I can't solve this
$$\int_{0}^{5}{\sqrt{1+\left(\dfrac{\pi}{2}\cos(10 \pi x)\right)^2}dx}$$
My approach: If $10\pi x =u \to 10\pi dx=du$, so $$\dfrac{1}{10\pi}\int_{0}^{50\pi}{\sqrt{1+\left(\dfrac{\pi}{2}\cos(u)\right)^2}dx}=\dfrac{1}{10\pi}\int_{0}^{50\pi}{\sqrt{1+\dfrac{\pi^{2}}{8}(\cos(2u)+1)}dx}$$
And I don't know how continue this, and another thing is maybe elliptical integral...
On
As Lucian is stating, the reason you cannot calculate this integral is because it so happens to be a type of elliptic integral. To put things in general, the elliptic integral is not one you can find the antiderivative in terms of "elementary functions", like polynomials, exponentials, sine, cosine, or even rational functions. Therefore, to evaluate it, we just "make up" an anti-derivative called "EllipticE(z)", as evaluated by Maple by user64494, or whatever you want to call it. These "artificial" anti-derivatives are constructed such that they do satisfy the fundamental theorem of calculus and all that you know about them. Another great example is evaluating $$\int_{0}^{\infty} e^{{-{x^2}}} dx$$. The solution to this integral is actually $\frac{\sqrt(\pi)}{2}$, but in order to get this answer, you would actually use what is known as the error function, which is $Erf(z)$, defined as
$Erf(t) = $$\frac{2}{\sqrt(\pi)}$$\int_{0}^{t} e^{{-{s^2}}} ds$
Of course, there are other ways of doing this, but if you go by just finding an anti-derivative as in Calculus I, then you would need to "construct" one, since there isn't one in terms of elementary functions. What is then left for you to do is either use numerical quadrature (evaluate using numerical methods), or in the case of this simple example, use polar coordinates. Note that if we take off the limits of integration off, then since we aren't evaluating the integral, then to find the antiderivative, we are left with no real choice but to go with the method I just highlighted. Thus, the general antiderivative would be:
$\frac{\sqrt(\pi)}{2}Erf(x)$ + C, where C is an arbitrary constant of integration.
Hope this helps to explain things somewhat.
On
$$\int_{0}^{5}{\sqrt{1+\left(\dfrac{\pi}{2}\cos(10 \pi x)\right)^2}dx}=\left[\frac{\sqrt{4+\pi^2}E\left(10\pi x|\frac{\pi^2}{4+\pi^2}\right)}{20\pi}\right]_{0}^{5}=$$
$$\left(\frac{\sqrt{4+\pi^2}E\left(10\pi*5|\frac{\pi^2}{4+\pi^2}\right)}{20\pi}\right)-\left(\frac{\sqrt{4+\pi^2}E\left(10\pi*0|\frac{\pi^2}{4+\pi^2}\right)}{20\pi}\right)=$$
$$\frac{5\sqrt{4+\pi^2}E\left(10\pi|\frac{\pi^2}{4+\pi^2}\right)}{20\pi}$$
No wonder you can't, since you are basically trying to evaluate the arc length of the sine function, which is one of the main questions that gave birth historically to the notion of a special class of functions, called elliptic integrals.