\textbf{Substitution}

Let's say we want to integrate \[f(x)\] in order obtain its primitve function \[F(x)\] over some interval \[[a,b]\]:

,,\qquad \int f(x)\mathrm{d}x = F(x) + C,\quad x\in[a, b]

Now let's suppose we can identify a surjective function  \[g(x)\] that is continuous on \[[a,b]\] whose derivative on that interval is also continuous.
We can then write. 

,,\qquad \int f(g(x))\cdot g'(x)\mathrm{d}x = \int f(g(x))\ \mathrm{d}g(x) = F(g(x)) + C

In a nushell, we've substituted \[x\] with \[g(x)\] and solved the integral in the usual way.
The tricky part is identifying \[g'(x)\]. Without it, it would not be possible to 


\textbf{Example:}
Ok, the \textbf{based and redpilled} way to do this is to notice from the definition of \[\mathrm{d}\] in the OP is \[\mathrm{d}f(x) = f'(x)\mathrm{d}x\].
That means we could write \[g'(x)\mathrm{d}x\] as \[\mathrm{d}g(x)\]:

,,\qquad \int f(g(x))\cdot g'(x)\mathrm{d}x = \int f(g(x))\ \mathrm{d}g(x)

So a funny thing based engineers like to do (which makes mathfags seethe and cope) is this:

,,\qquad\int\sin(x)\cos(x)\mathrm{d}x = \int\sin(x)\ \mathrm{d}\sin(x) = \frac{\sin^2(x)}{2} + C 

Basically, treating \[\sin(x)\] like \[x\] and using \[\int x\mathrm{d}x = \frac{x^2}{2} + C\]. 

The \textbf{cringe and bluepilled} way of solving this, of course, is recognizing \[\sin(2x) = 2\sin(x)\cos(x)\]

,,\qquad\int\sin(x)\cos(x)\mathrm{d}x = \frac{1}{2}\int\sin(2x)\mathrm{d}x = \frac{-\cos(2x)}{4} + C = \frac{2\sin^2(x) - 1}{4} + C = \frac{\sin^2(x)}{2} + \underbrace{\left(-\frac{1}{4} + C\right)}_{\text{constant}} = \frac{\sin^2(x)}{2} + C