TITLE: Evaluate $3\int_{0}^{2\pi} \sin(t) \cos(t) \,{\rm d}t$
QUESTION [11 upvotes]: $$3\int_{0}^{2\pi} \sin(t) \cos(t) \,{\rm d}t$$
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,
$$3\int_{u=0}^{u=0}u \,{\rm d}u=0$$

REPLY [6 votes]: I thought I'd create a separate answer to just try and explain the confusion people have a bit more. @Gribouillis has given you a really good answer, so please give him the solution check-mark.

It's not true that the substitution needs to be injective. You can see for yourself on the Wikipedia article the conditions required for $u$ substitution, and injectivity is not one of those requirements: https://en.wikipedia.org/wiki/Integration_by_substitution. In this case - your solution for this integral is completely valid (it is indeed $0$) and there is nothing wrong with your method.
There are some cases where if you blindly apply substitution, you can get the incorrect result. For example, consider
$${\int_{0}^{\pi}\sin(x)dx}$$
Let's substitute ${u=\sin(x)}$. Then the bounds become ${\int_{0}^{0} ...du=0}$. Does that mean the original integral is $0$? NO!. We have incorrectly applied the theorem of Integration by Substitution. All substitution says is that
$${\int_{a}^{b}f(\phi(x))\phi'(x)dx=\int_{\phi(a)}^{\phi(b)}f(u)du}$$
${\int_{0}^{\pi}\sin(x)dx}$ does not match this form. You can make it happen - so let's try it. We can write
$${\int_{0}^{\pi}\sin(x)\frac{\cos(x)}{\cos(x)}dx}$$
Now here's the problem - we need to write the bottom ${\cos(x)}$ in terms of ${\sin(x)}$ - we know ${\cos(x)=\pm\sqrt{1-\sin^2(x)}}$ - that's the problem - the ${\pm}$. If ${x \in \left[0,\frac{\pi}{2}\right)}$, we have ${\cos(x)=\sqrt{1-\sin^2(x)}}$, and if ${x \in \left(\frac{\pi}{2},\pi\right]}$ then ${\cos(x)=-\sqrt{1-\sin^2(x)}}$. So you will have to split up the integral into two halves, like so:
$${\int_{0}^{\pi}\sin(x)dx=\int_{0}^{\frac{\pi}{2}}\frac{\sin(x)}{\sqrt{1-\sin^2(x)}}\cos(x)dx+\int_{\frac{\pi}{2}}^{\pi}\frac{\sin(x)}{-\sqrt{1-\sin^2(x)}}\cos(x)dx}$$
And now we can apply substitution to the two individual integrals to get
$${=\int_{0}^{1}\frac{u}{\sqrt{1-u^2}}du-\int_{1}^{0}\frac{u}{\sqrt{1-u^2}}du=2\int_{0}^{1}\frac{u}{\sqrt{1-u^2}}du}$$
So as you can see - there are certain subtleties with using substitution - but injectivity is not a direct condition required for it. Your solution to the integral is absolutely fine.