# Change of variable in indefinite integrals

• quasar987
In summary, the conversation discusses the justification of a change of variables in indefinite integrals and proposes a solution using the fundamental theorem. The solution involves finding a primitive function and using the chain rule.
quasar987
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I've had this very basic question on the back of my mind for 2 almost years now and I think I've found a satisfactory answer. The question is this most simple one: how do we justify a change of variables such as

$$\int e^{ax}dx = \int \frac{1}{a}e^udu$$

in an indefinite integral? My "solution" is this... We are looking for a primitive $F(\beta)$ to $f(\beta)$. By one version of the fondamental thm, as soon as f is continuous, such a primitive is given by

$$F(\beta)=\int_{\phi(\alpha)}^{\phi(\beta)}f(x)dx = \int_{\alpha}^{\beta}(f\circ \phi)(s)\frac{d\phi}{ds}ds$$

where $\phi:(\alpha,\beta)\rightarrow (\phi(\alpha), \phi(\beta))$ is $C^1$. (Ths is just the change of variable thm)

This may be total nonsense, but my eyes are itching from tiredness so I wouldn't know. Any comments on the above or an opinion on how to justify change of variable in indefinite integrals will be warmly appreciate tomorrow morning.

Good night!

quasar987 said:
I've had this very basic question on the back of my mind for 2 almost years now and I think I've found a satisfactory answer. The question is this most simple one: how do we justify a change of variables such as

$$\int e^{ax}dx = \int \frac{1}{a}e^udu$$

in an indefinite integral? My "solution" is this... We are looking for a primitive $F(\beta)$ to $f(\beta)$. By one version of the fondamental thm, as soon as f is continuous, such a primitive is given by

$$F(\beta)=\int_{\phi(\alpha)}^{\phi(\beta)}f(x)dx = \int_{\alpha}^{\beta}(f\circ \phi)(s)\frac{d\phi}{ds}ds$$

where $\phi:(\alpha,\beta)\rightarrow (\phi(\alpha), \phi(\beta))$ is $C^1$. (Ths is just the change of variable thm)

This may be total nonsense, but my eyes are itching from tiredness so I wouldn't know. Any comments on the above or an opinion on how to justify change of variable in indefinite integrals will be warmly appreciate tomorrow morning.

Good night!
I'm not at all sure what you are asking. Certainly if
$$F(x)= \int e^{ax}dx$$ and
$$G(u)= \frac{1}{a}\int e^u du$$
then, by the chain rule, with u(x)= ax,
$$\frac{dG}{dx}= \frac{dG}{du}\frac{du}{dx}$$
$$= \left(\frac{1}{a}e^u\right)(a)$$
$$= e^u= e^{ax}= \frac{dF}{dx}$$
Since F and G have the same derivatives they are anti-derivatives (indefinite integrals) of the same thing.

"Substitution" in integrals is just the chain rule "in reverse".

ack, I had tried things with the chain rule, but never this. Yep, this is exactly what I was loking for, thanks Halls!

## 1. What is a change of variable in indefinite integrals?

A change of variable in indefinite integrals is a technique used to simplify the integration of complex functions by substituting a new variable for the existing one. This allows for easier integration and can often result in a more manageable expression.

## 2. Why is a change of variable useful in indefinite integrals?

A change of variable can be useful in indefinite integrals because it can transform a complicated integral into a simpler form that is easier to solve. It can also help to identify patterns and relationships between different functions.

## 3. How do you choose the appropriate variable for a change of variable in indefinite integrals?

The appropriate variable for a change of variable in indefinite integrals is usually chosen based on the form of the integral. In some cases, it may be helpful to choose a variable that cancels out or simplifies certain terms in the integral. It is also important to choose a variable that is easy to integrate.

## 4. Can a change of variable be used in all indefinite integrals?

No, a change of variable may not always be applicable to all indefinite integrals. It is most effective when the integrand contains a single variable and can be rewritten in terms of the new variable. In some cases, it may not lead to a simpler form or may even make the integral more complicated.

## 5. How do you know if a change of variable is successful in an indefinite integral?

A successful change of variable in an indefinite integral will result in a simplified expression that is easier to integrate. The integral should also be solvable in terms of the new variable. If the integral becomes more complicated or cannot be solved, the chosen variable may not have been appropriate for the change of variable.

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