help with integral?

Question

the integral is :

e^(-8000t) * [cos(6000t) + 2sin(6000t)] dt

thanks in advance!

Answer 1

let 6000 t = x : 8000t = x`+ x/3 = 4x/3

6000 dt = dx

dt = dx/6000

(1/6000)∫e^(4/3) x * [ cos x + 2sinx] dx

(1/6000) ∫e^(4/3) x cosx dx + 1/3000∫e^(4/3) sin x dx

Let us do one by one keeping the constant aside

∫e^(4/3) x cosx dx

integration by parts

u = e^(4/3) x : du = (4/3) e^(4/3)x

dv = cosx : v = sinx

∫e^(4/3) x cosx dx = uv - ∫v du

=>e^(4/3)x sinx - (4/3)∫ sinx e^(4/3)x dx

again integration by parts

u = e^(4/3) x : du =(4/3) e^(4/3) x

dv = sinx : v = - cosx

e^(4/3)x sinx - (4/3)( e^(4/3)x (- cosx)+(16/9)∫-cosx)e^(4/3)x dx

so
∫e^(4/3) x cosx dx =

e^(4/3)x sinx + (4/3)( e^(4/3)x cosx) -(16/9)∫cosx)e^(4/3)x dx

bringing -(16/9)∫cosx)e^(4/3)x dx to LHS

(25/9)∫e^(4/3) x cosx dx = e^(4/3)x sinx + (4/3)( e^(4/3)x cosx

∫e^(4/3) x cosx dx = (9/25) e^(4/3)x sinx+(12/25) e^(4/3)x cosx

similarly second integral

∫e^(4/3) sin x dx = -(9/25) e^(4/3)x cosx+(12/25)e^(4/3)x sinx

finally
(1/6000) ∫e^(4/3) x cosx dx + 1/3000∫e^(4/3) sin x dx =

(3/50000)e^(4/3) x sinx + (4/25000) e^(4/3)x cosx +

-(3/50000)e^(4/3) x cosx + (4/25000) e^(4/3)x sinx

=>
(11/50000)e^(4/3)x sinx +(1/10000)e^(4/3)x cosx + c

substituting back x = 6000t

=>(1/10000)e^(8000t) [(11/5)sin(6000t)+cos(6000t)] + c

Answer 2

It looks like an integration by parts type problem where U=[the trig term] and dV= exp(-8000t)dt. You may have to go through 2 iterations because you get changes of SIGN when you deal with integrating or differentiating cosine terms.