Integration Using Trigonometric Substitution

  • #1
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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  • #2
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:
  • #4
##1-\cos(\theta) = 2\sin^2(\frac\theta 2)##
 
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  • #5
martinbn said:
##1-\cos(\theta) = 2\sin^2(\frac\theta 2)##
Power reducing! I guess that makes it quite solvable indeed!

Thank You!
 
  • #8
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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