} } {\sin{(1.5πe^{ix})}} dx$ on WolframAlpha we get $-2i$.
Did I make a mistake here or is it the problem with the computation (which probably isn't the case)?
Yes, that integral is equal to $-2i$. The residue of $\frac1{\sin(\pi z)}$ at $0$ is $\frac1\pi$ and the residue at $\pm1$ is $-\frac1\pi$. Therefore,$$\oint_{|z|=\frac32}\frac1{\sin(\pi z)}\,\mathrm dz=2\pi i\left(-\frac1\pi+\frac1\pi-\frac1\pi\right)=-2i.$$