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Intigration problem

  1. Jan 26, 2006 #1
    I just cannot solve this intigrating problem, either i can't think straight, or it just cant be done..

    [tex] e^x cos(x) [/tex]

    If you intigrate by parts, its series that goes on and on... so is it even possible??
  2. jcsd
  3. Jan 26, 2006 #2


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    Integrate by parts twice and you'll get another integral that looks like your original one. Then you can add that to the original, so that the stuff that's left is equal to twice your original integral.
  4. Jan 27, 2006 #3
    the thing to this question is [tex] e^x cos(x) [/tex]
    you integrate it a few times and when you integrate it you will go back to this function [tex] e^x cos(x) [/tex]

    (integral [tex] e^x cos(x) [/tex] = blah blah blah blah blah +-(either plus or minus: say minus) integral [tex] e^x cos(x) [/tex]dx)
    so now you add the integral([tex] e^x cos(x) [/tex]
    ) to the left side and becomes 2*
    so then this becomes
    2*integral[tex] e^x cos(x) [/tex]=blah blah blah blah blah

    integral[tex] e^x cos(x) [/tex]=(blah blah blah blah blah)/2 now you solved it
    Last edited: Jan 27, 2006
  5. Jan 27, 2006 #4


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    No, it cannot be either plus or minus, it must be minus, if it's a plus then after rearraninging it, you will get something like:
    [tex]\int 0 \ dx = \mbox{constant}[/tex].
    And obviously, you don't want to get this result, right?
    @ Pearce_09, have you worked it out yet?
  6. Jan 27, 2006 #5


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    Another way...

    Another way to evaluate [tex]\int e^{x}\cos x dx[/tex] is:

    By Euler's equation, viz. [tex] e^{ix}= \cos x +i\sin x ,[/tex] and hence

    [tex] e^{x}e^{ix} = e^{x}\left( \cos x +i\sin x\right) = e^{x}\cos x +ie^{x}\sin x[/tex]

    so that we have

    [tex]\Re\left[ e^{x}e^{ix} \right] = e^{x}\cos x ,[/tex]

    where [itex]\Re\left[ z\right] [/itex] denotes the real part of z, (which for z=x+iy is x)

    and hence

    [tex]\int e^{x}\cos x dx = \Re\left[ \int e^{x}e^{ix} dx\right] = \Re\left[ \int e^{(1+i)x} dx\right] = \Re\left[ \frac{1}{1+i} e^{(1+i)x} + C\right] [/tex]
    [tex] = \Re\left[ \frac{1}{2}(1-i) e^{x}\left( \cos x +i\sin x\right) \right] + C_1= \Re\left[ \frac{1}{2}\left( e^{x}\cos x + e^{x}\sin x\right) + i\frac{1}{2}\left( -e^{x}\cos x + e^{x}\sin x\right)\right] + C_1 [/tex]
    [tex] = \frac{1}{2}\left( e^{x}\cos x + e^{x}\sin x\right) +C_1 [/tex]

    where [tex]C_1[/tex] is a real contant, and a 'free' corollary is:

    [tex]\int e^{x}\sin x dx = \Im\left[ \int e^{x}e^{ix} dx\right] = \frac{1}{2}\left( -e^{x}\cos x + e^{x}\sin x\right) +C,[/tex]

    where [itex]\Im\left[ z\right] [/itex] denotes the imaginary part of z, (which for z=x+iy is y.)
    Last edited: Jan 27, 2006
  7. Jan 27, 2006 #6
    Easier way is (as others say above) put

    [tex]S(x) = \int e^x \cos x dx = e^x \sin x - \int e^x \sin x dx[/tex]


    [tex]\int e^x \sin x dx = -e^x \cos x +\int e^x \cos x dx[/tex]

    Inserting this into the first formula,

    [tex]S(x) = e^x \sin x + e^x \cos x - S(x)[/tex]

    and we have

    [tex]S(x)=\frac {e^x} 2 (\sin x + \cos x)[/tex]
  8. Jan 27, 2006 #7


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    No complete solution, please!!!!!!!! :grumpy: :grumpy: :grumpy: :grumpy:
    You most let the OP think, that's for his own's shake. Unless you can see that he (the OP) has tried damn hard and still cannot figure out the problem, don't give out complete solution. READ HERE! :grumpy: :grumpy:
    Or you can try to post a new solution (like what benorin did) that the OP has not known yet, so after he works out the problem on his own way, he'll have 2 different ways to solve the problem! And that's good for him!
    And look at your solution again, where's the constant of integration????
  9. Jan 27, 2006 #8
    oh..sorry for that!!
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