# Find the exact length of the parametric curve(Not sure what I'm doing wrong)

As the title says, I'm not sure what I'm doing wrong. Any help would be greatly appreciated. Here's the problem with my solution.

Find the exact length of the parametric curve $(x,y)=(\theta+\sin \theta,−\cos θ)$, where $0\le θ \le \pi$.

Solution: $$\frac{dx}{d\theta}=1+\cos \theta$$ $$\frac{dy}{d\theta}=\sin \theta$$

Using the formula $$\int_0^\pi \sqrt{\left( \frac{dx}{d\theta} \right)^2+ \left( \frac{dy}{d\theta} \right)^2} \, d\theta$$

I have $$\int_0^\pi \sqrt{2+2\cos \theta} \, d\theta$$

and then $$\left[ \frac{(4+4\cos \theta)\sqrt{2+2\cos \theta}}{3} \right]_{0}^{\pi}$$

I end up with $\dfrac{16}{3}$!

• I'm not sure how you integrated, but try using the substitution $t=\tan\frac{\theta}{2}.$ Also, make what you're integrating with respect to more explicit please. – garserdt216 Mar 8 '17 at 4:35
• What I did was (2+2cos(theta))^(3/2))/(3/2) as my integration technique. – Infremo Mar 8 '17 at 4:41
• @Infremo Maybe check Wolfram alpha to see if your in-definite integral is right. (It is not. That's not a valid integration rule.) – spaceisdarkgreen Mar 8 '17 at 4:55
• @ Infremo It is a cycloid, obtainable by $x$ shift and $y$ reflection.. standard references would be of help, $\theta = 2 \phi$ substitution is also helpful – Narasimham Mar 8 '17 at 10:09
• @Infremo Yeah, you can't integrate (crud)$^n$ as (crud)$^{n+1}/(n+1)$, unless (crud)$=x$. There's some chain-rule-y stuff you have to unwind. – B. Goddard Mar 8 '17 at 14:09

As for how to do the integral, there's a popular trig identity involving $\sqrt{1+\cos(x)}$
Hint: $$2+2\cos\theta=2(1+\cos\theta)=2(2\cos^2\dfrac{\theta}{2})=4\cos^2\dfrac{\theta}{2}$$