How can I get arc-length of this polar function? $$ r= 4(1-\sin{\phi})$$ $$-\frac{\pi}{2}\leq\phi\leq\frac{\pi}{2}$$

I know that arc-length of polar function can get calculate by $$l=\int_\alpha^\beta\sqrt{r^2+(r')^2}d\phi$$ So: $$r'=-4\cos{\phi}$$ $$l=\int_\frac{-\pi}{2}^\frac{\pi}{2}\sqrt{4^2(1-\sin{\phi})^2+(-4\cos{\phi})^2}d\phi=\int_\frac{-\pi}{2}^\frac{\pi}{2}\sqrt{16-32\sin{\phi}+16\sin^2{\phi}+16\cos^2{\phi}}d\phi=\cdots$$ $$=\displaystyle\int_\frac{-\pi}{2}^\frac{\pi}{2}\sqrt{-32\sin{\phi}}d\phi$$ Does this really have to imply that I will need to calculate $\sqrt{32} \displaystyle\int_\frac{-\pi}{2}^\frac{\pi}{2} \sqrt{-\sin{\phi}}d\phi$? If so then how can I do that?

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    $\begingroup$ $16 + 16 \sin^2 \phi + 16 \cos^2 \phi = 32$. $\endgroup$ – Umberto P. Apr 8 '14 at 15:05

You should have $$16-32\sin\phi +16(\sin^2(\phi)+\cos^2(\phi))=32(1-\sin\phi)$$

For the integral, note that $$\int{\sqrt{1-\sin x}}dx=\int{\frac{\sqrt{1-\sin^2(x)}}{\sqrt{1+\sin x}}}dx=\int{\frac{\cos x}{\sqrt{1+\sin x}}}dx=2\sqrt{1+\sin x}+C$$


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