Definite Integration of F=ma

by Astrum
Tags: definite, integration
 P: 274 This isn't a HM question, and I'm asking for an explanation. This is "The effect of a Radio Wave on an Ionospheric Electron" The integration is weird, I don't follow what is being done. $$a=\frac{-eE}{m}$$ - reworking of F=ma $$\frac{-eE}{m}sin(\omega t$$ only interested in the x axis. $$\int\frac{dv}{dt}=\int^{t}_{0}a_{0}sin(\omega t) dt$$ This becomes: $$v(t)=v_{0}-\frac{a_{0}}{\omega}cos(\omega t-1)$$ - I don't get where this came from, I understand the indefinite integration, but not where the "ωt-1" came from. And the last step: $$\int\frac{dx}{dt}=\int^{t}_{0}[v_{0}-\frac{a_{0}}{\omega}cost(\omega t-1)]dt$$ = $$x_{0} + (v_{0}+\frac{a_{0}}{\omega})t-\frac{a_{0}}{\omega^{2}}sin(\omega t)$$ Not sure where the final answer comes from. Could't you just integrate it twice, then tack on the definite integral?
 Engineering Sci Advisor HW Helper Thanks P: 7,121 I think some parentheses are in the wrong place. $$\int_0^t a_0 \sin(\omega t)\, dt = \left[ -\frac{a_0}{\omega} \cos(\omega t)\right]_0^t$$ $$= - \frac{a_0}{\omega}(\cos(\omega t) - \cos 0)$$ $$= - \frac{a_0}{\omega}(\cos(\omega t) - 1)$$