# Trouble remembering integration

1. Sep 2, 2005

### LocationX

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

$$\int cos(60 \pi x)$$

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

2. Sep 2, 2005

### jamesrc

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

3. Sep 2, 2005

### 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).

4. Sep 2, 2005

### borisleprof

integration of cos...

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

5. Sep 3, 2005

### 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)
$$\frac{d}{{dx}}\left[ {\frac{{\sin \left( {60\pi x} \right)}}{{60\pi }} + C} \right] = \frac{{\cos \left( {60\pi x} \right) \cdot 60\pi }}{{60\pi }} = \cos \left( {60\pi x} \right)$$