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Indefinite integral problem

  1. Mar 31, 2014 #1
    1. The problem statement, all variables and given/known data

    Evaluate the indefinite integral of x*cos(3x)^2

    2. Relevant equations

    Integration by parts: [itex]\int(udv)[/itex]= uv - [itex]\int(vdu)[/itex]

    3. The attempt at a solution

    Im having trouble finding the antiderivative of cos(3x)^2 (which I designated as dv when doing integration by parts). I keep getting (1/9)(sin(3x)^3) using the power and chain rules, but that's not correct.

    Thanks!
     
  2. jcsd
  3. Mar 31, 2014 #2

    Mark44

    Staff: Mentor

    Is this your integral?
    $$\int x cos(9x^2)dx$$

    If so, you might be able to do it by integration by parts, but I wouldn't do it this way. An ordinary substitution will work just fine.
     
  4. Mar 31, 2014 #3

    LCKurtz

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    Or is it$$
    \int x\cos^2(3x)$$in which case a double angle formula and integration by parts would work.
     
  5. Mar 31, 2014 #4

    Mark44

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    Loopas,
    Obviously, what you wrote is ambiguous.
     
  6. Mar 31, 2014 #5

    HallsofIvy

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    Assuming you mean [itex]cos^2(3x)[/itex] use the identity [itex]cos(2a)= cos^2(a)- sin^2(a)= cos^2(a)- (1- cos^2(a)= 2cos^2(a) -1[/itex]. So [itex]cos^2(3x)= (cos(6a)+ 1)/2[/itex].
     
    Last edited by a moderator: Mar 31, 2014
  7. Apr 1, 2014 #6
    Ok, the integral is [itex]\int(xcos^{2}(3x))[/itex]

    Im trying to use integration by parts but I can't find the antiderivative of cos[itex]_{2}[/itex](3x)

    What exactly is the form of the double angle formula?
     
  8. Apr 1, 2014 #7

    Mark44

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    What you wrote is extremely difficult to read. I have fixed it up to make it clear.
     
  9. Apr 1, 2014 #8

    LCKurtz

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    $$\cos^2(3x)=\frac {1+\cos(6x)} 2$$
     
  10. Apr 1, 2014 #9
    So would the correct antiderivative be [itex]\frac{x}{2}[/itex]+[itex]\frac{sin(6x)}{12}[/itex]?
     
  11. Apr 1, 2014 #10

    Mark44

    Staff: Mentor

    It's easy enough to check for yourself. Just differentiate your answer and you should get the integrand.
     
  12. Apr 1, 2014 #11
    Ok so I can use the double angle formula to put cos[itex]^{2}[/itex](3x) into a form that I can find the antiderivative more easily. But I'm confused about how to use the double angle identity to get [itex]\frac{1+cos(6x)}{2}[/itex]
     
  13. Apr 1, 2014 #12

    HallsofIvy

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    The basic double angle formula for cosine says that [itex]cos(2\theta)= cos^2(\theta)- sin^2(\theta)[/itex]. But [itex]sin^2(\theta)= 1- cos^2(\theta)[/itex] so that can be written [itex]cos(2\theta)= cos^2(\theta)- (1- cos^2(\theta))= 2cos^2(\theta)- 1[/itex] From that [itex]2cos^2(\theta)= 1+ cos(2\theta)[/itex] and so [itex]cos^2(\theta)= \frac{1+ cos(2\theta)}{2}[/itex].

    Now replace "[itex]\theta[/itex]" with "3x": [itex]cos^2(3x)= \frac{1+ cos(6x)}{2}[/itex].
     
  14. Apr 1, 2014 #13
    Ahh I can see it now! Thanks guys this was all really helpful!
     
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