Evaluate the definite integral

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Homework Help Overview

The discussion revolves around evaluating the definite integral of the function f(((x^2)(sinx))/(1+x^6)) from a = (-pi/2) to (pi/2). Participants are exploring substitution methods and the implications of their choices in the context of integral calculus.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts a substitution method but is questioned on its effectiveness. Some participants suggest alternative substitutions and express confusion regarding the implications of taking the square root of the numerator and denominator.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's substitution setup and exploring alternative approaches. There is a recognition of mistakes made in the reasoning process, but no consensus has been reached on a definitive method.

Contextual Notes

Participants are navigating the complexities of substitution in integration and are addressing potential misunderstandings regarding the manipulation of the integral's components.

sapiental
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the definite integral:

from a = (-pi/2) to (pi/2) f(((x^2)(sinx))/(1+x^6))dx

this is the way it seems most logic to me to set it up using substitution:

u = x
du = dx

from a = (-pi/2) to (pi/2) f(((u^2)(sinu))/(1+u^6))du

= (((-cos(u))(1/3u^3))/(u+1/7u^7))+CI know how to evaluate it from here, I just need some feedback on my substitution setup.

Thanks in advance.
 
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sapiental said:
u = x
du = dx
That sort of substitution does nothing but change the letter denoting the variable.
 
Try sketching the graph. Notice anything?

courtrigrad said:
Take the square root of the numerator and denominator and use the substitution [tex]x^{3} = \tan \theta, x = \sqrt[3]{\tan \theta}[/tex] You end up getting [tex]\frac{1}{3} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \tan \theta[/tex]

I'm worried about what "Take the square root of the numerator and denominator" could possibly mean? I don't see how your substitution gives what you claim it gives either.
 
yeah, I made a mistake. I took the integral of the square root of the function instead of the actual function.
 

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