What is this equation $\int f(g(t)) g'(t) dt = \int f(x) dx$?

Question

What is this equation $\int f(g(t)) g'(t) dt = \int f(x) dx$ ?

I can understand a equation like $\int_\alpha^{\beta} f(g(t)) g'(t) dt = \int_{a}^{b} f(x) dx$ because both of the left hand side and the right hand side are real numbers.

I am confused.

$\int f(g(t)) g'(t) dt$ is a function whose derivative is $f(g(t)) g'(t)$.
$\int f(x) dx$ is a function whose derivative is $f(x)$.

If we wanna get a function $\int f(g(t)) g'(t) dt$ and $\int f(x) dx$ is easy to integrate, then $F(g(t))$ is what we want, where $F(x)$ is a $\int f(x) dx$.
I can understand this.

If we wanna get a function $\int f(x) dx$ and $\int f(g(t)) g'(t) dt$ is easy to integrate, then $F(g^{-1}(x))$ is what we want, where $F(t)$ is a $\int f(g(t)) g'(t) dt$. But what happens if $g^{-1}(x)$ doesn't exist?

Answer 1

It is more an identity than an equation, and must be accompanied by the substitution in use, $x=g(t)$.

$$\int f(g(t))\,g'(t)\,dt=\int f(g(t))\,dg(t)=\int f(x)\,dx=F(x)=F(g(t))$$

where $F$ is an antiderivative of $f$.

You could also invoke the chain rule directly,

$$\int f(g(t))\,g'(t)\,dt=F(g(t)).$$

---

The "trick" only works if the values taken by $x$ are the same as those taken by $f(t)$. In particular the endpoints of the integration domain must be matching, and if $g$ is not injective, you may have to decompose the domain in subdomains where it is.

Answer 2

Let $x=g(t)$, then $dx = g'(t) dt $ and thus $\int f(g) g'(t) dt = \int f(x) dx $

Answer 3

Notice that this is the key principle behind the technique of "U-Substitution". If we start of with the integral $\int f(x) \mathrm dx$. We may make the following subtitutions. $$\begin{bmatrix}x \\ \mathrm dx \end{bmatrix}=\begin{bmatrix}\ g(t) \\ g'(t)\mathrm dt \end{bmatrix} \tag1$$ Now plugging this information into our original indefinite integral, we get: $$\int f(x) \mathrm dx=\int f(g(t))g'(t)\mathrm dt$$

I can understand a equation like $\int_\alpha^{\beta} f(g(t)) g'(t) dt = \int_{a}^{b} f(x) dx$ because both of the left hand side and the right hand side are real numbers.

Note that for this equality to hold, $\alpha$, $\beta$ must be somehow related to $a$ and $b$ via the function $g(t)$. To get that relation, simply consider the definite integral $\int_{a}^{b} f(x) \mathrm dx$. Again using the substitution as mentioned in $(1)$, we get: $$\int_{a}^{b} f(x) \mathrm dx=\int_{g^{-1}(a)}^{g^{-1}(b)}f(g(t))g'(t)\mathrm dt \tag2$$ Comparing equality $(2)$ with $\int_\alpha^{\beta} f(g(t)) g'(t) dt = \int_{a}^{b} f(x) dx$, we get the following relation: $$g^{-1}(a)=\alpha \iff a=g(\alpha) \\ g^{-1}(b)=\beta \iff b=g(\beta) \tag3$$ It may get easier to comprehend if we start off with the definite integral $\int_{\alpha}^{\beta} f(g(t))g'(t)\mathrm dt$ and then transform it to $\int_{a}^{b}f(x)\mathrm dx$.