Confusion in Proof of Noether's theorem This question is related to this Noether's theorem under arbitrary coordinate transformation  and this  Transformation of $d^4x$ under translation disregarded?
To proof Noether's theorem every text book that i know assumes that the variation of the action under the trasformation $x\mapsto x^\prime$is given by
$$\delta S=\int \mathscr{L}'(\phi^\prime,\partial_\mu\phi^\prime)d^4x^\prime-\int \mathscr{L}(\phi,\partial_\mu\phi)d^4x$$
For me this variation should always be zero since this is just a change of variables so there is nothing new.
Suppose that in one dimension  flat space we have $\phi(x)=\sin(x)$ and our action is given by
$$S=\int{\sin(x)dx}$$
under a trasnlation $x\mapsto x+a$ we should have $\phi(x) \mapsto \phi'(x)=\sin(x-a)$.
Under a change of coordinates $x\mapsto y=x+a$,  $\phi(x) \mapsto \phi'(y)=\sin(y-a)$ , since $x$ and $y$ are dummy index the action is the same in both case.
Now assuming we have $\sqrt g= x^2$. Translating will give us an action
$$S=\int{ x^2\sin(x-a)dx}$$
But change of variables will give us an action
$$S=\int{ (y-a)^2\sin(y-a)dx}$$ 
witch is  different.
Shouldn't the variation of the action be written like this $$\delta S=\int \mathscr{L}'(\phi^\prime,\partial_\mu\phi^\prime)d^4x^-\int \mathscr{L}(\phi,\partial_\mu\phi)d^4x$$
 A: This issue has come up many times on this site. It's one of those things where the standard textbook presentation is severely lacking.
For example, consider translational symmetry. Sometimes this is written as
$$x \to x' = x + a, \quad \phi \to \phi'(x') = \phi(x).$$
However, you don't implement this symmetry transformation by substituting every $x$ with an $x'$ and every $\phi$ with a $\phi'$, because that's just a trivial change of variables that always leaves the action invariant, much like how a $u$-substitution always leaves an integral invariant. In other words, obviously we have
$$\int_{-\infty}^\infty x^2 f(x) \, dx = \int_{-\infty}^\infty (x+a)^2 f(x+a) \, dx$$
even though $x^2$, which stands in for the Lagrangian here, is not actually translationally invariant. 
What they really mean is that we are considering a genuine, nontrivial transformation of the fields. We start with the field $\phi$, and end up with the field $\phi'$. Indeed in general, we have
$$\int_{-\infty}^\infty x^2 f(x) \, dx \neq \int_{-\infty}^\infty x^2 f(x+a) \, dx.$$
You can derive the expression for $\phi'$ by thinking "we map the point $x$ to $x'$" but you don't actually do that as well, because then you end up with a trivial transformation. Another way of saying this is that a symmetry means that a system stays the same if you change some things but not other things. In this case, we shifted the function $f$ relative to the fixed background $x^2$ in order to test translational symmetry, which didn’t hold. If you instead changed both, you wouldn't get any useful information.
A: In most physics textbooks (including Noether's own seminal 1918 paper) on Noether's theorem, the infinitesimal variation$^1$
$$ \delta S~:=~S_{V^{\prime}}[\phi^{\prime}]- S_{V}[\phi]\tag{A}$$
describes an active (as opposed to passive) infinitesimal transformation $$\phi(x)\to\phi^{\prime}(x^{\prime}), \tag{B}$$ $$ x\to x^{\prime}, \tag{C}$$ of the action functional $$ S_{V}[\phi]~=~\int_V \mathrm{d}^nx ~{\cal L}(\phi(x),\partial\phi(x),x).\tag{D}$$ 
The infinitesimal variation (A) needs not vanish in general. However, pure horizontal $x$-variation (C) without vertical $\phi$ variation does vanish cf. my Phys.SE answer here.
--
$^1$ The integration region $V\to V^{\prime}$ is conventionally moved according to the horizontal variation (C). 
A: Noether's theorem states the conservation laws as a function of (the symmetries of) the Lagrangian. The lagrangian is chosen to reflect the symmetries of the physical system it describes. So whether $\delta \cal L=0$ for a coordinate transformation is a choice. 
A: 
To proof Noether's theorem every text book that i know assumes that the variation of the action under the trasformation $x\mapsto x²$ is given by ...

I have here only one book where to search for Noether's theorem.
Itzykson-Zuber do not consider coordinate transformations. They write for space translations (eq. (1-94))
$$\mathscr L(x+a) \equiv 
\mathscr L(\phi_i(x+a), \partial_\mu\phi_i(x+a))$$
where $a$ may also depend on $x$. Variation of fields is written
$$\delta\phi_i = \delta a^\mu(x)\,\partial_\mu\phi_i(x)$$
etc. No variation of coordinates in the action integral.
A final note may help you to clarify the matter. Variational
formulation in relativistic field theory is but an extension of the
one in ordinary mechanics. The analogy runs as follows:
$$\matrix{
\hfill \rm time & \rightarrow & \rm spacetime\ coordinates \hfill\cr
\hfill \rm lagrangian\ coordinates & \rightarrow & \rm fields. \hfill \cr}$$
So the relevant variation is the one of fields, not of coordinates.
