How to evaluate the integral $\int_0^\pi{\sqrt{1+\cos^2x}}dx$ 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?
 A: 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.
A: 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
$$ \int_0^\pi\!dx\,\sqrt{1+\cos^2 x} \approx 2 \sqrt{2}\; 1.35064 = 3.8202.$$
