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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?

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  • $\begingroup$ 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. $\endgroup$ Mar 8, 2017 at 4:35
  • $\begingroup$ What I did was (2+2cos(theta))^(3/2))/(3/2) as my integration technique. $\endgroup$
    – Timmy Wick
    Mar 8, 2017 at 4:41
  • $\begingroup$ @Infremo Maybe check Wolfram alpha to see if your in-definite integral is right. (It is not. That's not a valid integration rule.) $\endgroup$ Mar 8, 2017 at 4:55
  • $\begingroup$ @ 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 $\endgroup$
    – Narasimham
    Mar 8, 2017 at 10:09
  • $\begingroup$ @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. $\endgroup$
    – B. Goddard
    Mar 8, 2017 at 14:09

2 Answers 2

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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 $\sqrt{1+\cos(x)}$

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

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