Evaluate int \ e^(-st)sint \ dt ?

Jan 28, 2018

$\int \setminus {e}^{- s t} \sin t \setminus \mathrm{dt} = - \frac{{e}^{- s t} \left(s \sin t + \cos t\right)}{{s}^{2} + 1} + C$

Explanation:

We seek the integral:

$I = \int \setminus {e}^{- s t} \sin t \setminus \mathrm{dt}$

We can then apply Integration By Parts:

Let $\left\{\begin{matrix}u & = \sin t & \implies \frac{\mathrm{du}}{\mathrm{dt}} & = \cos t \\ \frac{\mathrm{dv}}{\mathrm{dt}} & = {e}^{- s t} & \implies v & = - \frac{1}{s} {e}^{- s t}\end{matrix}\right.$

Then plugging into the IBP formula:

$\int \setminus \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dt}}\right) \setminus \mathrm{dt} = \left(u\right) \left(v\right) - \int \setminus \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dt}}\right) \setminus \mathrm{dt}$

We have:

$\int \setminus \left(\sin m t\right) \left({e}^{- s t}\right) \setminus \mathrm{dt} = \left(\sin t\right) \left(- \frac{1}{s} {e}^{- s t}\right) - \int \setminus \left(- \frac{1}{s} {e}^{- s t}\right) \left(\cos t\right) \setminus \mathrm{dt}$

$\therefore I = - \frac{1}{s} {e}^{- s t} \sin t + \frac{1}{s} \int \setminus {e}^{- s t} \cos t \setminus \mathrm{dt}$

Now consider the integral given by:

${I}_{2} = \int \setminus {e}^{- s t} \cos t \setminus \mathrm{dt}$

We will now need to apply IBP again:

Let $\left\{\begin{matrix}u & = \cos t & \implies \frac{\mathrm{du}}{\mathrm{dt}} & = - \sin t \\ \frac{\mathrm{dv}}{\mathrm{dt}} & = {e}^{- s t} & \implies v & = - \frac{1}{s} {e}^{- s t}\end{matrix}\right.$

Then plugging into the IBP formula we have::

$\int \setminus \left(\cos t\right) \left({e}^{- s t}\right) \setminus \mathrm{dt} = \left(\cos t\right) \left(- \frac{1}{s} {e}^{- s t} v\right) - \int \setminus \left(- \frac{1}{s} {e}^{- s t}\right) \left(- \sin t\right) \setminus \mathrm{dt}$

${I}_{2} = - \frac{1}{s} {e}^{- s t} \cos t - \frac{1}{s} \setminus \int \setminus {e}^{- s t} \sin t \setminus \mathrm{dt}$
${I}_{2} = - \frac{1}{s} {e}^{- s t} \cos t - \frac{1}{s} I$

And so combining the results we find that:

$I = - \frac{1}{s} {e}^{- s t} \sin t + \frac{1}{s} \left\{- \frac{1}{s} {e}^{- s t} \cos t - \frac{1}{s} I\right\}$

As $s \ne 0$, then by Multiplying by ${s}^{2}$ we get:

${s}^{2} I = - s {e}^{- s t} \sin t - {e}^{- s t} \cos t - I$

$\therefore {s}^{2} I + I = - s {e}^{- s t} \sin t - {e}^{- s t} \cos t$

$\therefore \left({s}^{2} + 1\right) I = - {e}^{- s t} \left(s \sin t + \cos t\right)$

$\therefore I = \frac{- {e}^{- s t} \left(s \sin t + \cos t\right)}{{s}^{2} + 1}$

And not forgetting the constant of integration,

$I = \frac{- {e}^{- s t} \left(s \sin t + \cos t\right)}{{s}^{2} + 1} + C$