What is wrong with the following u-substitution? We will calculate $\displaystyle\int^{2 \pi}_0 x \, dx$.  Let $u=\sin (x)$, and observe that $\sin(2 \pi)=0$ and $\sin(0)=0$.  We also have that $\frac{du}{dx}=\cos(x)=\sqrt{1-u^2}$. Hence,
$$
\int^{2 \pi}_0 x \, dx=\int^0_0 \frac{\sin^{-1}(u)}{\sqrt{1-u^2}} \, du = 0.
$$
This is very obviously wrong, but I am not sure how to explain the error formally.
Edit:
Thanks for the responses and in particular the link below to the related problem!  The error is indeed caused by the substitution $x=\sin^{-1}(u)$.  The integration is performed over $[0,2 \pi]$ which is outside the range of the $\sin^{-1}$ function.  
Remark
The error is slightly better disguised when calculating $\displaystyle\int^1_{-1}\frac{2x}{1+x^2} \, dx.$
Let $u(x)=1+x^2$, and observe that $u(1)=u(-1)=2$.  Then since $dx=\frac{1}{2x} du$, we have that
$$
\int^1_{-1} \frac{2x}{1+x^2} \, dx =  \int^2_2 \frac{1}{u} \, du=0.
$$
This time, no trigonometric substitution is used, but it is still an incorrect proof for the same reason as above.  A correct proof can be obtained by using the fact that $x \mapsto \displaystyle\frac{2x}{1+x^2}$ is odd.
This example is more disturbing because the procedures above are entirely intuitive and yield the correct result. It seems to me that students when taught integration by substitution of definite integrals should also be taught that great care be exercised in checking the range of integration, particularly when the (apparent) substituting function is not invertible in that range.
 A: The substituted formula must match the formula substituted on each interval. $\sin^{-1}(u)\in\left[-\frac\pi2,\frac\pi2\right]$ cannot cover all of $[0,2\pi]$. A safer way to substitute is to substitute on monotonic intervals of the substituted formula:
$$
\begin{align}
\int_0^{2\pi}x\,\mathrm{d}x
&=\overbrace{\int_0^1\color{#C00}{\sin^{-1}(u)}\,\mathrm{d}\sin^{-1}(u)}^{x=\sin^{-1}(u)\in\left[0,\frac\pi2\right]}
+\overbrace{\int_{-1}^1\left(\color{#090}{\pi}+\color{#00F}{\sin^{-1}(u)}\right)\,\mathrm{d}\sin^{-1}(u)}^{x=\pi+\sin^{-1}(u)\in\left[\frac\pi2,\frac{3\pi}2\right]}
+\overbrace{\int_{-1}^0\left(\color{#C90}{2\pi}+\color{#C9F}{\sin^{-1}(u)}\right)\,\mathrm{d}\sin^{-1}(u)}^{x=2\pi+\sin^{-1}(u)\in\left[\frac{3\pi}2,2\pi\right]}\\
&=\color{#C00}{\frac{\pi^2}8}\color{#090}{+\pi^2}\color{#00F}{+0}\color{#C90}{+\pi^2}\color{#C9F}{-\frac{\pi^2}8}\\[9pt]
&=2\pi^2
\end{align}
$$
A: *

*Theorem: integration by substitution (change of variable):
If $g'$ is integrable on $[a,b]$ and $f$ is integrable, and has an antiderivative, on $g[a,b],$ then, substituting $u=g(x)$ into the LHS, $$\int_a^bf\big(g(x)\big)\, g'(x)\,\mathrm{d}x=\int_{g(a)}^{g(b)}f(u)\,\mathrm{d}u.$$


*Contrary to your Remark, the integration-by-substitution in the example $\displaystyle\int^1_{-1}\frac{2x}{1+x^2} \, \mathrm{d}x\,$ is perfectly valid even though $u=1+x^2$ is not invertible on $[-1,1].$
Let $g(x)=1+x^2,\,\,f(u)=\frac1u,$ and $\,a=-1,b=1$.
Then $f(g(x))=\frac1{1+x^2},\,\,g'(x)=2x,$ and $\,g(a)=g(b)=2$.
Although $g$ is not injective on $[-1,1],$ the above theorem does validly apply: $$\int^1_{-1}\frac{2x}{1+x^2}\,\mathrm{d}x=\int_2^2\frac1u\,\mathrm{d}u=0.$$ Similarly, this is valid: $$\int_{-4}^3\frac{2x}{1+x^2}\,\mathrm{d}x =\int_{17}^{10}\frac1u\,\mathrm{d}u=-0.531.$$


*Changing variable during integration does not inherently require monotonicity or invertibility or even injectivity, except when obtaining the new integration limits after making an implicit substitution (i.e., one in which the new variable $u$ is an implicit function of the starting variable $x$). To wit: $$h_1(u)=h_2(x)\;\implies\; u=g(x)=h_1^{-1}h_2(x)$$ requires $h_1$ to be invertible, typically by restricting it to its principal domain.
For example, given the integrand $\frac1{x^2\sqrt{4-x^2}},$ we can make the substitution $x=2\sin\theta$ with $\theta\in\left[-\frac\pi2,\frac\pi2\right].$


*The mistake in your main example arises not from non-invertibility per se (that is, there is no need to separately ensure that $g$ is invertible), but from neglecting that $$\sqrt{\cos^2(x)}\not\equiv\cos(x)\tag1$$ and that $$u=\sin(x)\kern.6em\not\kern-.6em\implies x=\arcsin(u).\tag2$$  Due to $(1)$ and $(2),$ the given integrand $x$ is expressible on its integration domain $[0,2\pi]$ as $f(g(x))\,g'(x),$ with $g=\sin,$ only piecewise; that is, the above theorem is applicable only piecewise; so, the substitution must also be piecewise, as follows.
Let $u=\sin(x),$ so $\,\mathrm{d}u=\cos(x) \,\mathrm{d}x.$
Now, for $x\in\left[0, \frac{\pi}{2}\right],$ \begin{align}x&=\arcsin(u),\\\cos(x) &= \sqrt{1-u^2};\end{align} for $x\in\left[\frac{\pi}{2},\frac{3\pi}{2}\right],$ \begin{align}x&=\pi -\arcsin (u),\\\cos(x) &= -\sqrt{1-u^2};\end{align} for $x\in\left[\frac{3\pi}{2}, 2\pi\right],$ \begin{align}x&=2\pi +\arcsin(u),\\\cos(x) &= \sqrt{1-u^2}.\end{align} Therefore, \begin{align}\int^{2\pi}_0 x \,\mathrm{d}x&=\int^\frac{\pi}{2}_0 x \,\mathrm{d}x + \int^\frac{3\pi}{2}_\frac{\pi}{2} x \,\mathrm{d}x + \int^{2\pi}_\frac{3\pi}{2} x \,\mathrm{d}x\\&=\int^1_0 \frac{\arcsin(u)}{\sqrt{1-u^2}}\mathrm{d}u + \int^{-1}_1 \frac{\pi-\arcsin(u)}{-\sqrt{1-u^2}} \mathrm{d}u + \int^0_{-1} \frac{2\pi + \arcsin(u)}{\sqrt{1-u^2}} \mathrm{d}u\\&=\pi\int^1_{-1} \frac{\,\mathrm{d}u}{\sqrt{1-u^2}}+2\pi\int^0_{-1} \frac{\,\mathrm{d}u}{\sqrt{1-u^2}}\\&=\pi\bigg[ \arcsin(u)\bigg]_{-1}^1+2\pi \bigg[\arcsin(u)\bigg]_{-1}^0\\&=2\pi^2.\end{align}


*The previous example actually fails to satisfy the condition $“f$ is integrable on $g[a,b]”$ of the above theorem, because $\displaystyle\int^1_0 \frac{\arcsin(u)}{\sqrt{1-u^2}}\mathrm{d}u,\; \int^{-1}_1 \frac{\pi-\arcsin(u)}{-\sqrt{1-u^2}} \mathrm{d}u,\;\int^0_{-1} \frac{2\pi + \arcsin(u)}{\sqrt{1-u^2}} \mathrm{d}u\,$ are not bound. But this technicality can be ignored, because they are convergent improper integrals.
The common version of the above theorem actually has a stronger condition, namely, that $f$ be continuous on $g[a,b];$ in this case, it's harder to brush aside this violation as a technicality.


*Two other types of examples where non-injectivity is not the root culprit of the initially perceived errors during integration by substitution.
A: I think that qwerty314 might have a better more direct answer to your bottom line question of why your particular method does not produce the desired result. That being said, it should be noted that you can actually get the desired result using trigonometric substitution by not changing the limits of integration, and instead back substituting the result then evaluating that at the original $x$ limits. To wit, given the indefinite integral
$$\int x\, dx,$$
let $$x=\sin u \Rightarrow dx=\cos u \, du.$$
Then
$$
\begin{align*}
\int x\, dx &= \int \sin u \cos u \, du \\
&= -\frac{\cos^2 u}{2}.
\end{align*}
$$
As we let $\frac{x}{1}=\sin u$ in our original substitution, we now form a right triangle with angle $u$, opposite side $x$, and hypotenuse $1$. Thus the adjacent side is $\sqrt{1-x^2}$. Reading from this triangle we have that
$$-\frac{\cos^2 u}{2}=-\frac{1-x^2}{2}.$$
Now if we evaluate this result at our limits we get
$$
\begin{align*}
-\frac{1-x^2}{2}\bigg|_0^{2\pi} &= -\frac{1-4\pi^2}{2}-\left( -\frac{1-0^2}{2} \right)\\
&=-\frac{1}{2}+\frac{4\pi^2}{2}+\frac{1}{2}\\
&=2\pi^2.
\end{align*}
$$
This is of course the result we would expect had we computed the definite integral in the standard manner.
A: substituting u=sin(x) will restrict the range of x to [-pi,pi], while x has range [-inf,inf].
