Question

How do you find the antiderivative of `(cosx)(e^x)`?

Answer 1

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

Explanation:

Let:

`I = int \ cosx \ e^x \ 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^xcosx + 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^xsinx - I`

Inserting this result into [A] we get:

`I = e^xcosx + e^xsinx - I + A`

`:. 2I = e^xcosx + e^xsinx + A`
`:. I = 1/2(e^xcosx + e^xsinx) + C`