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I want to evaluate this integral $$\int_0^{2\pi} \sqrt{1+\cos^2(x)}\ dx$$ But I cannot find a useful substitution/strategy. Could you please give me a hint?

I was thinking proving that this is equal to $$\int_0^{2\pi} \sqrt{1+\sin^2(x)}\ dx$$ but don't know how to proceed.


marked as duplicate by J. M. is a poor mathematician, Asaf Karagila, Brian M. Scott, Did, vonbrand Apr 3 '13 at 8:41

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    $\begingroup$ Mathematica returns $\int_0^{2 \pi} \sqrt{1+\cos^2(x)} dx = 4 \sqrt{2} E(\frac{1}{2})\approx 7.6404$, where E(x) is the Complete elliptic integral of the second kind... $\endgroup$ – PML Mar 28 '13 at 18:56
  • $\begingroup$ This wil not give you its value, but this integral is the arclength of the curve parameterized by $(\theta,\sin\theta)$ over $[0,2\pi]$. $\endgroup$ – Julien Mar 28 '13 at 18:56
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    $\begingroup$ Integrals such as these fall under the category of elliptic integrals and typically you cannot obtain any "nice" value as such. $\endgroup$ – user17762 Mar 28 '13 at 18:57
  • $\begingroup$ @julien Yes in fact the question is to find the length of the curve of graph of sin(x) under one period $\endgroup$ – mez Mar 28 '13 at 18:58
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    $\begingroup$ Note that, another form for the solution can be found in terms of the hypergeometric function as $$ E(k) = \int_{0}^{\pi/2} \sqrt{1-k^2 \sin^2(x) } dx = \frac{\pi}{2} F \left(\frac{1}{2},-\frac{1}{2};1; k^2 \right). $$ $\endgroup$ – Mhenni Benghorbal Mar 28 '13 at 22:28

Integrals such as these fall under the category of elliptic integrals and typically you cannot obtain any "nice" value as such. \begin{align} \int_0^{2\pi}\sqrt{1+\cos^2(x)} dx & = 4 \int_0^{\pi/2}\sqrt{1+\cos^2(x)}dx = 4 \int_0^{\pi/2}\sqrt{2-\sin^2(x)}dx\\ & = 4 \sqrt2 \int_0^{\pi/2} \sqrt{1-\dfrac{\sin^2(x)}2} dx = 4 \sqrt2 E\left(\dfrac1{\sqrt2}\right) \end{align} where $$E(k) = \displaystyle \int_0^{\pi/2} \sqrt{1-k^2 \sin^2(x)} dx \tag{$\star$}$$ and is referred to as the complete elliptic integral of second kind.

This doesn't prevent you from coming up with approximations to the integral. It can be shown, by using the binomial theorem for fractional powers to expand $(\star)$ and swapping the integral and summation, that $E(k)$ has the power series expansion given by $$E(k) = \dfrac{\pi}2 \sum_{l=0}^{\infty} \left(\dfrac{\dbinom{2l}l}{4^l} \right)^2 \dfrac{k^{2l}}{1-2l}\tag{$\heartsuit$}$$ Truncating $(\heartsuit)$ gives us an approximation to the elliptic integral.

  • $\begingroup$ @Cameron Buie: Thanks. I didn't know that the command \tag existed. Good thing to learn. $\endgroup$ – user17762 Mar 28 '13 at 19:29
  • $\begingroup$ No problem. Glad I can return the favor, since I learned from you that the card suits existed. ;) $\endgroup$ – Cameron Buie Mar 28 '13 at 19:33
  • $\begingroup$ So that's why it looked familiar... anyway, you really should stick to a single convention; you use the modulus in the definition, but the parameter in your answer. $\endgroup$ – J. M. is a poor mathematician Apr 3 '13 at 6:11
  • $\begingroup$ @J.M. I do not understand your comment. Do you mean I should have $E(1/sqrt2)$ and not $E(1/2)$? $\endgroup$ – user17762 Apr 3 '13 at 15:07
  • $\begingroup$ Well, if you stick with the modulus convention like you have in your definition, then yes, you should take the square root. $\endgroup$ – J. M. is a poor mathematician Apr 3 '13 at 15:08

It seems you can't get an exact result using elementary functions.

If you are interested in a series form of this integral, you can use the Taylor series for the square root: $$\sqrt{1+t}=\sum_{n=0}^\infty \frac{(-1)^n (2n)!}{(1-2n) n!^2 4^n} t^n $$ and integrate term by term.

I'll also note that this integral has geometric meaning: It is the length of the sine curve over one complete period.


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