Evaluating $\int_0^{2\pi} \sqrt{1+\cos^2(x)}\ dx$ 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.
 A: 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.
A: 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.
