How can I evaluate this integral using basic integration techniques?

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Homework Help Overview

The discussion revolves around evaluating the integral \(\int \frac{1}{\sqrt{8x-x^2}} dx\) using basic integration techniques. Participants are exploring methods to simplify the integrand and apply integration strategies.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to multiply the integrand by \(\frac{\sqrt{8x-x^2}}{\sqrt{8x-x^2}}\) but finds it unhelpful. They question whether completing the square is a viable approach. Some participants suggest that completing the square is a good starting point and propose a substitution afterward. Others express doubt about the effectiveness of completing the square in this context.

Discussion Status

Participants are actively discussing various approaches to the integral, including completing the square and substitution methods. There is recognition of the challenges faced with the negative term in the expression, and some guidance has been offered regarding substitutions and the consideration of inverse trigonometric functions.

Contextual Notes

There is an ongoing discussion about the appropriateness of certain algebraic manipulations and the implications of negative terms in the square root. The participants are navigating the constraints of the problem while adhering to basic integration techniques.

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Homework Statement


I am reviewing some basic integration techniques. how would i evaluate this?


Homework Equations


[itex]\displaystyle\int {\frac{1}{\sqrt{8x-x^2}} dx}[/itex]


The Attempt at a Solution


i've tried multiplying by

[itex]\frac{\sqrt{8x-x^2}}{\sqrt{8x-x^2}}[/itex] but that isn't getting me anywhere.

should i use something like completing the square?
 
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Completing the square is a good way to begin. Afterwards, a clever substitution can help.
 
okay. I am stuck with the subtracted term.

[itex]\frac{1}{-\sqrt{(x^2 - 8x + 16) - 16}}[/itex]

[itex]\frac{1}{-\sqrt{(x-4)^2 - 16}}[/itex]
 
i don't think completing the square works here.
 
whatlifeforme said:
okay. I am stuck with the subtracted term.

[itex]\frac{1}{-\sqrt{(x^2 - 8x + 16) - 16}}[/itex]

[itex]\frac{1}{-\sqrt{(x-4)^2 - 16}}[/itex]

Looks like you pulled a negative outside of the square root. You can't do that! Try again.
 
Once you've fixed up the negative, try the substitution u=(x-4) and then consider what the derivatives of the inverse trigonometric functions look like.
 
great. thanks phosgene.

so we are looking at:

answer:
[itex]arcsin(\frac{x-4}{4}) + c[/itex]
 

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