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This question was in my textbook:

$$L=\int_0^\pi{\sqrt{1+\cos^2x}}dx$$

The textbook says $L{\approx}3.82$ using technology. I put it into a online integral calculator, but that said the antiderivative could not be found. I'd like to know how to solve it exactly if that is possible?

Jack
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    It is an Elliptic integral $E(z)$, see http://functions.wolfram.com/EllipticIntegrals/EllipticE/07/01/01/ – Pierpaolo Vivo Jan 19 '16 at 11:54
  • https://www.wolframalpha.com/input/?i=integrate+sqrt%281%2Bcos%28x%29^2%29+from+0+to+pi&t=efni01 gives the numerical value 3.820197789… – Lutz Lehmann Jan 19 '16 at 11:57
  • The antiderivative is not an elementary function, which means that it's not a combination of $+$, $-$, $\times$, $\div$, exponentiation, roots, $\log$, and trig functions. This happens occasionally when dealing with integrals; not every function has an elementary antiderivative. – Akiva Weinberger Jan 19 '16 at 12:54
  • I don't know how to prove that a function's antiderivative is not an elementary function. I know the proof exists, but it uses pretty advanced mathematics. – Akiva Weinberger Jan 19 '16 at 12:56
  • A related question. – Lucian Jan 19 '16 at 14:28

2 Answers2

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The online integral calculator was wrong: indeed you can find the antiderivative to be $$\sqrt{2} \text{ Elliptic}\left(x, \frac{1}{2}\right)$$ where $\text{Elliptic}(x,y)$ is the Elliptic integral.

sirfoga
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    I am not familiar with the elliptic integral. How can I solve it exactly using this? – Jack Jan 19 '16 at 12:00
  • Since it is not an elementary function, nor a composition of any, you can only get an estimate using numerical methods, like Newton or Monte Carlo .. the approximate value is $3.820197789$ – sirfoga Jan 19 '16 at 12:05
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The result of the question can be written in terms of the complete elliptic integral $$ \int_0^\pi\!dx\,\sqrt{1+\cos^2 x}= 2 \sqrt{2} E(1/\sqrt{2}).$$ Note that I use the convention as in the article on wikipedia.

Geometrically, the result is the circumference of an ellipse with semi major axis $a=1/\sqrt{2}$ and eccentricity $\epsilon=1/\sqrt{2}$.

The value of the integral $E(x)$ can be obtained by the following method:

We define the arithmetic geometric mean $M(x,y)$ of $x$ and $y$ as the limit of the sequence $$ a_1 = \tfrac12 (x+y), \qquad g_1=\sqrt{xy}, \\ a_{n+1} = \tfrac12(a_n+g_n), \qquad g_{n+1}= \sqrt{a_n g_n}$$ with $M(x,y) = \lim_{n\to\infty} a_n =\lim_{n\to\infty} g_n$.

Additionally, we define the modified arithmetic geometric mean $N(x,y)$ as $$ \alpha_1 =x, \qquad b_1 =y, \qquad c_1 =0,\\ \alpha_{n+1} =\tfrac12(\alpha_n + b_n) , \qquad b_{n+1} = c_n + \sqrt{(\alpha_n-c_n)(b_n-c_n)}, \qquad c_{n+1} = c_n -\sqrt{(\alpha_n-c_n) (b_n-c_n)}$$ via $N(x,y) = \lim_{n\to\infty} \alpha_n =\lim_{n\to\infty} b_n$.

We then have that $$E(x) = \frac{\pi N(1,x^2)}{2 M(1,x)}.$$

In your example, we need $x=1/\sqrt{2}$. So, we calculate $M(1,1/\sqrt{2})$ and $N(1,1/2)$. Using three iterations, i.e., $n=4$, we obtain $$M(1,1/\sqrt{2}) \approx g_4 = 0.847213.$$ Similarly, we obtain $$N(1,1/2) \approx b_4= 0.728473$$ and thus $$E(1/\sqrt{2})\approx \frac{\pi b_4}{2 g_4} = 1.35064.$$

Finally, the value of the original integral is $$ \int_0^\pi\!dx\,\sqrt{1+\cos^2 x} \approx 2 \sqrt{2}\; 1.35064 = 3.8202.$$

Fabian
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