Question

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

Answer 1

`int \ e^(-st)sint \ dt = -(e^(-st)(s sint +cost ))/(s^2+1) + C`

Explanation:

We seek the integral:

`I =int \ e^(-st)sint \ dt`

We can then apply Integration By Parts:

Let `{ (u,=sint, => (du)/dt,=cost), ((dv)/dt,=e^(-st), => v,=-1/se^(-st) ) :}`

Then plugging into the IBP formula:

`int \ (u)((dv)/dt) \ dt = (u)(v) - int \ (v)((du)/dt) \ dt`

We have:

`int \ (sinmt)(e^(-st)) \ dt = (sin t)(-1/se^(-st)) - int \ (-1/se^(-st))(cos t) \ dt`

`:. I = -1/s e^(-st)sint + 1/s int \ e^(-st)cos t \ dt`

Now consider the integral given by:

`I_2 = int \ e^(-st)cos t \ dt`

We will now need to apply IBP again:

Let `{ (u,=cost, => (du)/dt,=-sint), ((dv)/dt,=e^(-st), => v,=-1/se^(-st) ) :}`

Then plugging into the IBP formula we have::

`int \ (cost)(e^(-st)) \ dt = (cost)(-1/se^(-st)v) - int \ (-1/se^(-st))(-sint) \ dt`

`I_2 = -1/s e^(-st)cost - 1/s \ int \ e^(-st)sint \ dt`
`I_2 = -1/s e^(-st)cost - 1/s I`

And so combining the results we find that:

`I = -1/s e^(-st)sint + 1/s {-1/s e^(-st)cost - 1/s I}`

As `s != 0`, then by Multiplying by `s^2` we get:

`s^2I = -s e^(-st)sint - e^(-st)cost - I`

`:. s^2I +I = -s e^(-st)sint - e^(-st)cost`

`:. (s^2+1)I = -e^(-st)(s sint +cost )`

`:. I = (-e^(-st)(s sint +cost ))/(s^2+1)`

And not forgetting the constant of integration,

`I = (-e^(-st)(s sint +cost ))/(s^2+1) + C`