find-the-length-r-2-1-cos-if-between-pi-2-to-pi-

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

f i n d t h e l e n g t h r = 2 / 1 - c o s θ i f θ b e t w e e n p i / 2 t o p i

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

l = \int_{\frac{π}{2}}^{π} \sqrt{{(\frac{d r}{d θ})}^{2} + r^{2}} d θ

\frac{d r}{d θ} = \frac{2 s i n θ}{{(1 - c o s θ)}^{2}}

\Rightarrow \int_{\frac{π}{2}}^{π} \sqrt{\frac{4 {s i n}^{2} θ}{{(1 - c o s θ)}^{4}} + \frac{4}{{(1 - c o s θ)}^{2}}} d θ = \int_{\frac{π}{2}}^{π} \sqrt{\frac{4 {s i n}^{2} θ + 4 {(1 - c o s θ)}^{2}}{{(1 - c o s θ)}^{4}}} d θ =

\int_{\frac{π}{2}}^{π} \sqrt{\frac{8 (1 - c o s θ)}{{(1 - c o s θ)}^{4}}} d θ = \int_{\frac{π}{2}}^{π} \sqrt{\frac{8}{{(1 - c o s θ)}^{3}}} d θ =

\int_{\frac{π}{2}}^{π} \sqrt{\frac{8}{8 {s i n}^{6} \frac{θ}{2}}} d θ = \int_{\frac{π}{2}}^{π} \frac{1}{{s i n}^{3} \frac{θ}{2}} d θ = \int_{\frac{π}{2}}^{π} \frac{s i n \frac{θ}{2}}{{s i n}^{4} \frac{θ}{2}} d θ =

\int_{\frac{π}{2}}^{π} \frac{s i n \frac{θ}{2}}{{(1 - {c o s}^{2} \frac{θ}{2})}^{2}} d θ

s e t u = c o s \frac{θ}{2}, i t i s e a s y n o w .