original problem: $\displaystyle \int e^xcos(2x)dx$ I've used int. by parts twice so far and I have this: $\displaystyle e^xcos(2x) + 2e^xsin(2x) - 4\int e^xcos(2x)dx$ Where do I go from here? I'm lost...
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Originally Posted by saiyanmx89 original problem: $\displaystyle \int e^xcos(2x)dx$ I've used int. by parts twice so far and I have this: $\displaystyle e^xcos(2x) + 2e^xsin(2x) - 4\int e^xcos(2x)dx$ Where do I go from here? I'm lost... $\displaystyle \int e^xcos(2x)dx= e^xcos(2x) + 2e^xsin(2x) - 4\int e^xcos(2x)dx$ so $\displaystyle \int 5e^xcos(2x)dx= e^xcos(2x) + 2e^xsin(2x) $ so $\displaystyle \int e^xcos(2x)dx= \frac{e^xcos(2x) + 2e^xsin(2x)}{5} $
where did you get 5e^x from? that's where I get lost....
$\displaystyle \textcolor{red}{\int e^xcos(2x)dx}= e^xcos(2x) + 2e^xsin(2x) \textcolor{red}{- 4\int e^xcos(2x)dx}$ note the like terms ... now move $\displaystyle \textcolor{red}{- 4\int e^xcos(2x)dx}$ to the left side and combine.
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