Integration Using Trigonometric Substitution

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    Calculus Integral
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Discussion Overview

The discussion revolves around the integration of the function $$ \int \frac{d \theta}{ \sqrt{1 - \cos \theta}} $$ using trigonometric substitution techniques. Participants explore various methods, including Weierstrass substitution and power-reducing identities, while expressing uncertainty about the integrability of the function.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant proposes a trigonometric substitution involving a right triangle to simplify the integral, but later questions the validity of their derivative calculations.
  • Another participant suggests that the integral may be non-elementary, as it does not appear in their integration tables.
  • A participant recommends Weierstrass substitution as a potential technique for solving the integral.
  • Power-reducing identities are mentioned as a method to simplify the expression $$1 - \cos(\theta)$$, which could make the integral more manageable.
  • Some participants express curiosity about the solution found on Wolfram Alpha, noting that it did not appear straightforward.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method for solving the integral, and multiple competing techniques are proposed. There is uncertainty regarding the integrability of the function.

Contextual Notes

Participants express limitations in their knowledge of integration techniques and the complexity of the integral, indicating that it may not be easily solvable with standard methods.

erobz
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I've got this integral I'm trying to find:

$$ \int \frac{d \theta}{ \sqrt{1 - \cos \theta}} $$

To me it smells like trig sub, so I investigate the right triangle:

1723165771053.png


Such that:

$$ \cos u = \sqrt{1-cos \theta} $$

we also have from the same triangle:

$$ \sin u = \sqrt{\cos \theta} $$

Square both sides and differentiate w.r.t ## \theta##

$$ \sin ^2 u = \cos \theta$$

$$ \frac{d}{d \theta} \sin ^2 u = \frac{d}{d \theta} \sqrt{ 1 - \sin^2 \theta } $$

$$ \implies 2 \sin u \cos u \frac{du}{d \theta} = -2 sin \theta \cos\theta $$

$$ \implies \sin ( 2 u ) \frac{du}{d\theta} = \sin (- 2 \theta ) $$

$$ \implies \frac{du}{d\theta} = -1 $$

Which should make the integral:

$$ - \int \frac{du}{\cos u} $$

Is that legitimate?
 
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Rats...I see my tom foolery already in the derivative. Never mind!

I'll let it stand in case someone has a technique. My guess is it's in fact non-elementary as I don't see it in my integration tables...
 
Last edited:
##1-\cos(\theta) = 2\sin^2(\frac\theta 2)##
 
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martinbn said:
##1-\cos(\theta) = 2\sin^2(\frac\theta 2)##
Power reducing! I guess that makes it quite solvable indeed!

Thank You!
 
fresh_42 said:
Yes, it is. But, honestly, I have looked at the solution on WA, and it did not look nice.
What is WA?
 

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