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

Question

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 $\dfrac{16}{3}$!

However, Wolfram is saying it's wrong. Please help?

Answer 1

Hint

First recall that integration is harder than differentiation and that you can't just use the power rule when there's a cosine running around inside (remember the chain rule? It's not as easy in reverse.)

As for how to do the integral, there's a popular trig identity involving $\sqrt{1+\cos(x)}$

Answer 2

Hint: $$2+2\cos\theta=2(1+\cos\theta)=2(2\cos^2\dfrac{\theta}{2})=4\cos^2\dfrac{\theta}{2}$$