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Homework Help: Integration by parts, I need help on this problem.

  1. Aug 23, 2008 #1
    1. The problem statement, all variables and given/known data
    ∫ (theta)^3 *cos(theta)^2


    2. Relevant equations
    integration by parts ∫u dv= uv- ∫v du



    3. The attempt at a solution
    u=theta^3 dv=cos(theta)^2
    du=3theta^2 v=sin(theta)^2

    here's the problem do i use cos(theta)^2 equal to u, because i'm not sure if the antiderivative i got is right

    this is where i'm stuck
     
  2. jcsd
  3. Aug 23, 2008 #2

    rock.freak667

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    If you wrote the problem correctly, then you should put [itex]u=cos(\theta^2)[/itex] since the anti-derivative of it can't be expressed in a way which will help you.
     
  4. Aug 23, 2008 #3
    ok so u=cos(theta)^2

    du=2(theta) *sin(theta)^2
     
    Last edited: Aug 23, 2008
  5. Aug 23, 2008 #4

    rock.freak667

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    wait...was your question


    [tex]\int \theta^3 cos(\theta^2) d\theta[/tex]

    OR

    [tex]\int \theta^3 (cos\theta)^2 d\theta[/tex]
     
  6. Aug 23, 2008 #5
    my question was the 1st one you put
     
  7. Aug 23, 2008 #6

    rock.freak667

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    well in that case. [itex]dv=\theta^3 d\theta[/itex]. v=?. Then put it in the formula.
     
  8. Aug 23, 2008 #7
    ok i think i'm stuck again now that i have my equation it's
    (1/4 cos) (theta)^2 * (theta)^4 - 1/2 ∫theta^4 *(theta)sin(theta)^2
    =(1/4) cos (theta)^4 * (theta)^4 -1/2 ∫ (theta)^4 *(theta) sin(theta)^2


    i edited it
     
    Last edited: Aug 23, 2008
  9. Aug 23, 2008 #8
    v= (1/4) theta^4
     
  10. Aug 23, 2008 #9

    rock.freak667

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    [tex]u=cos(\theta^2)[/tex]

    Let [itex]t=\theta^2 \Rightarrow \frac{dt}{d\theta}=2\theta[/itex]

    [tex]u=cos(t) \Rightarrow \frac{du}{dt}=-sin(t)[/tex]

    [tex]\frac{du}{d\theta}=\frac{du}{dt}*\frac{du}{d\theta}[/tex]

    Check back your du.
     
    Last edited: Aug 23, 2008
  11. Aug 23, 2008 #10
  12. Aug 23, 2008 #11
    i forgot to put sin
     
  13. Aug 23, 2008 #12
    ok so u=cos (theta)^2
    du= - 2(theta) sin theta^2
     
  14. Aug 23, 2008 #13

    rock.freak667

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    You might need to another integration by parts and see if it simplifies.
     
  15. Aug 23, 2008 #14
    =(1/4) cos (theta)^4 * (theta)^4 -1/2 ∫ (theta)^5 * sin(theta)^2

    so i take this part ∫ (theta)^4 * sin(theta)^2
    u=sin (theta)^2
    du= 2 theta * cos (theta)^2
    dv=(theta)^5
    v=(1/6) (theta^6)

    it looks like i'll keep on integrating
     
  16. Aug 23, 2008 #15
    sin (theta)^2 *1/6) (theta^6)-∫ 1/6) (theta^6*2 theta * cos (theta)^2
     
  17. Aug 23, 2008 #16
    afcwestwarrior, that's not necessary.

    Notice that you can split the (I'm going to make x = theta) x^3 cos (x^2) into 0.5(x^2)(2x cos (x^2)). You solve the problem by integrating the 2x cos (x^2) term and differentiating the 0.5x^2 term.
     
  18. Aug 23, 2008 #17
    how'd u make it into 0.5(x^2)(2x cos (x^2)).
     
  19. Aug 23, 2008 #18
    Break the x^3 term into x^2 and x.

    When you differentiate x^2 you get 2x, so what you need to do is to create a 2x term. So the integral breaks down into 0.5x^2 (2x cos (x^2)).
     
  20. Aug 23, 2008 #19
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