Question

How do you integrate `(e^x)(cosx) dx`?

Answer 1

This integral is a cyclic one and has to be done two times with the theorem of the integration by parts, that says:

`intf(x)g'(x)dx=f(x)g(x)-intg(x)f'(x)dx`

We can assume that `f(x)=cosx` and `g'(x)dx=e^xdx`,

`f'(x)dx=-sinxdx`

`g(x)=e^x`,

so:

`I=e^xcosx-inte^x(-sinx)dx=`

`=e^xcosx+inte^xsinxdx=(1)`.

And now...again:

`f(x)=sinx` and `g'(x)dx=e^xdx`,

`f'(x)dx=cosxdx`

`g(x)=e^x`.

So:

`(1)=e^xcosx+[e^xsinx-inte^xcosxdx]=`

`=e^xcosx+e^xsinx-inte^xcosxdx`.

So, finally, we can write:

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

`2I=e^xcosx+e^xsinx-IrArr`

`I=e^x/2(cosx+sinx)+c`.