Trouble remembering integration

  • Thread starter LocationX
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  • #1
LocationX
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Hi, I'm having trouble remembering integration...

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

this becomes... sin(60 pi x)/60 pi?
 

Answers and Replies

  • #2
jamesrc
Science Advisor
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477
1
Yes, that's correct... plus a constant if you're looking for the indefinite integral.
 
  • #3
quasar987
Science Advisor
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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).
 
  • #4
borisleprof
36
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integration of cos...

Yes it's ok. Remember how to derive sin and cos and it will surely be easier for you
 
  • #5
bomba923
760
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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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