Question

Evaluate the integral? `int e^(2x) cosx dx`

Answer 1

See below.

Explanation:

Using the de Moivre's identity

`e^(ix)=cosx+isinx` we have

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

`=1/(2+i) e^(2x)(cosx+isin x)+C=`

`(2-i)/(2^2+1)(cos x+isinx) +C =`

`1/5(2cosx+sinx+i(2sinx-cosx))e^(2x)+C`

Taking the real part of this integral we have

`int e^(2x)cosx dx = 1/5(2cosx+sinx)e^(2x) + C`

Answer 2

I have solved with another way:

Answer 3

`int \ e^(2x) \cosx \ dx = 2/5e^(2x)cosx + 1/5e^(2x)sinx + C`

Explanation:

Let:

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

We can use integration by parts:

Let `{ (u,=cosx, => (du)/dx,=-sinx), ((dv)/dx,=e^(2x), => v,=1/2e^(2x) ) :}`

Then plugging into the IBP formula:

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

gives us

`int \ (cosx)(e^(2x)) \ dx = (cosx)(1/2e^(2x)) - int \ (1/2e^(2x))(-sinx) \ dx`
`:. I = 1/2e^(2x)cosx + 1/2int \ e^(2x) \ 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^(2x), => v,=1/2e^(2x) ) :}`

Then plugging into the IBP formula, gives us:

`int \ (sinx)(e^(2x)) \ dx = (sinx)(1/2e^(2x)) - int \ (1/2e^(2x))(cosx) \ dx`
`:. int \ e^(2x) \ sinx \ dx = 1/2e^(2x)sinx - 1/2I`

Inserting this result into [A] we get:

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

`:. I = 1/2e^(2x)cosx + 1/4e^(2x)sinx - 1/4I + A`

`:. 4I = 2e^(2x)cosx + e^(2x)sinx - I + 4A`

`:. 5I = 2e^(2x)cosx + e^(2x)sinx + 4A`

`:. I = 2/5e^(2x)cosx + 1/5e^(2x)sinx + C`