# Integration by parts

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 dvYou 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!

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

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