Question

How do you find the antiderivative of `e^( -x)*cos2x`?

Answer 1

`=-1/5 e^(-x) ( cos 2x - 2 sin 2x) + C`

Explanation:

`int dx \ e^(-x) cos 2x`

`= mathbb(Re) ( int dx \ e^(-x) e^(2 i x) )`

`= mathbb(Re) ( 1/(- 1 + 2 i) e^((- 1 + 2 i) x) )`

`=e^(-x) mathbb(Re) ( (- 1 - 2 i)/(5) (cos 2x + i sin 2x) )`

`=1/5e^(-x) ( - cos 2x + 2 sin 2x) + C`

`=-1/5 e^(-x) ( cos 2x - 2 sin 2x) + C`

Answer 2

`int \ e^(-x) cos(2x) \ dx = e^(-x)/5{2sin(2x)-cos(2x)} + C`

Explanation:

Let:

`I = int \ e^(-x) \ cos(2x) \ dx`

We can use integration by parts:

Let `{ (u,=cos2x, => (du)/dx=-2sinx), ((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 \ (cos2x)(e^(-x)) \ dx = (cos2x)(-e^(-x)) - int \ (-e^(-x))(-2sin2x) \ dx`
`:. I = -e^(-x) cos2x - 2 \ int \ e^(-x) \ sin2x \ dx` .... [A]

At first it appears as if we have made no progress, as now the second integral is similar to `I`, having exchanged `sin2x` for `sin2x`, but if we apply IBP a second time then the progress will become clear:

Let `{ (u,=sin2x, => (du)/dx=2cos2x), ((dv)/dx,=e^(-x), => v=-e^(-x) ) :}`

Then plugging into the IBP formula, gives us:

`int \ (sin2x)(e^(-x)) \ dx = (sin2x)(-e^(-x)) - int \ (-e^(-x))(2cos2x) \ dx`
`:. int \ e^(-x) \ sin2x \ dx = -e^xsin2x + 2I`

Inserting this result into [A] we get:

`I = -e^(-x) cos2x - 2 (-e^xsin2x + 2I) + A`

Now we can solve fior `I`

`I = -e^(-x) cos2x + 2e^xsin2x - 4I + A`

`:. 5I = e^x{2sin2x - cos2x} + A`

`:. I = e^x/5{2sin2x - cos2x} + A`