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Integrate $$\int (1+\alpha^{2})^{-3/2} \sin \theta d \theta $$where $\alpha = \cos \theta + a \sin \theta $ with a constant $a$.

How could I possibly do that? Trigonometrical manipulations? Or integration parts?

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Hint Making the change of variables $ \theta=\arctan(t) $ casts the integral to the form

$$ \int \!{\frac {t}{ \left( 2+2\,at+({a}^{2}+1){t}^{2} \right) ^{3/2 }}}{dt}=\frac{1}{\alpha}\int \!{\frac {t}{ \left( (t+\frac{a}{\alpha})^2+\frac{a^2+2}{\alpha} \right) ^{3/2 }}}{dt}\,.$$

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+1 Elite answer. – Babak S. Oct 30 '12 at 14:01

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