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How do I find the integral of sin^3 (t) cos(tà

  1. Feb 9, 2005 #1
    How would you do: [itex] \int \sin^{3} t cos t [/itex]? Would [itex] u = t\cos t [/itex]?

    Thanks
     
  2. jcsd
  3. Feb 9, 2005 #2

    dextercioby

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    Isn't your integral just
    [tex] \int \sin^{3}t \ \cos t \ dt [/tex]

    ...?

    Daniel.
     
  4. Feb 9, 2005 #3
    yes. so would [itex] u = t, dt = du [/tex]

    Also for [tex] \int (x-1)e^{x^{2} - 2x} [/tex] woud [tex] u = x^{2} - 2x [/tex]? Because I know that the answer is [tex] \frac{1}{2}(x-1)e^{x^{2} - 2x} [/tex].

    Thanks
     
  5. Feb 9, 2005 #4

    dextercioby

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    For the first part,that's not a valid change of variable.

    For the second,it's okay.

    Daniel.
     
  6. Feb 9, 2005 #5
    would i apply the identity [tex] sin^{2} x = 1 - cos^{2} x [/tex]? Also for the second one, once i make the substitution how do we get the [tex] \frac{1}{2} [/tex] in the front?

    Thanks :smile:
     
  7. Feb 9, 2005 #6

    dextercioby

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    [tex] \int sin^{3}x \cos x \ dx=\int \sin^{3}x d(\sin x) =\frac{\sin^{4}x}{4} +C [/tex]

    As for the second,the 1/2 comes from the cancelation of the 2 which would result from the derivative of the exponent.

    Daniel.
     
  8. Feb 9, 2005 #7
    ok thanks a lot. So you substituted [tex] d(sin x) = cos x [/tex] so we get our elementary integrations.

    Thanks
     
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