Substitutions for some trig integrals?

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SUMMARY

The discussion focuses on finding substitutions for the integrals involving the expression \((a^2 - b^2 \cos\theta)^{-3/2}\). The user initially struggles with three specific integrals but realizes a substitution for the second integral after posing the question. The integrals in question are \(\int (a^2 - b^2 \cos\theta)^{-3/2} d\theta\), \(\int (a^2 - b^2 \cos\theta)^{-3/2} \sin\theta d\theta\), and \(\int (a^2 - b^2 \cos\theta)^{-3/2} \cos\theta d\theta\). The user also expresses interest in the integral \(\int (a^2 - b^2 \cos\theta)^{-3/2} e^{i\theta} d\theta\

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Peeter
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Does anybody know substitutions that can be used for the following integrals:

[tex] \int (a^2 - b^2 \cos\theta)^{-3/2} d\theta[/tex]
[tex] \int (a^2 - b^2 \cos\theta)^{-3/2} \sin\theta d\theta[/tex]
[tex] \int (a^2 - b^2 \cos\theta)^{-3/2} \cos\theta d\theta[/tex]

I also had a look for these in my table of integrals, but didn't see anything appropriate, but am also a bit rusty, and could just be overlooking the obvious.

EDIT: oops. fixed sign in square root.
 
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Funny, how asking the question is enough to see the answer. I see what to do with the second one now, and it gets me halfway to:

[tex] \int (a^2 - b^2 \cos\theta)^{-3/2} e^{i\theta} d\theta[/tex]

which is actually what I was after for the second two. However, it's still not obvious to me what to do with the first nor third integrals.
 

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