# 3.1 Integration by Parts

### Learning Objectives

- Recognize when to use integration by parts.
- Use the integration-by-parts formula to evaluate indefinite integrals.
- Use the integration-by-parts formula for definite integrals.

By now we have a fairly thorough procedure for how to evaluate many basic integrals. However, although we can integrate by using the substitution, something as simple looking as defies us. Although there is no product rule for integration, there is a technique based on the product rule for differentiation that allows us to exchange one integral for another. We call this technique integration by parts.

### The Integration-by-Parts Formula

If then by using the product rule, we obtain

We rewrite this formula as and then integrate both sides of the last equation:

Since is an antiderivative of , we have

By making the substitutions and which in turn make and we have the more compact form

### Integration by Parts

Let and be functions with continuous derivatives. Then, the integration-by-parts formula for the integral involving these two functions is:

The advantage of using the integration-by-parts formula is that we can exchange one integral for another, possibly easier, integral. The following example illustrates its use.

### Using Integration by Parts

Use integration by parts with and to evaluate

#### Solution

By choosing we have Since we can take to be any antiderivative of , and the simplest choice would be It is handy to keep track of these values as follows:

Applying the integration-by-parts formula (*) results in

## Analysis

At this point, there are probably a few items that need clarification. First of all, you may be curious about what would have happened if we had chosen and If we had done so, then we would have and Thus, after applying integration by parts, we would get Unfortunately, with the new integral, we are in no better position than before. It is important to keep in mind that when we apply integration by parts, we may need to try several choices for and before finding a choice that works.

Second, you may wonder why, when we find as an antiderivative of we do not use To see that it makes no difference, we can rework the problem using

As you can see, it makes no difference in the final solution.

Last, we can verify that our antiderivative is correct by differentiating

Therefore, the answer we obtained is correct.

Evaluate using the integration-by-parts formula with and

#### Answer

#### Hint

Find and and use the previous example as a guide.

The natural question to ask at this point is: How do we know how to choose and Sometimes it is a matter of trial and error; however, the acronym LIATE can often help to take some of the guesswork out of our choices. This acronym stands for **L**ogarithmic Functions, **I**nverse Trigonometric Functions, **A**lgebraic Functions, **T**rigonometric Functions, and **E**xponential Functions. This mnemonic serves as an aid in determining an appropriate choice for

The type of function in the integral that appears first in the list should be our first choice of For example, if an integral contains a logarithmic function and an algebraic function, we should choose to be the logarithmic function, because L comes before A in LIATE. The integral in has a trigonometric function and an algebraic function Because A comes before T in LIATE, we chose to be the algebraic function. When we have chosen is selected to be the remaining part of the function to be integrated, together with

Why does this mnemonic work? Remember that whatever we pick to be must be something we can integrate. Since we do not have integration formulas that allow us to integrate simple logarithmic functions and inverse trigonometric functions, it makes sense that they should not be chosen as values for Consequently, they should be at the head of the list as choices for Thus, we put LI at the beginning of the mnemonic. (We could just as easily have started with IL, since these two types of functions won’t appear together in an integration-by-parts problem.) The exponential and trigonometric functions are at the end of our list because they are fairly easy to integrate and make good choices for Thus, we have TE at the end of our mnemonic. (We could just as easily have used ET at the end, since when these types of functions appear together it usually doesn’t really matter which one is and which one is ) Algebraic functions are generally easy both to integrate and to differentiate, and they come in the middle of the mnemonic.

### Using Integration by Parts

Evaluate

#### Solution

Begin by rewriting the integral:

Since this integral contains the algebraic function and the logarithmic function choose since L comes before A in LIATE. After we have chosen we must choose

Next, since we have Also, and so we take . Summarizing,

Substituting into (*) gives

Evaluate

#### Answer

#### Hint

Use and

In some cases, as in the next two examples, it may be necessary to apply integration by parts more than once.

### Applying Integration by Parts More Than Once

Evaluate

#### Solution

Using LIATE, choose and Thus, and which means we can take . Therefore,

Substituting into (*) produces

We still cannot integrate directly, but the integral now has a lower power on We can evaluate this new integral by using integration by parts again. To do this, choose and Thus, and which means we can take . Now we have

Going back to the previous equation and using (*) yields

After evaluating the last integral and simplifying, we obtain

### Applying Integration by Parts When LIATE Doesn’t Quite Work

Evaluate

#### Solution

If we use a strict interpretation of the mnemonic LIATE to make our choice of we end up with and Unfortunately, this choice won’t work because we are unable to evaluate However, since we can evaluate , we can try choosing and We then have

and so we can take

With these choices we have

Thus, we obtain

### Applying Integration by Parts More Than Once

Evaluate

#### Solution

This integral appears to have only one function—namely, —however, we can always use the constant function 1 as the other function. In this example, let’s choose and (The decision to use is easy. We can’t choose because if we could integrate it, we wouldn’t be using integration by parts in the first place!) Consequently, and we can take as an antiderivative of 1. After applying integration by parts to the integral and simplifying, we obtain

Unfortunately, this process leaves us with a new integral that is very similar to the original. However, let’s see what happens when we apply integration by parts again. This time let’s choose and making and, again, Substituting, we have

After simplifying, we obtain

The last integral is now the same as the original. It may seem that we have simply gone in a circle, but now we can actually evaluate the integral. To see how to do this more clearly, let be a particular antiderivative of . Substituting instead of into the above equality, we obtain the following equation for .

To find , add to both sides of the equation:

and then divide by 2

Since was a particular antiderivative of , we have

where is an arbitrary constant.

## Analysis

If this method feels a little strange at first, we can check the answer by differentiation:

Evaluate

#### Answer

#### Hint

This is similar to from one of the previous examples.

### Integration by Parts for Definite Integrals

Now that we have used integration by parts successfully to evaluate indefinite integrals, we turn our attention to definite integrals. The integration technique is really the same, only we add a step to evaluate the integral at the upper and lower limits of integration.

### Integration by Parts for Definite Integrals

Let and be functions with continuous derivatives on Then

### Finding the Area of a Region

Find the area of the region bounded above by the graph of and below by the *x*-axis over the interval

#### Solution

This region is shown in Figure 1 below. To find the area, we must evaluate

For this integral, let’s choose and thereby making and After applying the integration-by-parts formula (**) for definite integrals, we obtain

We use a substitution of to evaluate . We have that , and hence . Also, when , , and when , . It follows that

Therefore,

### Finding a Volume of Revolution

Find the volume of the solid obtained by revolving the region bounded by the graph of the *x*-axis, the *y*-axis, and the line about the *y*-axis.

#### Solution

We use cylindrical shells method to solve this problem. Begin by sketching the region to be revolved, along with a typical rectangle (see the following graph).

According to the formula, we must evaluate

To do this, let and These choices lead to and as an antiderivative of . Using integration by parts, we obtain

Evaluate

#### Answer

#### Hint

Use integration by parts with and

### Key Concepts

- The integration-by-parts formula allows the exchange of one integral for another, possibly easier, integral.
- Integration by parts applies to both definite and indefinite integrals.

### Key Equations

**Integration by parts formula**

**Integration by parts for definite integrals**

### Exercises

In using the technique of integration by parts, you must carefully choose which expression is *u.* For each of the following exercises, use the guidelines in this section to choose *u.* Do **not** evaluate the integrals.

**1.**

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**2.**

**3.**

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**4.**

**5.**

#### Answer

or ; this example requires two applications of integration by parts, and it is important to choose in the same way in both of them.

For the following exercises, evaluate the integral using integration by parts.

**6.**

**7.**

(Hint: is equivalent to )

#### Answer

**8.**

**9.**

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**10.**

**11.**

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**12.**

**13.**

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**14.**

**15.**

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**16.**

**17.**

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**18.**

**19.**

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**20.**

**21.**

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**22.**

**23.**

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**24.**

(Hint: you can simplify the integrand using properties of logarithms.)

**25.**

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**26.**

**27.**

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**28.**

**29.**

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**30.**

**31.**

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**32.**

**33.**

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For the following exercises, compute the given definite integral using integration by parts.

**34.**

**35.**

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**36.**

(Hint: make a substitution )

**37.**

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**38.**

**39.**

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**40.**

**41.**

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**42.**

**43.** Evaluate

#### Answer

Assume that *n* is a positive integer. Derive the following formulas using integration by parts. These formulas are called *reduction formulas* because the exponent in the *x* term has been reduced by one in each case.

**44.**

**45.**

#### Answer

Take and .

**46.**

**47.** Integrate using two methods:

- Using parts, letting
- Substitution, letting

#### Answer

State whether you would use integration by parts to evaluate the integral. If so, identify *u* and *dv*. If not, describe the technique you would use. You don’t have to evaluate the integral.

**48.**

**49.**

#### Answer

No need to use integration by parts. Make a substituion .

**50.**

**51.**

#### Answer

No need to use integration by parts. Make a substitution .

**52.**

**53.**

#### Answer

No need to use integration by parts. Make a substitution .

For the following exercises, sketch the region bounded by the given curves and find its area using integration by parts.

**54.**

**55*.**

(Hint: take and .)

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**56.**

**57.**

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**58.**

For the following exercises, find the volume generated by rotating the region bounded by the given curves about the specified axes.

**59.** about the *x*-axis.

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**60.** about the y-axis

**61.** , about

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**62.** about the *x*-axis.

**63.** about the *y-*axis.

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**64.** A particle moving along a straight line has a velocity of after *t* sec. How far does it travel in the first 2 sec?

### Glossary

- integration by parts
- a technique of integration that allows the exchange of one integral for another using the formula