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Integrals, oh yeah

  1. Feb 26, 2005 #1
    integrals, oh yeah......

    hey whats up,
    Ok so....i am having a little bit of a problem on solving the following integral...

    [tex]\int_{0}^{12} 50 + 14 sin\frac{\pi t}{12} dt [/tex]

    would i use....trig substitution or by parts?.... :uhh: yeah...i need some help..thanks.
     
  2. jcsd
  3. Feb 26, 2005 #2

    dextercioby

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    No,need for part integration,a simple,obvious substitution would do it.Pay attention to the change of limits (of integration).

    Daniel.
     
  4. Feb 26, 2005 #3
    wait maybe i dont see it....would i let u= sin pi t/12? and then get du......?
     
    Last edited: Feb 26, 2005
  5. Feb 26, 2005 #4
    The [tex]\int (a + b) = \int a + \int b[/tex]

    So the only thing that seems tricky is the second part.

    Let [tex] u = \frac{\pi{t}}{12} , du = \frac{\pi}{12}[/tex]

    Set [tex] \frac{\pi}{12} = 14dx [/tex] and put the new integral in the form of

    [tex] C*\int sin(u)du[/tex]
     
  6. Feb 26, 2005 #5

    BobG

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    You don't need any substitution. You have the following:

    The second part is:

    [tex]14 \int \sin (kt) dt[/tex]

    where [tex]k=\frac{\pi}{12}[/tex]

    The anti-derivative of sin kt is [tex]\frac{-\cos kt}{k}[/tex]
     
  7. Feb 26, 2005 #6
    Or you could do it like that. Good point. :rolleyes: :tongue2:
     
  8. Feb 26, 2005 #7
    great...thanks a lot guys...i got the right answer for the problem. i just used the substitution of [tex] u = \frac{\pi{t}}{12} [/tex] and brought out all of the constants. Thanks again.
     
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