- #1
erobz
Gold Member
- 4,445
- 1,840
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:
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?
$$ \int \frac{d \theta}{ \sqrt{1 - \cos \theta}} $$
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?