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Could anyone tell me what change of variables I can use to make $$\int_0^\pi \sin^4 x ({1\over a^2} \sin^4 x+ \sin^2 x \cos^2 x)^{-3/2} dx$$ into $\int_0^1 (1+(a^2-1)y^2)^{-3/2} dy$? Thank you very much!

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Your answer is missing an $a^3$ in the front – Kirthi Raman Mar 12 '12 at 19:05

Observe that $$\sin^4 x\left( \frac{1}{a^2} \sin^4x + \sin^2 \cos^2 x \right)^{-3/2}=a^3\frac{\sin^4 x}{\sin^3 x}\left(\sin^2 x + a^2 \cos^2 x \right)^{-3/2}$$ this expression so it's also equal to $$a^3 \sin x \left( 1 + (a^2-1)cos^2 x\right )$$ and so letting be $y = -\cos x$ you have that $dy = \sin x \ dx$ and thus you get your substitution, of course you have to restrict the domain of integration to $\left[0,\frac{\pi}{2}\right]$in order to use this substitution, otherwise this is not invertible.

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