Question

How to solve for `inte^xcosxdx` ?

Answer 1

`int\ e^x cos(x)\ "d"x=1/2e^x(sin(x)+cos(x))+C`

Explanation:

`I=int\ e^x cos(x)\ "d"x`

We will be using integration by parts, which states that `int\ u\ "d"v=uv-int\ v\ "d"u`.

Use integration by parts, with `u=e^x`, `du=e^x\ "d"x`, `"d"v=cos(x)\ "d"x`, and `v=sin(x)`:

`I=e^xsin(x)-int\ e^xsin(x)\ "d"x`

Use integration by parts again to the second integral, with `u=e^x`, `"d"u=e^x\ "d"x`, `"d"v=sin(x)\ "d"x`, and `v=-cos(x)`:

`I=e^xsin(x)+e^xcos(x)-int\ e^xcos(x)\ "d"x`

Now, recall we defined `I=int\ e^x cos(x)\ "d"x`. Thus, the above equation becomes the following (remembering to add a constant of integration):

`I=e^xsin(x)+e^xcos(x)-I+C`

`2I=e^xsin(x)+e^xcos(x)+C=e^x(sin(x)+cos(x))+C`

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

Answer 2

See below.

Explanation:

Using de Moivre's identity

`e^(ix) =cos x + i sin x` we have

`int e^x cos x dx = "Re"[int e^x(cos x + i sin x)dx] = "Re"[int e^(x + ix) dx]`

but `int e^((1+i)x) dx = 1/(1+i)e^((1+i)x) = (1-i)/2 e^x e^(ix) =`

`= (1-i)/2e^x(cos x+isinx) = 1/2e^x(cosx+sinx)+i1/2e^x(sinx -cosx)`

and finally

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