# How to compute $\int e^{-st} \sin(2t) dt$

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?

• Try writing $\sin(2t)$ in exponential form using Euler's formula.
– SDiv
Oct 21, 2014 at 12:46
• Oct 21, 2014 at 14:15

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?

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

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,

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