More obvious pattern recognition

Here’s another one:

    \[\int\sin(x)\cos(x)\mathrm{d}x\]

It’s not too difficult to substitute:
if u=\sin(x) then \mathrm{d}u=\cos(x)\mathrm{d}x, and so

    \[\int\sin(x)\cos(x)\mathrm{d}x=\int u\mathrm{d}u=\tfrac{1}{2}\sin^2(x)+C\]

Were we not in these fortunate circumstance to recognize it, we would have to resort to something like partial integration.
In that case we would choose u=\sin(x) and u'=\cos(x), next to v'=\cos(x) and v=\sin(x) which is ridiculous if you’re not awake by now.
Nevertheless we end up with

    \[\int uv'=uv-\int u'v\]

and get

    \[\sin^2(x)-\int\cos(x)\sin(x)\mathrm{d}x\]

and the integral repeats itself.
Luckily

    \[\int\sin(x)\cos(x)\mathrm{d}x=\sin^2(x)-\int\cos(x)\sin(x)\mathrm{d}x\]

means

    \[2\int\sin(x)\cos(x)\mathrm{d}x=\sin^2(x)\]

so

    \[\int\sin(x)\cos(x)\mathrm{d}x=\tfrac{1}{2}\sin^2(x)+C\]

Well, integration is integration, not only in both ways, but in every way.