arc length of the polar curve $r^2= \sin2\theta$

Question

given curve is $r^2 = \sin2\theta $

I got $L= \int_0^{2\pi} \sqrt{r^2+ ({\dfrac{dr}{d\theta}})^2}\ d\theta$ = $\int_0^{2\pi} \sqrt{\dfrac{1}{r^2}} d\theta = \int_0^{2\pi} \sqrt{\dfrac{1}{\sin2\theta}} d\theta = 2\int_0^{\dfrac{\pi}{2}} \sqrt{1 \over \sin2\theta} d\theta$ (by the polar graph)

but for 2$\int_0^{\pi \over 2} \sqrt{1 \over \sin2\theta} d\theta$, i don't have any idea to proceed

Answer 1

arc length = $4I = 4\int_0^{\pi/4} \sqrt{1 \over \sin 2 \theta}\ d\theta.$ make a change of variable $u = \tan \theta, du = (1+u^2) d\theta.$

$$I = \int_0^{\pi/4} \sqrt{1 \over \sin 2 \theta} \ d \theta = \int_0^1 \sqrt{1+u^2 \over 2u} \ {du \over 1 + u^2} = {1 \over \sqrt 2} \int_0^1 {1 \over \sqrt{u(1+u^2)}} \ du$$

i will have to think about how to proceed from here.