University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{-1}{7} (\dfrac{\sqrt{1-r^2}}{r})^7 +C$
Let us consider $r=\sin \theta \implies dr = \cos \theta d\theta$ Now, the given integral can be written as: $\int\dfrac{(1-\sin^2 \theta)^{5/2} \cos \theta d\theta}{\sin^8 \theta} =-\dfrac{1}{7} \cot^7 \theta+C$ or, $=\dfrac{-1}{7} (\dfrac{\sqrt{1-r^2}}{r})^7 +C$
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