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I found this integral in a book:

link here

The integrated function is a phase function or just a probability density function.

I don't understand why instead of dω' = sinθ dφ dθ in this case dω' = dφ dcosθ.

1 Answers 1

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Hint: $$ \frac{d \cos\theta}{d\theta} = -\sin \theta, $$ and I suppose the minus sign is absorbed in $p(\cos \theta)$.

If the above is incorrect, please comment more details pertaining to the question.

Thanks!

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    Thanks a lot! That seems to explain it. – 2017-02-17
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    The minus sign is absorbed by the switching of integration order $$\int_0^{\pi} \sin{\theta} \mathrm d \theta = \int_1^{-1} - \mathrm d \cos{\theta} = \int_{-1}^1 \mathrm d \cos{\theta}$$ – 2017-02-17
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    Yes. Saw that now. Thanks. – 2017-02-17