Question

How do you integrate `(e^x)sin4x 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)=sin4x` and `g'(x)dx=e^xdx`,

`f'(x)dx=cos4x*4dx`

`g(x)=e^x`,

so:

`I=e^xsin4x-inte^xcos4x*4dx=`

`=e^xsin4x-4inte^xcos4xdx=(1)`.

And now...again:

`f(x)=cos4x` and `g'(x)dx=e^xdx`,

`f'(x)dx=-sin4x*4dx`

`g(x)=e^x`.

So:

`(1)=e^xsin4x-4[e^xcos4x-inte^x(-sin4x*4)dx]=`

`=e^xsin4x-4e^xcos4x-16inte^xsin4xdx`.

So, finally, we can write:

`I=e^xsin4x-4e^xcos4x-16IrArr`

`17I=e^xsin4x-4e^xcos4xrArr`

`I=e^x/17(sin4x-4cos4x)+c`.