Evaluate $3\int_{0}^{2\pi} \sin(t) \cos(t) \,{\rm d}t$ $$3\int_{0}^{2\pi} \sin(t) \cos(t) \,{\rm d}t$$
My calculus is a bit rusty and I can not find where I get it wrong. Setting $u= \sin(t)$, I get ${\rm d} u=\cos(t) \,{\rm d} t$ and, thus,
$$3\int_{u=0}^{u=0}u \,{\rm d}u=0$$
The substitution is correct. If $f$ is a function with an antiderivative $F$, one has by the fundamental theorem of calculus
\begin{equation}
\int_a^b f(u(t))u'(t)dt = \int_a^b(F(u(t)))' dt = F(u(b)) - F(u(a)) = \int_{u(a)}^{u(b)}f(x) d x
\end{equation}
A sufficient hypothesis is that $u'$ is continuous in $[a,b]$ and that $f$ is continuous on an interval that contains $u([a,b])$.
In your case $u(t) = \sin(t)$ and $f(t)=t$.
I'm amazed that so many people are puzzled by this simple application of the fundamental theorem of calculus.
Let us take the case of the integral
\begin{equation}
I = \int_0^\pi \sin(x) d x
\end{equation}
and let $u(x) = - \cos(x)$ as suggested in the comments, with $f(x)= 1$. The above substitution formula gives
\begin{equation}
I = \int_{-\cos(0)}^{-\cos(\pi)} dx = \int_{-1}^1 d x = 2
\end{equation}
which is the correct result.
There is no contradiction with this counterexample because in the counterexample, the invalid substitution is $u(x) = \sin(x)$. It would imply $f(u) = \pm\frac{u}{\sqrt{1-u^2}}$ and the original integral must be split at $\pi/2$ to choose between $+$ and $-$. It does not invalidate the above formula with the continuity condition on $f$.
The result is correct, because the function $x \mapsto \sin(x) \cos(x)$ is $\pi$-periodic.
However, this was more of a lucky strike here, because the sub is not appropriate.
Another (bad) example would be the integral $\displaystyle\int_0^\pi \sin(x) \mathrm{d}x$. If you set $u = \sin(x)$, then you get $\mathrm{d}x = \dfrac{\mathrm{d}u}{\sqrt{1 - u^2}}$, and after substitution, you end up with $\displaystyle\int_0^0 \dfrac{\mathrm{d}u}{\sqrt{1 - u^2}} = 0$, which is clearly false. So something went wrong somewhere.
The main problem, in both cases, is that the inverse sine function is only defined to give you outputs in the &quot;rightmost&quot; part of the trigonometric circle (in other words, the image of $\sin^{-1}$ is $[-\pi/2, \pi, 2]$).
So, since in your example, your range is $[0, 2\pi]$, it messes up as soon as $x$ goes beyond $\pi/2$.
That's why you need to be careful when making trig subs. To do it properly, you'd have to split your intervals in parts where $\sin$ is injective, so that you can get a well defined inverse.