# Solve Integral Using Integration by Parts

• Matty R
In summary, I'm struggling to integrate by parts for a question that I understand. Can you help me out?
Matty R
Hello

I was hoping someone could help me with this integral.

## Homework Statement

$$I=\int{(x^2sin(5x^3-3))}dx$$

## Homework Equations

$$\int{(u.\frac{dv}{dx})}dx=[uv]-\int{(v.\frac{du}{dx})}dx$$

$$\frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dx}$$

3a. The first attempt at a solution

$$u=x^2, \frac{dv}{dx}=sin(5x^3-3)$$

$$v=-\frac{cos(5x^3-3)}{15x^2}, \frac{du}{dx}=2x$$

$$I=[\frac{x^2.-cos(5x^3-3)}{15x^2}]-\int{((\frac{-cos(5x^3-3)}{15x^2}).2x)}dx$$

$$So, I=-[\frac{cos(5x^3-3)}{15}]+\int{((\frac{2cos(5x^3-3)}{15x}))}dx$$

3b. The second attempt at a solution

$$u=sin(5x^3-3), \frac{dv}{dx}=x^2$$

$$v=\frac{x^3}{3}, \frac{du}{dx}=15x^2cos(5x^3-3)$$

$$I=[sin(5x^3-3).\frac{x^3}{3}]-\int(\frac{x^3}{3}.15x^2cos(5x^3-3))$$

$$So, I=[\frac{x^3sin(5x^3-3)}{3}]-[\frac{x^4sin(5x^3-3)}{12}]$$

That is as far as I've got in both situations.

I've been trying to get :

$$I=[something]\pm J$$
$$J=[something]\pm I$$
$$So, I = [something]\pm[something]\pm I$$

I just can't seem to get there.

So I was wondering if someone could have a look through what I've done and see if I've gone wrong somewhere. I got through the rest of the work pretty quickly, but I'm really stuck with this one.

Thank you

Don't integrate by parts. Just do the substitution u=5x^3-3.

Yes, remember that u-substitution is basically like reversing the chain rule. When you do the chain rule you multiple the outside by the derivative of the inside (in loose terms). Notice in your integral that you have the function 5x3-3 on the inside of sin and its derivative x2 on the outside (you're only missing a constant, which is no big deal). This is what tells you to use u-substitution. You use it when you notice a composition of functions, and you use integration by parts when you see a product of functions (because integration by parts is the product rule in reverse).

Thanks for the replies.

Dick said:
Don't integrate by parts. Just do the substitution u=5x^3-3.

I've done it now. I overcomplicate things far too often. :shy:

$$-\frac{cos(5x^3-3)}{15}$$

Thank you very much.

n!kofeyn said:
Yes, remember that u-substitution is basically like reversing the chain rule. When you do the chain rule you multiple the outside by the derivative of the inside (in loose terms). Notice in your integral that you have the function 5x3-3 on the inside of sin and its derivative x2 on the outside (you're only missing a constant, which is no big deal). This is what tells you to use u-substitution. You use it when you notice a composition of functions, and you use integration by parts when you see a product of functions (because integration by parts is the product rule in reverse).

I hadn't thought of things in that way. I knew that Integration was the reverse of Differentiation, but I didn't see that u-substitution was the reverse of the Chain rule, or that Integration by parts was the opposite of the Product rule. That makes so much sense.

However, I'm still a bit confused by this question.

I saw it as x2 multiplied by sin(5x3-3), so I thought it was to be done using Integration by parts.

I've done examples like x2cosx and excosx in the past using Integration by parts. I don't understand the difference between those and my original question.

The difference between x2sin(5x3 - 3) and your other two examples is that sin(5x3 - 3) is a composite function and the other two are just products of functions that aren't composite. When you antidifferentiate a composite function, the first thing you would try would be an ordinary substitution. With a product of two functions, a natural thing to try would probably be integration by parts.

Mark44 said:
The difference between x2sin(5x3 - 3) and your other two examples is that sin(5x3 - 3) is a composite function and the other two are just products of functions that aren't composite. When you antidifferentiate a composite function, the first thing you would try would be an ordinary substitution. With a product of two functions, a natural thing to try would probably be integration by parts.

Hello again.

I'm sorry. I'm still struggling. I can do the maths, but its knowing which method to use that is getting me.

I saw that integral as a function multiplied by a composite function, so two functions multiplied together.

I understand using substitution for the composite function, but the x2 made it look like two functions multiplied together. So I used substitution on the composite function followed by parts for the non-composite and integrated composite functions combined.

If I have a non-composite function multiplied by a composite function, I use substitution, and if I have two non-composite functions multiplied together, I integrate by parts?

I'm sorry if I'm missing something simple. It wouldn't be the first, or last, time.

Matty R said:
Hello again.

I'm sorry. I'm still struggling. I can do the maths, but its knowing which method to use that is getting me.

I saw that integral as a function multiplied by a composite function, so two functions multiplied together.
And the key here is that one of the functions is composite.
Matty R said:
I understand using substitution for the composite function, but the x2 made it look like two functions multiplied together. So I used substitution on the composite function followed by parts for the non-composite and integrated composite functions combined.
You are confused here. You don't use one technique on part of a product and another technique on another part. You are correct in thinking that you want to try a substitution, but are incorrect in thinking that you want to use integration by parts on only a part of the integrand. At least that's what I think you are saying.

As you learn different techniquest for integration it becomes more difficult to know which technique to try for a given integral. As a general rule, start with the simpler techniques first, and if they don't work, then try the more complicated techniques. When you have a product of two functions and one of them is a composite function, your very first thought should be to try a substitution. For you problem the integrand was x2sin(5x3 - 3). The most obvious substitution is u = 5x3 - 3, which means that du = 10x2dx. All that's missing in your integrand is that factor of 10, which you can easily insert and compensate for by another factor of 1/10 outside the integral.

With this substitution your integral is
$$(1/10)~\int sin(u)du$$

For this problem, the obvious substitution worked, but that won't always be the case. After you have exhausted the possibilities for substitutions you can move on to other techniques. Even though you haven't found the antiderivative, substitutions can be done fairly quickly, so you haven't wasted much time.
Matty R said:
If I have a non-composite function multiplied by a composite function, I use substitution, and if I have two non-composite functions multiplied together, I integrate by parts?
Yes, pretty much.
Matty R said:
I'm sorry if I'm missing something simple. It wouldn't be the first, or last, time.

Mark44 said:
And the key here is that one of the functions is composite.

You are confused here. You don't use one technique on part of a product and another technique on another part. You are correct in thinking that you want to try a substitution, but are incorrect in thinking that you want to use integration by parts on only a part of the integrand. At least that's what I think you are saying.

As you learn different techniquest for integration it becomes more difficult to know which technique to try for a given integral. As a general rule, start with the simpler techniques first, and if they don't work, then try the more complicated techniques. When you have a product of two functions and one of them is a composite function, your very first thought should be to try a substitution. For you problem the integrand was x2sin(5x3 - 3). The most obvious substitution is u = 5x3 - 3, which means that du = 10x2dx. All that's missing in your integrand is that factor of 10, which you can easily insert and compensate for by another factor of 1/10 outside the integral.

With this substitution your integral is
$$(1/10)~\int sin(u)du$$

For this problem, the obvious substitution worked, but that won't always be the case. After you have exhausted the possibilities for substitutions you can move on to other techniques. Even though you haven't found the antiderivative, substitutions can be done fairly quickly, so you haven't wasted much time.

Thanks for the explanation. I think I understand it now.

I guess it comes down to practise and experience. I'll be doing plenty of practise over the coming month, so I'll put your help to good use.

Thanks again.

## 1. What is integration by parts?

Integration by parts is a method used in calculus to solve integrals by breaking them down into simpler integrals that are easier to solve. It involves using the product rule from differentiation to rearrange the integral into a form that can be easily solved.

## 2. When should I use integration by parts?

Integration by parts is useful when the integral involves a product of two functions, or when it is in the form of uv, where u and v are functions. It can also be used to solve integrals that follow a specific pattern, such as ln x or ex.

## 3. How do I use integration by parts?

To use integration by parts, follow the formula: ∫ u dv = uv - ∫ v du. Identify u and dv in the integral, and use the product rule to find v and du. Then, substitute these values into the formula and solve the simpler integral on the right side.

## 4. Can integration by parts be used to solve definite integrals?

Yes, integration by parts can be used to solve both indefinite and definite integrals. When solving a definite integral, make sure to substitute the limits of integration into the final solution.

## 5. What are some common mistakes when using integration by parts?

Some common mistakes when using integration by parts include choosing the wrong u and dv, forgetting to apply the product rule correctly, and not simplifying the integral before solving. It is also important to double check the final solution by differentiating it to ensure it is equivalent to the original integral.

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