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Integration by parts with e and sine

  1. Apr 13, 2014 #1
    1. The problem statement, all variables and given/known data

    Evaluate the integral.

    2. Relevant equations

    [tex] \int e^{2x} sin(3x) dx [/tex]

    3. The attempt at a solution

    I began by using integration by parts.

    [tex] u = sin(3x) [/tex]

    [tex] v = \frac {e^{2x}} {2} [/tex]

    [tex] du = 3 cos(3x) [/tex]

    [tex] dv = e^{2x} dx [/tex]

    but I get stuck after that because the integral keeps breaking down into more and more integrals, and I never arrive at the correct answer. I also may be making this too complicated. Any ideas?
     
    Last edited: Apr 13, 2014
  2. jcsd
  3. Apr 13, 2014 #2

    Zondrina

    User Avatar
    Homework Helper

    I believe you meant to write your integral as:

    $$I = \int e^{2x} sin(3x) dx $$

    Notice I've called the above integral ##I##, the reason for this will become apparent soon.

    Make the substitution ##u = sin(3x)## and ##dv = e^{2x} dx##. What is the result of the following integral:

    $$uv - \int v du$$
     
  4. Apr 13, 2014 #3

    Mark44

    Staff: Mentor

    You need to do integration by parts once more. If you use similar substitutions, you'll end up with another integral whose integrand is e2xsin(3x). At that point, you can solve algebraically for this integral.
     
  5. Apr 13, 2014 #4
    When you do parts twice, the same function will pop up inside the integral on both sides:

    [itex]\int e^{2x}Sin3x dx = \frac{1}{2} e^{2x} Sin3x - \frac{3}{4}e^{2x} Cos3x - \frac{9}{4} \int e^{2x} Sin3x dx[/itex]

    Notice how the integrand is the same on both sides, this means that you can move your 9/4 over to the left using algebra. If you call your integral I in this case, all you need to do is solve for I.

    You should put this special case in your memory bank - it always happens with trig functions and e since neither will be reduced by using parts.
     
  6. Apr 14, 2014 #5
    I like to do these problems in the following way as it gives the answer quickly.

    Consider:
    $$A=\int e^{2x}\cos (3x)\,dx$$
    $$B=\int e^{2x}\sin (3x)\,dx$$
    Hence,
    $$A+iB=\int e^{2x}e^{i3x}\,dx=\int e^{(2+3i)x}\,dx=\frac{e^{(2+3i)x}}{2+3i}+C$$
    To obtain the given integral, compare the imaginary parts.
     
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