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Trouble remembering integration

  1. Sep 2, 2005 #1
    Hi, I'm having trouble remembering integration...

    [tex]\int cos(60 \pi x) [/tex]

    this becomes... sin(60 pi x)/60 pi?
  2. jcsd
  3. Sep 2, 2005 #2


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    Yes, that's correct... plus a constant if you're looking for the indefinite integral.
  4. Sep 2, 2005 #3


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    Basically, if you remember derivation, there is nothing to remember about integration except that it is the inverse of derivation. If you doubt that the integral of cos(60x) is sin(60x)/60, just verify that the derivative of sin(60x)/60 is cos(60x).
  5. Sep 2, 2005 #4
    integration of cos...

    Yes it's ok. Remember how to derive sin and cos and it will surely be easier for you
  6. Sep 3, 2005 #5
    Just remember that nice substitution rule...
    [tex] \int {\cos \left( {60\pi x} \right)dx} = \frac{1}{{60\pi }}\int {\cos \left( 60\pi x \right)d\left( {60\pi x} \right)} = \boxed{\frac{{\sin \left( {60\pi x} \right)}}{{60\pi }} + C} [/tex]

    ...and check via derivative (chain rule here) :smile:
    [tex] \frac{d}{{dx}}\left[ {\frac{{\sin \left( {60\pi x} \right)}}{{60\pi }} + C} \right] = \frac{{\cos \left( {60\pi x} \right) \cdot 60\pi }}{{60\pi }} = \cos \left( {60\pi x} \right) [/tex]
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