# Integral help

## Homework Statement

trying to integrate this:
$$\int^{\theta}_{\theta_{0}} \sqrt{\frac{1-cos(\theta)}{cos(\theta_{0}) - cos(\theta)}d\theta$$

## Homework Equations

My book tells me to let theta = pi - 2gamma and then simplify from there but I'm just not seeing that ! any hints? is there a trig identity that I'm missing ? (the only thing I see is to change the sign of the cosines but I don't see where that gets me).

## The Attempt at a Solution

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dextercioby
Homework Helper
Your integrand kind of blows up in the lower limit. Make the following substitutions and see what you get

$$\cos \theta_0 = A \, , \, \tan\frac{\theta}{2} = x$$

Do you get a simplification ?

SammyS
Staff Emeritus
Homework Helper
Gold Member
Get the dθ out of the radicand.

$$\int^{\theta}_{\theta_{0}} \sqrt{\frac{1-\cos(\theta)}{\cos(\theta_{0}) - \cos(\theta)}}\ d\theta$$

What do you get if you use the hint, $$\text{Let }\theta=\pi-2\,\gamma\,?$$

$$\cos(\theta)=\cos(\pi-2\,\gamma)=\cos(\pi)\cos(2\,\gamma)+\sin(\pi)\sin(2\,\gamma)=\quad?$$

Then you have a choice for $$\cos(2\,\gamma).\quad\quad\cos(2x)=2\cos^2(x)-1=1-2\sin^2(x)$$

Btw... the upper limit should be pi...

And I forgot about my double angle formula... just now working through simplifying with that

$$\cos(\theta)=\cos(\pi-2\,\gamma)=\cos(\pi)\cos(2\,\gamma)+\sin(\pi)\sin( 2\,\gamma)= -cos(2\gamma)$$