How to Integrate sqrt(x/2-x)dx Using Trigonometric Substitution?

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Discussion Overview

The discussion revolves around the integration of the function sqrt(x/2-x)dx, specifically focusing on the correct interpretation of the integral and the application of trigonometric substitution techniques. Participants explore different substitution methods and clarify the expression to be integrated.

Discussion Character

  • Mathematical reasoning, Technical explanation, Debate/contested

Main Points Raised

  • One participant questions whether the integral is \int\sqrt{\frac{x}{2}-x}\space dx or \int\sqrt{\frac{x}{2-x}}\space dx.
  • Another participant assumes the latter interpretation and proposes a substitution method involving u= \sqrt{2-x} to transform the integral.
  • There is a repeated assertion of the substitution method, with a participant seeking clarification on the transition from u= \sqrt{2-x} to the integral -2 \int \sqrt{2-u^2} du.
  • A participant explains the relationship between u and x, stating that if u=\sqrt{2-x}, then \sqrt{x}=\sqrt{2-u^2} and provides the derivative for the substitution.
  • Discussion includes a question about the nature of trigonometric substitution, with one participant referencing a standard form involving a^2-x^2.
  • Another participant refines the trigonometric substitution concept, suggesting a specific substitution involving u^2=2sin^2\theta.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the interpretation of the integral or the substitution methods, with multiple views and clarifications presented throughout the discussion.

Contextual Notes

There are unresolved aspects regarding the assumptions made about the integral's form and the steps involved in the substitution process. The discussion reflects varying interpretations and approaches without definitive conclusions.

DigiDigi
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How do I integrate sqrt(x/2-x)dx?
 
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Is the problem:

\int\sqrt{\frac{x}{2}-x}\space dx or \int\sqrt \frac{x}{2-x}\space dx ?
 
I'm assuming it is the later. \int \sqrt{\frac{x}{2-x}} dx= \int \frac{\sqrt{x}}{\sqrt{2-x}} dx

Use the substitution u= \sqrt{2-x} to turn the integral into -2 \int \sqrt{2-u^2} du, which you can do by trig substitution.
 
HS-Scientist said:
I'm assuming it is the later. \int \sqrt{\frac{x}{2-x}} dx= \int \frac{\sqrt{x}}{\sqrt{2-x}} dx

Use the substitution u= \sqrt{2-x} to turn the integral into -2 \int \sqrt{2-u^2} du, which you can do by trig substitution.

Yes,it this one.I know U substitution, but I think I miss out something.How did you change u= \sqrt{2-x} to turn the integral into -2 \int \sqrt{2-u^2} du
 
If u=\sqrt{2-x}, then \sqrt{x}=\sqrt{2-u^2} and du=-\frac{dx}{2\sqrt{2-x}}.
 
When you said trigonometric substitution, you mean a^2-x^2=1-sin^2(x)?
 
More like the a^2-x^2=a^2(1-sin^2{\theta}) that you will get when you set u^2=2sin^2\theta
 

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