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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Try writing $\sin(2t)$ in exponential form using Euler's formula. – SDiv Oct 21 '14 at 12:46
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possible duplicate of Name of this identity? $\int e^{\alpha x}\cos(\beta x) \space dx = \frac{e^{\alpha x} (\alpha \cos(\beta x)+\beta \sin(\beta x))}{\alpha^2+\beta^2}$ – Hans Lundmark Oct 21 '14 at 14:15
3 Answers
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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
Dr. Sonnhard Graubner
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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,
$\int e^{-st}\sin(2t) dt = \frac{-e^{-st}\left(2\cos(2t) + s\sin(2t)\right)}{4+s^2}$.
G Tony Jacobs
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