Integration question (e^x Cos2x)

How do I integrate:
(e^x)(Cos2x)dx?

Parts wouldn't help, since both halves go on forever...right?

(Answer was something like (1/5)(e^x)(Cos2x+2Sin2x), which sort of looks like something parts would come up with, but I'm not sure how to do it...)

Thanks :suith:

Reply 1

meef cheese

By parts twice.

Reply 2

meef cheese

let I = e^x cos2x

u=e^x => u' = e^x
v' = cos2x => v = 1/2 sin2x

I = 1/2 e^x sin2x - 1/2 INT e^x sin2x

p=e^x => p'=e^x
q'=sin2x => q= -1/2 cos2x

I = 1/2 e^x sin2x - 1/2 [-1/2 e^x cos2x + 1/2 INT e^x cos2x]
= 1/2 e^x sin2x + 1/4 e^x cos 2x -1/4 I
=> 5/4 I = 1/2 e^x sin2x + 1/4 e^x cos 2x

Blah, blah, blah. Dunno if that's right. But the main thing is:

By parts twice to get I = f(x) + aI where a is some fraction

Reply 3

bobin

Personally, I prefer complex numbers over parts for these types of integral, but that's just me. I find parts all fidlly with trying to keep track of minus signs.

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Reply 4

Rabite

Many thanks to both of you ^ ^;;
Makes me feel a little stupid, not seeing the first method. Although I still have a while to learn and practise everything.

I never thought of doing it either way - and I've never even seen it been done by the latter method.
Although exp(ix) stuff is FP3, isn't it.

Thanks again.