Inserting a function into dx

Question

Sometimes when we calculate an indefinite integral in class we insert a function into $dx$, we always insert the primitive of the function into dx. ($f'(x)dx=d(f(x))$)

A simple example is:

$\int {\cot xdx} = \int {\frac{{\cos x}}{{\sin x}}dx} = \int {\frac{{d\left( {\sin x} \right)}}{{\sin x}}} = \left[ {t = \sin x} \right] = \int {\frac{{dt}}{t}} = - \frac{1}{{{t^2}}}+C = - \frac{1}{{{{\sin }^2}\left( x \right)}}+C$

I'm not sure I understand the meaning of doing such a thing? Can it always be done? I would like to understand $dx$ once and for all.

Answer 1

The reason this works and is not just a trick, but formally correct, is that you can think of $d$ as being the exterior derivative operator. The way $d$ works on a differentiable function of one variable is just as you have written, $d(f(x))=\frac{df}{dx} \cdot$dx

Answer 2

Suppose that $g(y)$ is a continuous function of $f$, and $f(x)$ is a continuously differentiable function of $x$.

Now, we propose that

$$\int g(f)\,df=\int g(f(x))\frac{df(x)}{dx}\,dx \tag 1$$

The left-hand side of $(1)$ represents a function whose derivative with respect to $f$ is $g(f)$. The right-hand side of $(1)$ represents a function whose derivative with respect to $x$ is $g(f(x))\,\frac{df(x)}{dx}$.

But, from the chain-rule, we have

$$\begin{align} \frac{d}{dx}\left(\int g(f)\,df\right)&=\left.\frac{d}{df}\left(\int g(f)\,df\right)\right|_{f=f(x)}\left(\frac{df}{dx}\right)\\\\ &=g(f(x))\left(\frac{df}{dx}\right)\\\\ &=\frac{d}{dx}\left(\int g(f(x))\frac{df(x)}{dx}\,dx \right) \tag 2 \end{align}$$

Comparing the left-hand and right-hand sides of $(2)$, we have

$$\int g(f)\,df=\int g(f(x))\frac{df(x)}{dx}\,dx+C$$

Inasmuch as the integration constant is arbitrary, we have established the proposed equivalence of $(1)$ and therefore, $\frac{df(x)}{dx}\,dx=d(f(x))$ when $f(x)$ is a continuously differentiable function of $x$.