Integrating (9-x^2)^{5/2}: Solving Integration Frustration

  • Thread starter negatifzeo
  • Start date
In summary: Thanks for pointing out my error.You are, of course, correct. I should have been more careful. Thanks for pointing out my error.In summary, the conversation discusses the integration of the function \int (9-x^2)^{5/2} dx using the substitution method. It is suggested to let x=3sin\theta and dx=3 cos \theta to simplify the integral. After some algebraic manipulation, the integrand becomes cos^6 \theta. It is then recommended to use a reduction formula or look up the integral in a table of integrals. There is also a note to be careful with the exponent of the cosine term.
  • #1
negatifzeo
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Homework Statement


[itex]\int (9-x^2)^{5/2} dx [/itex]


Homework Equations





The Attempt at a Solution


Letting[tex] x=3sin\theta[/tex] and [tex] dx=3 cos \theta[/tex],

[itex] 243\int(1-sin^2\theta)^{5/2}3 cos \theta d\theta[/tex]

Taking the 3 out gets

[tex] 729 \int (1-sin^2 \theta)^{5/2} cos \theta d\theta [/tex]

I'm not sure where to go from here. Do I combine everything algebraically and then do a u sub? Or do I integrate by parts here?
 
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  • #2
negatifzeo said:

Homework Statement


[itex]\int (9-x^2)^{5/2} dx [/itex]


Homework Equations





The Attempt at a Solution


Letting[tex] x=3sin\theta[/tex] and [tex] dx=3 cos \theta[/tex],

[itex] 243\int(1-sin^2\theta)^{5/2}3 cos \theta d\theta[/tex]

Taking the 3 out gets

[tex] 729 \int (1-sin^2 \theta)^{5/2} cos \theta d\theta [/tex]

I'm not sure where to go from here. Do I combine everything algebraically and then do a u sub? Or do I integrate by parts here?

Well, you could remember that [tex] (1 - sin^2 \theta) = cos^2 \theta [/tex] so that your integrand becomes [tex] cos^6 \theta [/tex] and then look that integral up in a table of integrals, or look up a reduction formula for [tex] cos^6 \theta [/tex]. You'll be able to integrate the expression that results from the reduction formula.
 
  • #3
AEM said:
Well, you could remember that [tex] (1 - sin^2 \theta) = cos^2 \theta [/tex] so that your integrand becomes [tex] cos^6 \theta [/tex] and then look that integral up in a table of integrals, or look up a reduction formula for [tex] cos^6 \theta [/tex]. You'll be able to integrate the expression that results from the reduction formula.

Careful; [tex](\cos^2 \theta)^{5/2}=|\cos^5 \theta|\neq\cos^5 \theta[/tex] in general.
 
  • #4
gabbagabbahey said:
Careful; [tex](\cos^2 \theta)^{5/2}=|\cos^5 \theta|\neq\cos^5 \theta[/tex] in general.

You are, of course, correct. I should have been more careful.
 

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