Integrals Archives | Page 2 of 2 | Jake's Math Lessons

Integration by parts is another common technique used to find complex antiderivatives. This method tends to be a little more straight forward in its application than u-substitution. The main reason for this is that it requires the use of a formula, and if you can follow the formula you should be able to work through the rest.

First let’s introduce the formula, then I’ll explain how to use it. If you already know how to do these and you’re looking for extra practice problems, click here.

$$\int u \ dv = uv- \int v \ du$$

All this formula is really saying is that if we need to integrate some function which can be thought of as the product of two pieces, u and dv, then we can rewrite our integral in this other form. Notice we still would have an integral to solve after using this formula. But the hope is that $\int v \ du$ is easier to find than $\int u \ dv$ .

But how do you use the formula?

Using the integration by parts formula can be broken down into 3 simple steps and is going to start out somewhat similarly to integrating with u-substitution.

1. Picking u and dv

The first thing we need to do to use this formula is decide which piece of our function will be called u and which piece will be called dv. As we work through this problem, we will eventually need to work with the derivative of u and the antiderivative of dv. Therefore, to decide which piece we want to be u and dv, we should also consider the derivative and antiderivative of the pieces.

Let’s consider the following integral which we will find using integration by parts.

$$\int xsin(10x) \ dx$$

Clearly we can see that we are being asked to integrate some function which is the product of two smaller functions. It is the product of x and sin(10x). Therefore, between x and sin(10x), we will need to call one of these u and the other will be dv.

Does it matter which is which?

Yes, it does matter. You will need to take the derivative of u and the antiderivative of dv. So you want to pick one to be u and the other to be dv so that the derivative of u and the antiderivative of dv are easiest to work with.

Consider this: the sin(10x) term can be either u or dv. The reason for this is that whether you take the derivative or the antiderivative of sin(10x), the result will be some constant multiplied by cos(10x). As a result, it doesn’t make much of a difference whether we call sin(10x) the u or the dv.

Let’s think about the x term. If we call it u and have to work with its derivative, we’ll make things pretty easy on ourselves. I say this because the derivative of x is just 1. Alternatively, if we make x be dv and take it’s antiderivative, we will need to work with an $\mathbf{x^2}$ term (due to the power rule). Therefore, it will be a lot easier to work with the derivative of x than it will be to work with it’s antiderivative. This tells us that it’ll be easiest to call x the u piece.

Since it doesn’t matter what we call the sin(10x) term, but it’ll be a lot easier to make x be the u piece, we will say

$$u=x$$

$$dv=sin(10x) \ dx.$$

2. Finding v and du

Now that we have determined our u and dv, we need to use these to calculate v and du. To find du we just need to take the derivative of u.

$$\frac{du}{dx} = \frac{d}{dx} \big[ x \big]$$

$$\frac{du}{dx} = 1$$

Now we can just imagine multiplying both sides by dx to find

$$du=dx.$$

And to find v we just need to take the antiderivative, or the integral, of dv. You can do this using u-substitution with $u = 10x$ , but I will use WolframAlpha.

$$v = \int sin(10x) \ dx$$

$$v = – \frac{1}{10} cos(10x)$$

3. Plugging it all into the formula

Once you have laid out all four of the pieces we need, we can plug them all into the integration by parts formula. Just so we have everything in one place, let’s list out everything we have up to this point.

$$u=x$$

$$du=dx$$

$$v= – \frac{1}{10} cos(10x)$$

$$dv= sin(10x) \ dx$$

Now going back to the integration by parts formula I mentioned earlier, we can plug all of these in to the formula.

$$\int u \ dv = uv- \int v \ du$$

$$\int xsin(10x) \ dx = (x)\bigg( – \frac{1}{10} cos(10x) \bigg) – \int \bigg(- \frac{1}{10} cos(10x) \bigg) \ dx$$

Before integrating, let’s simplify this as much as we can by pulling the constant out of the integral.

$$\int xsin(10x) \ dx = \ – \frac{1}{10} xcos(10x) + \frac{1}{10} \int cos(10x) \ dx$$

Notice, the integral we need to compute now is much simpler than the integral we started with. This will be similar to the integral we computed to find v earlier. We can use u-substitution to find this by using $u=10x$ . I’m not going to show these steps, but I encourage you to work this out on your own!

$$\int xsin(10x) \ dx = \ – \frac{1}{10} xcos(10x) + \frac{1}{10} \bigg( \frac{1}{10} sin(10x) \bigg) + c$$

$$\int xsin(10x) \ dx = \ – \frac{1}{10} xcos(10x) + \frac{1}{100} sin(10x) + c$$

Some additional comments

And that’s it! With some, more complex, integration by parts problems you may have to apply this formula more than one time. Once you get to step 3, you might find that the simpler integral is still somewhat complicated and requires the use of integration by parts again.

In these cases, you can simply treat this integral like a sub-problem to find our main integral and go through these 3 steps with that integral. You can see an example of this here.

Overall, integration by parts isn’t terribly complicated once you know the formula and understand how to apply it. Take a look at some of my other lessons about integrals for some extra practice. And don’t forget, email me at jakesmathlessons@gmail.com if you can’t find the lesson or problem you’re looking for!

Extra Examples

Also, I can send you updates anytime I have a new problem solution or lesson ready to post so you can keep practicing the material. All you have to do is put your name and email below to get content updates straight to your inbox as well as a FREE bonus calculus 1 study guide to help you prepare for exams!

U-substitution is one of the more common methods of integration. It allows us to find the anti-derivative of fairly complex functions that simpler tricks wouldn’t help us with. The best way to think of u-substitution is that its job is to undo the chain rule.

That’s all we’re really doing.

It’s not too complicated when you think of it that way. Although, the execution isn’t always that simple. But I’ll show you 6 simple steps that will help you solve any u-substitution problem!

1. Picking our u

A u-substitution problem will start out similarly to an integration by parts problem. With any u-substitution problem the first thing you will need to do is decide what piece of the function you will call u. This is the most important piece of the process, and really the only part where there are options to choose from. However, there’s a simple trick to make sure you’re selecting the u correctly.

When deciding which part of your function to call u, you will want to look for a piece of your function that you can see that piece’s derivative somewhere else in the function. That sounds a little strange, but let me give you an example.

Say we have some function like $f(x)=x(x^2+5)^3$ . We want to look for a small piece of this function that also has its own derivative somewhere else in the function.

So we might say $u=x^2+5$ because the derivative of $x^2+5$ is just 2x. Notice, our function contains an x, not a 2x, but it’s fine if the derivative differs by a constant like this. It’s easy to deal with the constant here, but it’s important that it’s an x term.

A quick note on substitution

Choosing the correct u in these problems is the most challenging part. It’s not always simple to see what the u should be, so it’s important to be willing to try different things and see what happens.

You may end up needing to pick a u to go a few steps into the problem and realize it won’t work, then go back and pick another u. I know this process can be frustrating at times, we’ve all been there, but sometimes trial and error is required in learning new math concepts. So I urge you to stay persistent and keep picking different parts of the function for u.

If you try calling every possible piece of the function u, and work through the next few steps only to find that none of them will work, you likely need to either use some uncommon trick or find the derivative using another method besides u-substitution. I will get into some of the uncommon tricks in a later post.

2. Finding du

Once you have decided which piece of your function will be u, you then need to calculate du. This should be fairly simple.

All you have to do to find du is take the derivative of u then multiply it all by dx. This will sometimes require the use of the chain rule, product rule, or quotient rule, but usually you will just need the power rule.

Let’s think back to our previous example, finding the antiderivative of $y=x(x^2+5)^3$ . Remember, we decided that we would use $u=x^2+5$ . Therefore, to find du we should take it’s derivative and multiply by dx. This means if

$$u=x^2+5, \ then$$

$$du=2x \ dx.$$

3. Solve for dx

Now we have determined which part of our function we will call u and we found du. However, what we really want is to find dx. This will be easier to work with when we do our substitution into the original function.

All we need to do to find dx is take our equation for du and isolate the dx. In this example, this will be very easy.

$$du=2x \ dx$$

$$\frac{du}{2x}=dx$$

4. Substitute back into the original function

Going back to the function we are trying to find the antiderivative of, we will first write this in integral form.

$$\int x(x^2+5)^3 dx$$

Now we just need to substitute our u and dx back into this integral. This requires replacing the $x^2+5$ with u, and replacing dx with $\frac{du}{2x}$ . This gives us

$$\int x(u)^3 \frac{du}{2x}$$

After making these two substitutions we should be able to do some simplifying that will cancel out any remaining x‘s. Simplifying this integral should leave us with

$$\int \frac{1}{2}u^3 du.$$

And since we can pull constants out of an integral, this can also be written as

$$\frac{1}{2} \int u^3 du.$$

5. Integrate with respect to u

Looking at the above integral, we can see that we no longer have an x in the problem. We have rewritten everything in terms of u. Since our integral contains only u and du, instead of x and dx, we can integrate with respect to u. This simply means that we are taking the antiderivative of the function $g(u)=u^3$ where u is our variable.

Notice, this integral is much simpler than the one we started with. All we need to find this one is the power rule for antiderivatives.

$$\frac{1}{2} \int u^3 du$$

$$\frac{1}{2} \cdot \frac{1}{4}u^4$$

$$\frac{1}{8} u^4$$

Now the hard part is over, we found the antiderivative. But unfortunately we aren’t quite done yet. We can’t say that $\frac{1}{8}u^4$ is the antiderivative of $f(x)=x(x^2+5)^3$ . This doesn’t really have any meaning because they are using two different variables. Instead we need to write the answer in terms of x, since that’s what we started with.

6. Substitute x back in

We are almost done now that we found the antiderivative. We just need to write it in terms of x so that our answer actually is the antiderivative of the function we started with. Since we already found the antiderivative in terms of u, and we know u in terms of x, we can simply substitute in for u.

We decided back in step 1 that

$$u=x^2+5$$

and we also found out that our antiderivative in terms of u is

$$\frac{1}{8}u^4.$$

Therefore, we can plug $x^2+5$ in for u to find that the antiderivative of $f(x)=x(x^2+5)^3$ is

$$F(x)= \frac{1}{8}(x^2+5)^4 + c.$$

And that’s it! You can apply these 6 steps to solve any u-substitution problem.

I hope this lesson helps, but if there’s still a topic you’d like to learn about take a look at some of my other lessons and problem solutions. If you can’t find the topic or question you’re looking for just let me know by emailing me at jakesmathlessons@gmail.com!

I also encourage you to join my email list! Just enter your name and email below and I’ll let send you my calc 1 study guide as a bonus to help you survive calculus!

Category: Integrals

Integration by parts