TITLE: What is wrong with the following u-substitution?
QUESTION [20 upvotes]: 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.

REPLY [3 votes]: 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:
$$
\begin{align}
\int_0^{2\pi}x\,\mathrm{d}x
&=\overbrace{\int_0^1\color{#C00}{\sin^{-1}(u)}\,\mathrm{d}\sin^{-1}(u)}^{x=\sin^{-1}(u)\in\left[0,\frac\pi2\right]}
+\overbrace{\int_{-1}^1\left(\color{#090}{\pi}+\color{#00F}{\sin^{-1}(u)}\right)\,\mathrm{d}\sin^{-1}(u)}^{x=\pi+\sin^{-1}(u)\in\left[\frac\pi2,\frac{3\pi}2\right]}
+\overbrace{\int_{-1}^0\left(\color{#C90}{2\pi}+\color{#C9F}{\sin^{-1}(u)}\right)\,\mathrm{d}\sin^{-1}(u)}^{x=2\pi+\sin^{-1}(u)\in\left[\frac{3\pi}2,2\pi\right]}\\
&=\color{#C00}{\frac{\pi^2}8}\color{#090}{+\pi^2}\color{#00F}{+0}\color{#C90}{+\pi^2}\color{#C9F}{-\frac{\pi^2}8}\\[9pt]
&=2\pi^2
\end{align}
$$