How Do You Integrate Cosine Powers in Calculus?

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Homework Help Overview

The discussion revolves around the integration of cosine powers, specifically the integral of the form \(\int \cos^{2n+1}(x) \, dx\). Participants are exploring methods to approach this integral within the context of calculus.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • One participant attempts to rewrite the integral using the identity \(\cos^2(x) + \sin^2(x) = 1\) and expresses it in terms of sine. Others suggest using the binomial theorem and discuss the implications of expressing the result in terms of hypergeometric functions.

Discussion Status

Participants are actively engaging with the problem, sharing their attempts and discussing potential approaches. Some guidance has been provided regarding the use of the binomial theorem, but there is no explicit consensus on a single method or solution.

Contextual Notes

There is mention of a lack of a simple formula for the integral, and participants are navigating through the complexities of expressing the result in different forms. Additionally, there are informal notes about formatting in LaTeX, indicating a learning curve in the discussion.

golanor
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Homework Statement



[itex]\int { { cos }^{ 2n+1 }(x)dx }[/itex]

Homework Equations


[itex]{ cos }^{ 2 }+{ sin }^{ 2 } = 1[/itex]

The Attempt at a Solution


i got to here:
[itex]\int { { (1-{ sin }^{ 2 }(x)) }^{ n }d(sin(x)) }[/itex]

Any help would be appreciated!
 
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So you got to find

[tex]\int (1-x^2)^ndx.[/tex]

Use the binomial theorem and linearity of the integral.
 
golanor said:

Homework Statement



[itex]\int { { cos }^{ 2n+1 }(x)dx }[/itex]

Homework Equations


[itex]{ cos }^{ 2 }+{ sin }^{ 2 } = 1[/itex]

The Attempt at a Solution


i got to here:
[itex]\int { { (1-{ sin }^{ 2 }(x)) }^{ n }d(sin(x)) }[/itex]

Any help would be appreciated!

There is no simple formula for the result, unless you want to express it in terms of a hypergeometric function. The answer is a polynomial of degree (2n+1) in ##\sin(x)##.

BTW: in [t e x] you should write "\sin(x)" instead of "sin(x)": the difference is ##\sin(x)## (nice) vs. ##sin(x)## (not nice).
 
Ray Vickson said:
There is no simple formula for the result, unless you want to express it in terms of a hypergeometric function. The answer is a polynomial of degree (2n+1) in ##\sin(x)##.

BTW: in [t e x] you should write "\sin(x)" instead of "sin(x)": the difference is ##\sin(x)## (nice) vs. ##sin(x)## (not nice).
Thanks for the tex hint..still getting the hang of it (used to microsoft equation editor =\)

micromass said:
So you got to find

[tex]\int (1-x^2)^ndx.[/tex]

Use the binomial theorem and linearity of the integral.
Thanks, i actually read about it, thought that i should use it but decided not to.

In case anyone wants the answer, it is:
[tex]\int { { (1-{ \sin }^{ 2 }(x)) }^{ n }d(\sin (x)) } =\sin (x)-\frac { n{ \sin }^{ 3 }(x) }{ 3 } +...+\frac { { (-1) }^{ n }{ \sin }^{ 2n+1 }(x) }{ 2n+1 } +C[/tex]
 

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