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Tricky trig integral (calc 1/2 level)

  1. Aug 12, 2007 #1
    Hello,

    I am an incoming EE freshman who completed Calculus BC one year ago and Multivariable Calculus last Fall. Right now I'm doing a self-study going over basic calculus again in one of my review books.

    Here is my problem:
    [tex]\int sin^3(x)cos^5(x) dx[/tex]

    and I solved it, with an answer my graphing calculator showed to be non-equivalent to the book answer:

    = [tex]\int sin(x)(\frac{1-cos(2x)}{2})(\frac{1+cos(2x)}{2})^2cos(x) dx[/tex]

    (expanding terms) = [tex]\int sin(x)cos(x)(\frac{1+cos(2x)-cos^2(2x)-cos^3(2x)}{8}) dx[/tex]

    using sin(2x) = 2sin(x)cos(x), I integrated:

    [tex]\frac{1}{8} \int \frac{sin(2x)}{2} + \frac{sin(2x)}{2}cos(2x) - \frac{sin(2x)}{2}cos^2(2x) - \frac{sin(2x)}{2}cos^3(2x) dx [/tex]

    u = 2x, du = 2dx
    u = sin(2x), du = 2cos(2x)dx
    u = cos(2x), du = -2sin(2x)dx

    My solution = [tex](1/16)sin^2(2x) - (1/32)cos(2x) + (1/24)cos^3(2x) + (1/32)cos^4(2x) + C[/tex]

    Their solution = [tex]-(1/6)cos^6(x) + (1/8)cos^8(x) + C[/tex]

    Like I said I graphed the solutions on my TI89 and they are not equivalent. What is also very difficult for me to understand is that I graphed my expansion of the integral vs the original integral, showing them to be equal. I then used the integration function to graph the integrals of the original and my expansion and they were shown to be different!

    If there is an easier way to do that problem I would love to see it but I would also like to understand what if anything is wrong with my approach. Thank you!
     
  2. jcsd
  3. Aug 12, 2007 #2

    dextercioby

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    [tex] \int \sin^3 x \cos^5 x {}dx =\int \sin^3 x \cos^4 x {}\cos x {} dx =\int \sin^3 x \left(1-\sin^2 x\right)^2 d\sin x [/tex]

    can you carry on from here ?
     
  4. Aug 12, 2007 #3

    arildno

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    What do you mean by "not equivalent"?

    DO remember that valid antiderivatives can differ by a constant.

    So, if your anti-derivative is a parallell curve to the one TI-89 spat out, you have done it right after all.
     
  5. Aug 12, 2007 #4
    dexter: I dont know why that didn't come to mind it is much simpler! (of course though you mean the last term to be cosx). thank you

    and arildno: I had just looked at the table of values instead of the graph, and actually both solutions have a constant difference of 1/48 so it is actually correct.

    thanks!! problem solved
     
  6. Aug 13, 2007 #5

    dextercioby

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    The last term is the differential one form [itex] d\sin x [/itex] which is nothing but the element of intergration.
     
  7. Aug 14, 2007 #6

    Gib Z

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    Alternatively, [tex]\int \sin^3 x \cos^5 x dx = \int \sin x \sin^2 x \cos^5 x dx = \int (\cos^5 x - \cos^7 x) \sin x dx[/tex]
     
  8. Aug 14, 2007 #7
    I am new to that notation. I suppose that would be the same as leaving dsinx as cosx and integrating with u-substitution?
     
  9. Aug 14, 2007 #8

    HallsofIvy

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    It would be the same as d(sin x)= cos x dx.
     
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