Question

How do you evaluate the integral `int e^-xcosx`?

Answer 1

`int \ e^(-x) \cosx \ dx = 1/2(e^(-x)sinx - e^(-x)cosx) + C`

Explanation:

Let:

`I = int \ e^(-x) \cosx \ dx`

We can use integration by parts:

Let `{ (u,=cosx, => (du)/dx,=-sinx), ((dv)/dx,=e^(-x), => v,=-e^x ) :}`

Then plugging into the IBP formula:

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

gives us

`int \ (cosx)(e^(-x)) \ dx = (cosx)(-e^(-x)) - int \ (-e^(-x))(-sinx) \ dx`
`:. I = -e^(-x)cosx - int \ e^(-x) \ sinx \ dx` .... [A]

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,=sinx, => (du)/dx,=cosx), ((dv)/dx,=e^(-x), => v,=-e^(-x) ) :}`

Then plugging into the IBP formula, gives us:

`int \ (sinx)(e^(-x)) \ dx = (sinx)(-e^(-x)) - int \ (-e^(-x))(cosx) \ dx`
`:. int \ e^(-x) \ sinx \ dx = -e^(-x)sinx + I`

Inserting this result into [A] we get:

`I = -e^(-x)cosx + e^(-x)sinx - I + A`

`:. 2I = e^(-x)sinx - e^(-x)cosx + A`
`:. \ \ I = 1/2(e^(-x)sinx - e^(-x)cosx) + C`