How to understand the definition of an inverse substitution when finding the primitive function?

Question

From textbook:

Inverse substitutions: Let $f$ be a function defined on an interval $I$. Let $g$ be a function from an interval $J$ into an interval $I$ which is differentiable on $J$, and let $h$ be a function from $I$ into $J$ such that $g(h(x)) = x$ on $I$. If $G$ is an antiderivative of $f (g)g′$ on $J$, then $G(h)$ is an antiderivative of $f$ on $I$:

$$\int f(x)\,dx = \int f(g(t))g'(t)\,dt|_{t=h(x)}$$

Note: The assumptions are most often satisfied by taking $g$ that is one-to-one on $J$ with the range $I$; therefore there exists an inverse function on $J$ which will do as $h$.

For me this whole definition is a bit difficult to understand, so to make sure I understand it well, I have a few questions.

First: "Let $g$ be a function from an interval $J$ into an interval $I$" Can anybody explain what it means the have a function from an interval to another one? Does that mean that the if we take a function, let's say $g=sin(x)$ (which is defined in $R$) and we got the interval $I$ as for eg. $(-2,1)$ and we want to find the interval $J$ to $I$ we would have to evaluate the function $g$ for $x∈(-2,1)$ which would find us the interval $J$ (in this case it would be $(something, \frac{π}{2})$?

Another question:"let $h$ be a function from $I$ into $J$ such that $g(h(x)) = x$ on $I$" why do we care about the function $h$, what would happen (change) if this wasn't written there?

And lastly: "...there exists an inverse function on $J$ which will do as $h$" why do we need the function $g$ to be one-to-one? Why do we need the existence of the inverse function?

I know I should ask only one question but I thought it would be better to have the whole context here.

Note: Feel free to edit this question so that people understand it because my english is not awesome. Thanks

Answer 1

For your first question: "What does it mean to have a function from and interval $J$ to an interval $I$?"

This simply means that if you restrict the function $g(x)$ to only values of $x$ in the interval $J$, then the range (outputs) will be the values in the interval $I$. Here are a few examples:

• If $J = \left[0,\frac{\pi}{2}\right]$, and $g(x) = \sin(x)$, then $I=[0,1]$.
• If $J = [-1,2]$ and $g(x) = 2x+3$, then $I = [1,7]$
• If $J = [0,1]$ and $g(x) = e^x$, then $I = [1,e]$

In practice, you most often use this for what are called "trigonometric substitutions". For example, you might take $x = g(t) = \sin(t)$, and then $dx = g'(t) \, dt = \cos(t) \, dt$. An example of how you might use this is the following:

$$ \int \frac{1}{\sqrt{1-x^2}} \, dx = \int \frac{\cos(t)}{\sqrt{1-\sin^2(t)}} \, dt = \int \, dt = t $$

The reason you might want to know $h(x)$ is to invert and go back to the original $x$-variable. The result of the integral was $t$, but we have to invert the equation $x = \sin(t)$, and re-write as $t = \sin^{-1}(x)$ in order to express the result in terms of the original variable $x$.