Integral e^xcos(2x)dx
If you had done some theory of complex numbers this would be trivial, but
as we must assume otherwise you can do this using integration by parts.
I = Integral e^xcos(2x)dx
put dv=e^x and u=cos(2x), the the integration by parts rule gives:
I = e^x cos(2x) + 2 integral e^x sin(2x) dx
Now we do this again with the integral on the right, with dv=e^x and u=sin(2x),
to get:
I = e^x cos(2x) + 2 [e^x sin(2x) - 2 integral e^x cos(2x) dx]
or:
I = e^x cos(2x) + 2 e^x sin(2x) - 4 I,
so:
5I = e^x cos(2x) + 2 e^x sin(2x)
or:
I = [e^x cos(2x) + 2 e^x sin(2x)]/5
and finally add in a constant of integration:
I = [e^x cos(2x) + 2 e^x sin(2x)]/5 + C
RonL
Integral e^xcos(2x)dx
Integrand is a product of two unrelated functions, so try to integrate by parts:
INT[u]dv = uv -INT[v]du ---------------***
INT.[(e^x)cos(2x)]dx -------------------------(0)
= INT.[cos(2x)][e^x dx] -----(i)
Let u = cos(2x) --------------And, dv = e^x dx
So, du = -2sin(2x) dx --------------v = e^x
= cos(2x) *e^x -INT.[e^x][-2sin(2x) dx]
= (e^x)cos(2x) +(2)INT.[sin(2x)][e^x dx] -------(ii)
The integral part of (ii) is by parts again.
Let r = sin(2x) ------------And, ds = e^x dx
So, dr = 2cos(2x) dx ------------s = e^x
= (e^x)cos(2x) +2{sin(2x) *ex -INT.[e^x][2cos(2x) dx]}
= (e^x)cos(2x) + 2(e^x)sin(2x) -(4)INT.[(e^x)cos(2x)]dx ----------(iii)
The integral part of (iii) is (-4) times the original integral (0), so collect like terms,
INT.[(e^x)cos(2x)]dx +(4)INT.[(e^x)cos(2x)]dx = (e^x)cos(2x) +2(e^x)sin(2x)
(5)INT.[(e^x)cos(2x)]dx = (e^x)cos(2x) +2(e^x)sin(2x)
Divide both sides by 5,
INT.[(e^x)cos(2x)]dx = (1/5)(e^x)cos(2x) +(2/5)(e^x)sin(2x) +C -----------answer.