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Intergral of sin^3(x)cos(x)+cos(x)

  1. Jan 24, 2009 #1
    Hi everyone,

    I have worked the following integral two different ways and received two different answers.

    ∫sin^3 xcosx dx+ ∫cosx dx

    The first method
    = ∫sin^3 xcosx dx+ ∫cosx dx
    Let u = sinx
    du/dx = cosx
    dx = 1/cosx du
    = ∫u^3 cosx*(1/cosx) du + ∫cosx dx
    = ∫u^3 du + ∫cosx dx
    = (1/4) u^4+ sinx+C
    = (1/4)sin^4 x+sinx+C

    The second method
    = ∫sinx(sin^2x)cosx dx + ∫cosxdx
    = ∫sinx(1-cos^2x)cosx dx + ∫cosx dx
    Let u = cosx
    du/dx = -sinx
    dx = (-1/sinx) du
    = ∫sinx(1-u^2 )u* (-1/sinx) du + ∫cosx dx
    = -∫(u-u^3)du+∫cosx dx
    = (-1/2)u^2+ (1/4)u^4 + sinx + C
    = (-1/2)cos^2x + (1/4)cos^4x + sinx + C

    If I graph the two answers they are the same graph, but one is shifted vertically by 1/2. Now as an indefinite integral it seems as if this may only be different values of C. But if this were a definite integral wouldn't we drop the C? Meaning that these two equations are not equal? Have I done something wrong in the second equation?


    Thank You, James
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 24, 2009 #2

    Hurkyl

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    You're missing a key idea, either about what C represents, or about the relationship between definite integrals and antidifferentiation.

    But no matter -- that can be fixed by experimentation! Pick your favorite value for C, and compute the definite integral from 0 to 1. Now pick another value for C, and compute again....

    (approximate values are fine!)
     
  4. Jan 24, 2009 #3

    nrqed

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    They are indeed equivalent. If you would have a definite integral, the final result would also be the same because the shift between the two expressions would cancel out. You can check with some numbers and see that it works out.
     
  5. Jan 25, 2009 #4
    Thank You both for your response. My instructor told me the same thing but I had not accepted it yet. I did work it out both ways from 0 to pi and got the same answers. There is still something that I am uneasy about. I will spend some more time with it and see if I can get my question more organized. Thanks again!
     
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