
#1
Mar2812, 07:34 AM

P: 2,453

If I have [itex] \int e^{2x}sin(x)sin(2x) [/itex]
And then I use Eulers formula to substitute in for the sine terms. So I have [itex] \int e^{2x}e^{ix}e^{2ix} [/itex] then I combine everything so i get [itex] e^{(2+3i)x} [/itex] so then we integrate the exponential, so we divide by 2+3i and then i multiply by the complex conjugate. now since sine is the imaginary part of his formula I took the imaginary part when I back substituted for e^(3i) but I didn't get the correct answer doing this, so am i not using Eulers formula correctly? 



#2
Mar2812, 08:32 AM

P: 44

e^i3x = sin3x+isin2x , so the imaginary part is different from i(sinxsin2x)




#3
Mar2812, 04:11 PM

P: 2,453

why does e^i3x = sin3x+isin2x , i guess im not seeing it off hand I probably should look at it more and try to manipulate it more.




#4
Mar2812, 09:29 PM

P: 44

Integral using Euler's formula.
sry typos , its sin3x




#5
Mar2812, 09:41 PM

P: 2,453

how come one part is not cos(3x)




#6
Mar2812, 10:16 PM

P: 44

another typos , sry =='




#7
Mar2912, 04:28 AM

P: 2,453

But we could get it in the form of
[itex] sin(x)e^{2ix}=isin(2x)sin(x)+cos(2x)sin(x) [/itex] Do we need to get an expression where we have just exponentials on the left hand side and then isin(x)sin(2x)+cos(2x)cos(x) 



#8
Mar2912, 06:32 AM

P: 44

but then ur integral cant become e^i3x now , can it




#9
Mar2912, 06:01 PM

P: 2,453

ok, im not sure exactly what you mean, How do you recommend I approach the problem.




#10
Mar2912, 06:45 PM

P: 44

sina*sinb = 0.5[cos(a+b)+cos(ab)] , then u have 2 solvable integrals




#11
Mar3012, 05:25 AM

P: 2,453

oh i see thanks for your answer.



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