Question

How do you integrate `int e^6xcos5xdx`?

Answer 1

Hello !

I propose another solution, shorter (?), using complex numbers.

`e^{6x} cos(5x) = e^(6x)\mathfrak{Re}(e^{5ix}) = \mathfrak{Re}(e^{(6+5i)x})`.

Now integrate easily :

`int e^{6x} cos(5x)dx = \mathfrak{Re} \int e^{(6+5i)x} d x = \mathfrak{Re} (e^((6+5i)x))/(6+5i) + c`

where `c \in CC` is a constant.

To take the real part, you have to write :

`(e^{(6+5i)x})/(6+5i) = e^(6x)((cos(5x) + i sin(5x))(6-5i))/((6+5i)(6-5i)) = e^(6x)((cos(5x) + i sin(5x))(6-5i))/(36+25)`

so,
`\mathfrak{Re} (e^((6+5i)x))/(6+5i) = e^(6x)(6 cos(5x) + 5 sin(5x))/61`.

Finally,

`int e^{6x} cos(5x) dx =e^(6x)(6/61 cos(5x) + 5/61 sin(5x))+c`