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Question Number 66520 by mhmd last updated on 16/Aug/19

$$findthelengthr=2/1-cos\theta if\theta betweenpi/2topi$$

Commented by kaivan.ahmadi last updated on 16/Aug/19

$$l={\int }_{\frac{\pi }{2}}^{\pi }\sqrt{{\left(\frac{dr}{d\theta }\right)}^2+r^2}d\theta$$$$\frac{dr}{d\theta }=\frac{2sin\theta }{{(1-cos\theta )}^2}$$$$\Rightarrow {\int }_{\frac{\pi }{2}}^{\pi }\sqrt{\frac{4{sin}^2\theta }{{(1-cos\theta )}^4}+\frac{4}{{(1-cos\theta )}^2}}d\theta ={\int }_{\frac{\pi }{2}}^{\pi }\sqrt{\frac{4{sin}^2\theta +4{(1-cos\theta )}^2}{{(1-cos\theta )}^4}}d\theta =$$$${\int }_{\frac{\pi }{2}}^{\pi }\sqrt{\frac{8(1-cos\theta )}{{(1-cos\theta )}^4}}d\theta ={\int }_{\frac{\pi }{2}}^{\pi }\sqrt{\frac{8}{{(1-cos\theta )}^3}}d\theta =$$$${\int }_{\frac{\pi }{2}}^{\pi }\sqrt{\frac{8}{8{sin}^6\frac{\theta }{2}}}d\theta ={\int }_{\frac{\pi }{2}}^{\pi }\frac{1}{{sin}^3\frac{\theta }{2}}d\theta ={\int }_{\frac{\pi }{2}}^{\pi }\frac{sin\frac{\theta }{2}}{{sin}^4\frac{\theta }{2}}d\theta =$$$${\int }_{\frac{\pi }{2}}^{\pi }\frac{sin\frac{\theta }{2}}{{(1-{cos}^2\frac{\theta }{2})}^2}d\theta$$$$setu=cos\frac{\theta }{2},itiseasynow.$$$$$$

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