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.
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Theorem: integration by substitution (change of variable):
If $g'$ is integrable on $[a,b]$ and $f$ is continuous on $g[a,b],$ then, substituting $u=g(x)$ into the LHS,$$\int_a^bf\big(g(x)\big)g'(x)\,\mathrm{d}x=\int_{g(a)}^{g(b)}f(u)\,\mathrm{d}u.$$
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Consider the separate implicit substitutions $$x=h_1(u)$$ and $$h_1(u)=h_2(x).$$ Notice that $\,g=h_1^{-1}\,$ and $\,g=h_1^{-1}h_2,\,$ respectively.
In these cases, the function $h_1$ is required to be invertible; typically, it is restricted to some choice of principal domain.
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However, integration by substitution does not inherently or generally require substitution functions to be invertible/bijective or monotonic—or even injective—on the interval of integration.
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Contrary to your Remark, the integration-by-substitution in the example $\displaystyle\int^1_{-1}\frac{2x}{1+x^2} \, \mathrm{d}x\,$ is actually perfectly valid despite $u=1+x^2$ not being 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 $u=g(x)$ 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: $$\int_{-4}^3\frac{2x}{1+x^2}\,\mathrm{d}x =\int_{17}^{10}\frac1u\,\mathrm{d}u=-0.531.$$
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In the main example, your mistake was in neglecting that $$\sqrt{\cos^2(x)}\not\equiv\cos(x),\tag1$$ and that outside $\sin(x)$'s principal domain, $$u=\sin(x)\kern.6em\not\kern-.6em\implies x=\arcsin(u);\tag2$$ here's the correction:
Let $u=\sin(x),$ so $\,\mathrm{d}u=\cos(x) \,\mathrm{d}x.$
Now, for $x\in[0, \frac{\pi}{2}],$ \begin{align}\cos(x) &= \sqrt{1-u^2},\\x&=\arcsin(u);\end{align} for $x\in[\frac{\pi}{2},\frac{3\pi}{2}],$ \begin{align}\cos(x) &= -\sqrt{1-u^2},\\x&=\pi -\arcsin (u);\end{align} for $x\in[\frac{3\pi}{2}, 2\pi],$ \begin{align}\cos(x) &= \sqrt{1-u^2},\\x&=2\pi +\arcsin(u).\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}
Note that splitting up the integral into intervals here was necessitated by $(1)$ and $(2),$ and that having effectively created injective substitution functions anyway is merely a consequence of this.
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The previous example actually fails to satisfy the condition $“f$ is continuous on $g[a,b]”$ of the integration-by-substitution theorem. This is 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 improper integrals. However, they are convergent improper integrals, so this technicality can be ignored.
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: $$ \begin{align} \int_0^{2\pi}x\,\mathrm{d}x &=\overbrace{\int_0^1\color{#C00}{\sin^{-1}(u)}\,\mathrm{d}\sin^{-1}(u)}^{x=\sin^{-1}(u)\in\left[0,\frac\pi2\right]} +\overbrace{\int_{-1}^1\left(\color{#090}{\pi}+\color{#00F}{\sin^{-1}(u)}\right)\,\mathrm{d}\sin^{-1}(u)}^{x=\pi+\sin^{-1}(u)\in\left[\frac\pi2,\frac{3\pi}2\right]} +\overbrace{\int_{-1}^0\left(\color{#C90}{2\pi}+\color{#C9F}{\sin^{-1}(u)}\right)\,\mathrm{d}\sin^{-1}(u)}^{x=2\pi+\sin^{-1}(u)\in\left[\frac{3\pi}2,2\pi\right]}\\ &=\color{#C00}{\frac{\pi^2}8}\color{#090}{+\pi^2}\color{#00F}{+0}\color{#C90}{+\pi^2}\color{#C9F}{-\frac{\pi^2}8}\\[9pt] &=2\pi^2 \end{align} $$
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.