In a section on fourier transforms, my textbook contains these steps for an example:
$$f(x) = \int_{-\infty}^\infty \frac{\sin{\alpha}}{\pi \alpha}e^{i\alpha x}d\alpha$$ $$= \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\sin{\alpha}(\cos{\alpha x} + i \sin{\alpha x})}{\alpha}d\alpha$$ $$= \frac{2}{\pi}\int_{0}^{\infty}\frac{\sin{\alpha}\cos{\alpha x}}{\alpha}d\alpha$$
I understand why the coefficient changes to $\frac{2}{\pi}$ and the integral changes to $\int_0^\infty$, but why does the $\sin{\alpha}\cdot i\sin{\alpha x}$ disappear in the final step? Am I missing something obvious here?
$\dfrac1\alpha\sin\alpha\cdot i\sin\alpha x$ is an odd function of $\alpha$.