Trouble remembering integration

[LocationX]

Hi, I'm having trouble remembering integration...

\int cos(60 \pi x)

this becomes... sin(60 pi x)/60 pi?

 

[jamesrc]

Yes, that's correct... plus a constant if you're looking for the indefinite integral.

 

[quasar987]

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).

 

[borisleprof]

integration of cos...

Yes it's ok. Remember how to derive sin and cos and it will surely be easier for you

 

[bomba923]

Just remember that nice substitution rule...
\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}

...and check via derivative (chain rule here)
\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)