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How to evaluate $$\int \frac{\sin^3 x}{\cos^5x}dx\ ?$$

I've tried various substitutions with $\sin x = u$ or $\cos x = u$, I've tried using Euler's formula which result in too heavy calculations and I've tried using $\sin^2x + \cos^2x = 1$ in various forms without success.

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    $\begingroup$ You tried $u=\cos x$? What was the output? $\endgroup$ – Did May 2 '14 at 20:46
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Hint: $$\int \frac{\sin^{3}x}{\cos^{5}x}dx=\int \frac{(1-\cos^{2}x)\sin x}{\cos^{5}x}dx$$ and try substitution $$t=\cos x.$$

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Another approach: $$ \begin{align} \int \frac{\sin^3 x}{\cos^5x}dx&=\int \frac{\sin^3 x}{\cos^3x\cos^2x}dx\\ &=\int\left(\frac{\sin x}{\cos x}\right)^3\frac{dx}{\cos^2x}\\ &=\int\tan^3x\ d(\tan x)\\ &=\frac14\tan^4x+C. \end{align} $$

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Let $u=\cos x$ then $du=-\sin xdx$ so $$\int\frac{\sin^3x}{\cos^5x}dx=-\int\frac{1-u^2}{u^5}du=\frac{u^{-4}}{4}-\frac{u^{-2}}{2}+C=\frac{\cos^{-4}x}{4}-\frac{\cos^{-2}x}{2}+C$$

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