Question Number 66520 by mhmd last updated on 16/Aug/19
$${find}\:{the}\:{length}\:{r}=\mathrm{2}/\mathrm{1}−{cos}\theta\:\:\:\:\:\:\:\:\:{if}\:\theta\:{between}\:{pi}/\mathrm{2}\:{to}\:{pi} \\ $$
Commented by kaivan.ahmadi last updated on 16/Aug/19
$${l}=\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \sqrt{\left(\frac{{dr}}{{d}\theta}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} }{d}\theta \\ $$$$\frac{{dr}}{{d}\theta}=\frac{\mathrm{2}{sin}\theta}{\left(\mathrm{1}−{cos}\theta\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \sqrt{\frac{\mathrm{4}{sin}^{\mathrm{2}} \theta}{\left(\mathrm{1}−{cos}\theta\right)^{\mathrm{4}} }+\frac{\mathrm{4}}{\left(\mathrm{1}−{cos}\theta\right)^{\mathrm{2}} }}{d}\theta=\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \sqrt{\frac{\mathrm{4}{sin}^{\mathrm{2}} \theta+\mathrm{4}\left(\mathrm{1}−{cos}\theta\right)^{\mathrm{2}} }{\left(\mathrm{1}−{cos}\theta\right)^{\mathrm{4}} }}{d}\theta= \\ $$$$\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \sqrt{\frac{\mathrm{8}\left(\mathrm{1}−{cos}\theta\right)}{\left(\mathrm{1}−{cos}\theta\right)^{\mathrm{4}} }}{d}\theta=\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \sqrt{\frac{\mathrm{8}}{\left(\mathrm{1}−{cos}\theta\right)^{\mathrm{3}} }}{d}\theta= \\ $$$$\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \sqrt{\frac{\mathrm{8}}{\mathrm{8}{sin}^{\mathrm{6}} \frac{\theta}{\mathrm{2}}}}{d}\theta=\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \frac{\mathrm{1}}{{sin}^{\mathrm{3}} \frac{\theta}{\mathrm{2}}}{d}\theta=\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \frac{{sin}\frac{\theta}{\mathrm{2}}}{{sin}^{\mathrm{4}} \frac{\theta}{\mathrm{2}}}{d}\theta= \\ $$$$\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \frac{{sin}\frac{\theta}{\mathrm{2}}}{\left(\mathrm{1}−{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\right)^{\mathrm{2}} }{d}\theta \\ $$$${set}\:{u}={cos}\frac{\theta}{\mathrm{2}}\:,{it}\:{is}\:{easy}\:{now}. \\ $$$$ \\ $$