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## Homework Statement

As part of a larger problem, I need to compute the following integral (over [itex]-1<\theta<1[/itex]):

[tex]\int \sin \theta P_{l}(\cos \theta) d (\cos \theta)[/tex]

## Homework Equations

[tex]\int P_{l}(x) P_{l'}(x) dx= \frac{2}{2l+1} \delta_{l',l}[/tex]

Also, solutions are known to the following integrals:

[tex]\int x P_{l}(x) P_{l'}(x) dx[/tex]

[tex]\int x^{2} P_{l}(x) P_{l'}(x) dx[/tex]

## The Attempt at a Solution

Trig identities for converting [itex]\sin \theta[/itex] into [itex]\cos \theta[/itex] don't seem to help, here - which is why I'm confused.

I would presume to convert the integrand to a form that

*only*has cosines. I would then use the fact that [itex]P_{0}(x)=1[/itex] with orthogonality (listed above) to get the explicit solution.

How do I get rid of that [itex]\sin \theta[/itex]?!

*EDIT:*Alright, I just found another formula online that was not in my text:

[tex]\int g(x) P_{l} (x) dx = \frac{(-1)^{l}}{2^{l} l!} \int (x^{2} - 1)^{l} \frac{d^{l}}{dx^{l}} g(x) dx[/tex]

If I were to use this, I might first convert the sine using

[tex]\sin \theta = (1-\cos ^{2} \theta)^{\frac{1}{2}}[/tex]

and then make that my [itex]g(x)[/itex].

I don't think this would really help me, though, since that would leave me with an awkward expression involving the l'th derivative inside an integral. So I'm still stuck.

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