I find myself frustrated with the solution of this problem since profit not find it, I'm stuck in the middle of the problem I can not solve the integral, I'm stuck in the solution of the integral is the same as follows:
$$r= \frac{6}{1+\cos\theta} \;\;\;\; 0 \leq \theta \leq \frac{\pi }{2} $$
P.d : I would appreciate if anyone could recommend me some text or page in which you could learn to plot in polar and learn more about these problems.
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Hint: The length of a polar curve $r(\theta)$ with starting angle $\alpha$ and ending angle $\beta$ is $$\int_\alpha^\beta \sqrt{r^2(\theta)+(r'(\theta))^2}$$
With your situation, it would be $$\int_0^{\pi/2} \sqrt{\left(\frac{6}{1+\cos\theta}\right)^2+\left(\left(\frac{6}{1+\cos\theta}\right)'\right)^2}\, d\theta$$ $$\frac{6}{1+\cos\theta}=3\sec^2(\theta/2)\text{ so}$$ $$\int_0^{\pi/2} \sqrt{9\sec^4(\theta/2)+\frac{576\sin^8\theta}{\sin^6(\theta/2)}}$$ Now try a weierstrass substitution ($t=\tan (\theta/2)$)
As you know, you have to consider $$I=\int \sqrt{r^2(\theta)+(r'(\theta))^2}\,d\theta$$ In you case $$r(\theta)= \frac{6}{1 + \cos \theta}$$ $$r'(\theta)=\frac{6 \sin (\theta)}{(\cos (\theta)+1)^2}$$ $$r^2(\theta)+(r'(\theta))^2=9 \sec ^6\left(\frac{\theta}{2}\right)$$ So, we are basically left with $$I=3\int \sec^3\left(\frac{\theta}{2}\right)\,d\theta$$ Now using $t=\tan\left(\frac{\theta}{4}\right)$, the integral reduces to $$I=6\int\frac{ \left(t^2+1\right)^2}{\left(1-t^2\right)^3}\,dt$$ Now, using partial fractions $$\frac{ \left(t^2+1\right)^2}{\left(1-t^2\right)^3}=\frac{1}{4 (t+1)}-\frac{1}{4 (t+1)^2}+\frac{1}{2 (t+1)^3}-\frac{1}{4 (t-1)}-\frac{1}{4 (t-1)^2}-\frac{1}{2 (t-1)^3}$$ I am sure that you can take from here.