Question

What is `int e^(-st) cosbt e^(at) dt`?

Answer 1

See below

Explanation:

Using the de Moivre's identity

`e^(ibt) = cos(bt)+i sin(bt)` we have

`int e^((a-s)t)cos(bt)dt + i int e^((a-s)t)sin(bt)dt = int e^((a-s+ib)t)dt =`

`1/(a-s-ib)e^((a-s+ib)t) + C =`

`= (a-s-ib)/((a-s)^2+b^2) e^((a-s)t)(cos bt+i sin bt)+C =`

but

`(a-s-ib)(cos bt+i sin bt)=`

`(a-s)cos(bt)+b sin(bt)+i((a-s)sin(bt)-bcos(bt))`

so taking the real part

`int e^((a-s)t)cos(bt)dt = e^((a-s)t)/((a-s)^2+b^2)((a-s)cos(bt)+b sin(bt))+C`

Answer 2

`int \ e^(-st) \ cosbt \ e^(at) \ dt = (e^((a-s)t){(a-s)cosbt + bsinbt})/((a-s)^2+b^2) + C`

Explanation:

Let:

`I = int \ e^(-st) \ cosbt \ e^(at) \ dt`
`\ \ = int \ e^((a-s)t) \ cosbt \ dt`
`\ \ = int \ e^(At) \ cosbt \ dt`, say where `A=a-s`

We can use integration by parts:

Let `{ (u,=cosbt, => (du)/dt,=-bsinbt), ((dv)/dt,=e^(At), => v,=1/Ae^(At) ) :}`

Then plugging into the IBP formula:

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

gives us

`int \ (cosbt)(e^(At)) \ dt = (cosbt)(1/Ae^(At)) - int \ (1/Ae^(At))(-bsinbt) \ dt`

`:. I = 1/A cosbt e^(At) + b/A int \ e^(At)sinbt \ dt`

At first it appears as if we have made no progress, as now the second integral is similar to `I`, having exchanged `cosx` for `sinx`, but if we apply IBP a second time then the progress will become clear:

Let `{ (u,=sinbt, => (du)/dt,=bcosbt), ((dv)/dt,=e^(At), => v,=1/Ae^(At) ) :}`

Then plugging into the IBP formula, gives us:

`int \ (sinbt)(e^(At)) \ dt = (sinbt)(1/Ae^(At)) - int \ (1/Ae^(At))(bcosbt) \ dt`

`:. int \ e^(At)sinbt \ dt = 1/A e^(At)sinbt - b/A int \ e^(At)cosbt \ dt`

`:. int \ e^(At)sinbt \ dt = 1/A e^(At)sinbt - b/A I + c`

Inserting this result into [A] we get:
`:. I = 1/A e^(At)cosbt + b/A {1/A e^(At)sinbt - b/A I + c}`

`:. AI = e^(At)cosbt + b/A e^(At)sinbt - b^2/A I + c`

`:. A^2I = Ae^(At)cosbt + be^(At)sinbt - b^2 I + cA`

`:. (A^2+b^2)I = Ae^(At)cosbt + be^(At)sinbt + cA`

`:. I = 1/(A^2+b^2) (Ae^(At)cosbt + be^(At)sinbt) + (cA)/(A^2+b^2)`

`:. I = (e^((a-s)t){(a-s)cosbt + bsinbt})/((a-s)^2+b^2) + C`