# Integration by Parts: Struggling with Homework

• the_dialogue
In summary, the author is trying to solve for v_bar in an integration by parts problem, but is having trouble understanding the proper notation. They attempted using the traditional integration by parts learned in undergraduate calculus courses, but are not getting anything close to the solution. A help getting started would be appreciated.

## Homework Statement

Here is a question I'm struggling with. I encountered it in a paper, and although a solution is provided I'm not so sure I understand where they're coming from.

## Homework Equations

$$\int_{r_1}^{r_2} \overline{v}\frac{1}{r}\frac{d}{dr}(r\frac{du}{dr})rdr$$

where $$\overline{v}$$ is the conjugate of $$v$$

## The Attempt at a Solution

I tried using the traditional integration by parts learned in undergraduate calculus courses, but I'm not getting anything close to the solution.

Any help getting started would be appreciated.

I would simplilfy it first before trying to integrate.
$$\frac{d}{dr}(r\frac{du}{dr})~=~r\frac{d^2 u}{dr^2} + \frac{du}{dr}$$
Here I'm using the product rule.

Using the differentiated expression above, and doing some other simplification (cancelling the 1/r and r), your first integral becomes
$$\int_{r_1}^{r_2} \bar{v}(r\frac{d^2 u}{dr^2} + \frac{du}{dr})dr$$

Now split the above integral into two integrals. That should make the integration a little easier.

Mark44 said:
I would simplilfy it first before trying to integrate.
$$\frac{d}{dr}(r\frac{du}{dr})~=~r\frac{d^2 u}{dr^2} + \frac{du}{dr}$$
Here I'm using the product rule.

Using the differentiated expression above, and doing some other simplification (cancelling the 1/r and r), your first integral becomes
$$\int_{r_1}^{r_2} \bar{v}(r\frac{d^2 u}{dr^2} + \frac{du}{dr})dr$$

Now split the above integral into two integrals. That should make the integration a little easier.

That's right -- I got that far. At this point I'm not sure how to go on. I have some trouble understanding the proper notation.

Could someone assist in continuing?

Is v_bar a constant in the integration or does it depend on r?

Mark44 said:
Is v_bar a constant in the integration or does it depend on r?

V_bar has dependence on r.

What's the expression you're trying to derive?

Well the result should be:

$$\left[\overline{v}r\frac{du}{dr}\right]_{r_1}^{r_2}-\left[\overline{r}u\frac{\overline{dv}}{dr}\right]_{r_1}^{r_2}+\int_{r_1}^{r_2}(\overline{\frac{d^2 v}{dr^2}+\frac{1}{r}\frac{dv}{dr})rudr$$

Sorry the latex isn't updating, or at least it hasn't yet. Look at the code to see the rest of the equation...

OK, they integrated by parts twice. The first time, they used

$$f=\overline{v} \mbox{ and } dg=\frac{d}{dr}(ru') dr$$

That gets you the first term and the integral of $ru'\overline{v}'$. For the second integration by parts, they used

$$f=r\overline{v}' \mbox{ and } dg=u' dr$$

vela said:
OK, they integrated by parts twice. The first time, they used

$$f=\overline{v} \mbox{ and } dg=\frac{d}{dr}(ru') dr$$

That gets you the first term and the integral of $ru'\overline{v}'$. For the second integration by parts, they used

$$f=r\overline{v}' \mbox{ and } dg=u' dr$$

For the second integration I get:

$$\left[\overline{v'}ru\right]_{r_1}^{r_2}-\int_{r_1}^{r_2}(u(\overline{v'}+r\frac{d\overline{v'}}{dr})dr$$Which is obviously different than what they got:

$$-\left[\overline{r}u\frac{\overline{dv}}{dr}\right]_{r_1}^{r_2}+\int_{r_1}^{r_2}(\overline{\frac{d^2 v}{dr^2}+\frac{1}{r}\frac{dv}{dr})rudr$$

Last edited:
Is this what you say they got?

$$\int_{r_1}^{r_2} (\overline{\frac{d^2 v}{dr^2}+\frac{1}{r}\frac{dv}{dr}}) ru dr$$

If so, it looks the same to me. Why do you think they're different?

Oh of course. Sorry about that. Many thanks.

## 1. What is integration by parts and how does it work?

Integration by parts is a mathematical technique used to find the integral of a product of two functions. It is based on the product rule of differentiation and involves breaking down a complicated integral into simpler parts. This can be done by choosing one function as the "u" and the other as the "dv" and then using a formula to solve for the integral.

## 2. Why is integration by parts important in calculus?

Integration by parts is an important tool in calculus because it allows us to solve integrals that would otherwise be difficult or impossible to solve. It is particularly useful when dealing with functions that involve products of trigonometric, logarithmic, or exponential functions. It is also essential in many real-world applications, such as in physics and engineering.

## 3. How do I know when to use integration by parts?

To determine when to use integration by parts, you should look for integrals that involve a product of two functions. Additionally, you can use the acronym "LIATE" to guide your decision. This stands for logarithmic, inverse trigonometric, algebraic, trigonometric, and exponential functions. The function that comes first in this list should be chosen as the "u" in the integration by parts formula.

## 4. What are some common mistakes to avoid when using integration by parts?

One common mistake when using integration by parts is choosing the wrong function as the "u." It is important to choose the "u" function based on the LIATE acronym to ensure the integral becomes simpler. Another mistake is not properly simplifying the integral after using the integration by parts formula. It is crucial to simplify the integral as much as possible before attempting to solve it.

## 5. How can I improve my skills in using integration by parts?

The best way to improve your skills in using integration by parts is to practice solving a variety of integrals using this technique. You can also look for online resources or textbooks that provide step-by-step solutions for integrals. It is also helpful to review the product rule of differentiation and understand how it is related to integration by parts. With practice and understanding, you can become proficient in using integration by parts to solve complex integrals.