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Wolfram Alpha shows me the result of $ \int e^{-st} \sin(2t) dt $ . However it doesn't let me see the step to step solution. Then I tried to do this by hand as the solution didn't look "too difficult", however couldn't do it. So can someone show me how to do it?

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3 Answers 3

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You can integrate by parts twice or use this method

$$\int e^{-st}\sin (2t)dt=\operatorname{Im}\int e^{(-s+2i)t}dt$$ Can you take it from here?

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using integration by parts we obtain $\int e^{-st}\sin(2t)dt=-e^{-st}\frac{cos(2t)}{2}-\int\frac{s}{2}e^{-st}\cos(2t)dt$ and now the same once more

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You start by using integration by parts two times, and then solve algebraically for the integral you want:

$\begin{align} \int e^{-st} \sin(2t) dt &= -\frac12 e^{-st}\cos(2t) -\frac{s}{2}\int e^{-st}\cos(2t) dt \\ &=-\frac12 e^{-st}\cos(2t) -\frac{s}{2}\left(\frac12 e^{-st}\sin(2t) + \frac{s}{2}\int e^{-st}\sin(2t) dt \right) \\ &=-\frac12 e^{-st}\cos(2t) -\frac{s}{4}e^{-st}\sin(2t) - \frac{s^2}{4}\int e^{-st}\sin(2t) dt \end{align}$

So, that's two applications of integration by parts. Now you can add that last term back over to the left-hand side:

$\int e^{-st} \sin(2t) dt + \frac{s^2}{4}\int e^{-st}\sin(2t) dt = -\frac12 e^{-st}\cos(2t) -\frac{s}{4}e^{-st}\sin(2t)$

Multiplying everything by $4$ and factoring appropriately, we obtain:

$(4+s^2)\int e^{-st}\sin(2t) dt = -e^{-st}\left(2\cos(2t) + s\sin(2t)\right)$

or,

$\int e^{-st}\sin(2t) dt = \frac{-e^{-st}\left(2\cos(2t) + s\sin(2t)\right)}{4+s^2}$.

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