- #1

erobz

Gold Member

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

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

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?