I'm having trouble figuring out how to find the integral of $$\int\limits_{}^{}e^{5x}\cos(4x)dx$$ any ideas how to solve it?

Question

anonymous · Answer

integrate by parts? gimme a sec to think about it

kainui · Answer

Yeah actually I think that'll work. I started doing it but ended up with the same integral with sine instead of cosine. But once I integrate by parts again I'll have the same term and I'll cancel it out I think.

anonymous · Answer

Oh i know its one of those stupid ones where you actually have to do it twice, then divide, ill do it though LOL

kainui · Answer

Nah, I got it thanks though haha.

anonymous · Answer

u=e^5x           dv = cos(4x)
du = 5e^5x     v = sin(4x)/4

uv - integral (vdu

(e^5x)(sin(4x)/4) - integral (sin(4x)/4)(5e^5x)

so integral (sin(4x)/4)(5e^5x) do parts again

u = 5e^5x           dv = sin4x/4
du = 25e^5x       v = -cos(4x)/16

uv-int vdu

(5e^5x)(-cos(4x)/16) - integral (-cos(4x)/16)(25e^5x)

soo from the first integral you add this in

(e^5x)(sin(4x)/4) - (5e^5x)(-cos(4x)/16) +(16*25) integral (cos(4x))(e^5x) = integral of (e^5x)(cos4x)

then do algebra so add the integral on the left to the right
and figure out what to do with the extra stuff 

so like i think the answer is (e^5x)(sin(4x)/4) - (5e^5x)(-cos(4x)/16) /2 or something

i dont know wolframalpha it

but i know you have to do 2 integration by parts then algebra

anonymous · Answer

i know i made a mistake somewhere, i just dont care to fix it, since im typing it and its not on paper LOL