Struggling with Trigonometric Substitution and Anti-Derivatives?

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Homework Help Overview

The discussion revolves around the topic of trigonometric substitution and anti-derivatives, specifically focusing on the integral involving the expression sqrt{1 + x^(-2/3)}. Participants are exploring methods to approach this integral and are sharing their thoughts on substitution techniques.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various substitution methods, including using trigonometric identities and the potential for multiple substitutions. There is a focus on transforming the integrand into a more manageable form for integration.

Discussion Status

Some participants have offered specific substitution suggestions, while others express confusion about the process and the implications of different substitutions. The conversation reflects a mix of ideas and approaches without a clear consensus on the best method to proceed.

Contextual Notes

There is mention of the original poster's uncertainty with calculus concepts, particularly regarding the use of trigonometric substitution and the manipulation of different variables. The discussion also hints at the complexity of the problem, suggesting that it may require careful consideration of multiple steps.

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anti derive sqrt{1 + x^(-2/3)}


So this isn't actually a homework problem... just a problem that's been bugging me. Supposedly it's possible to do this using u substitution, but I'm having quite a bit of trouble... I've tried making U = x^(-2/3) and I realize that's probably not the best thing to do... Either way I've been able to get everything to a u and get du in there, it's just in comes down to a odd multiplication thing. I'm not looking for an answer really, perhaps somehow to be led in the right direction. Or even just show me how to anti derive sqrt {x^2 + x}. That would help a bit since I'm stuck in a similar scenario.
 
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Pick x^{-2/3} = \tan(\theta)^2, and then \sqrt{1 + x^{-2/3}} = \sqrt{1 + \tan(\theta)^2} = \sqrt{\sec(\theta)^2}. You'll have to be sure to find dx = \frac{d}{d\theta} \tan(\theta)^{-3} and further work for the substitution there, and later some integration by parts for what shows up afterwards. But I feel that approach should work.
 
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The trick to using substitutions is to turn parts of the integrand into things that you already know how to integrate. You might also have to make multiple substitutions before you get something completely manageable. For instance,

\int \frac{du}{\sqrt{1+u^2}}

can be done by a substitution u=\tan v. In your case, you can manipulate your integrand to a form similar to this (though there's another factor of u to some power that might make things a bit harder.) I think that some substitutions of this form are helpful, but haven't worked the whole thing out myself.
 
You should try combining the terms under the root, see if that doesn't help a bit, at this point you should be looking for a substitution. Good luck.

Joe
 
So what you guys are saying is I can sub other things in for u? If \chi-2/3= U, can I really just make U = tan\Theta? I'm just a little confused since I'm semi new to calculus. And that makes things a little more complicated. how would I get d\Theta to replace du? Ahh I am confused. Either way, thank you for the help. I'm sure I'll figure it out eventually.
 
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