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integrate e^(-3t)cost dt. doable by hand?
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oh this is totally gonna be one of those things where you add the integral to both sides to solve start with integration by parts...
reoccurring integral
\[\int e^{-3t}\cos tdt=e^{-3t}\sin t+3\int e^{-3t}\sin tdt\]\[=e^{-3t}\sin t-3e^{-3t}\cos t-9\int e^{-3t}\cos tdt\]noticing that last integral is a multiple of our original we add it to both sides:\[10\int e^{-3t}\cos tdt=e^{-3t}\sin t-3e^{-3t}\cos t\]\[\int e^{-3t}\cos tdt=\frac1{10}(e^{-3t}\sin t-3e^{-3t}\cos t)+C\]
and I totally did not know the name of it, thanks unkle
i was about to say i might have made a mistake somewhere but no, this does seem a little familiar so i must be on the right track (part of a bigger problem). thanks
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