Integrating Sin(6θ): Am I Close?

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Homework Help Overview

The discussion revolves around the integration of the function sin(6θ), specifically the indefinite integral ∫sin(6θ) dθ. Participants are exploring the methods and reasoning behind finding the antiderivative of this trigonometric function.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to integrate sin(6θ) and questions if their approach is correct. Some participants suggest using a u-substitution with u=6θ, while others reference the relationship between derivatives and antiderivatives to clarify the integration process.

Discussion Status

Participants have provided guidance on the integration process, with some affirming the correctness of the original poster's approach. There is an ongoing exploration of different methods to arrive at the antiderivative, but no explicit consensus has been reached.

Contextual Notes

There is a mention of potential confusion regarding the integration process, as well as the need for clarity on the relationship between differentiation and integration in this context.

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Homework Statement



[tex]\int\sin(6\theta) d\theta[/tex]


Homework Equations





The Attempt at a Solution



[tex]\int6\cos6\theta[/tex]

Am I close?
 
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If it helps do a u-sub where u=6theta

Or you could think of it like this:

If we take the derivative of [tex]\frac{d}{d \theta}(-cos(\theta)) = sin(\theta)[/tex]. That's sort of like the anti-derivative which you seek.

If we take the derivative of [tex]\frac{d}{d \theta}(-cos(6 \theta)) = 6sin(6 \theta)[/tex].

So from there it's easy to see that[tex]\int\sin(6\theta) d\theta = \frac{-cos(6\theta)}{6}[/tex]
 
errr...

[tex]\frac{-\cos6\theta}{6}[/tex]
 
thanks :)
 
Yeah looks right^^
 

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