Trouble remembering integration

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Homework Help Overview

The discussion revolves around the topic of integration, specifically focusing on the integral of the cosine function, \(\int \cos(60 \pi x) \, dx\). Participants are exploring the relationship between integration and differentiation, as well as methods to remember integration techniques.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the integration of cosine functions and the importance of understanding the relationship between integration and differentiation. There are attempts to verify integration results through differentiation.

Discussion Status

Some participants have provided guidance on remembering integration techniques, emphasizing the connection to differentiation. Multiple interpretations of the integration process are being explored, with no explicit consensus reached.

Contextual Notes

There is an emphasis on recalling derivative rules to aid in remembering integration, and some participants mention the need for a constant in indefinite integrals. The discussion reflects a learning environment where assumptions about prior knowledge are being questioned.

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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?
 
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Yes, that's correct... plus a constant if you're looking for the indefinite integral.
 
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).
 
integration of cos...

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