In this explainer, we will learn how to use integration by parts to find the integral of a product of functions.

The fundamental theorem of calculus tells us that differentiation and integration are reverse processes to one other.

This means that any rule for differentiation can be applied as an integration rule in reverse. For example, consider the product rule for differentiating , where and are differentiable functions:

Rearranging this equation gives

Next, we integrate both sides of this equation with respect to :

Then, by the first part of the fundamental theorem of calculus, the first term on the right-hand side simplifies to , though the constant of integration will merge with the constant from the other indefinite integral. This gives us the formula for integration by parts:

### Theorem: Integration by Parts

For two differentiable functions and ,

This formula replaces an integral with another integral. The aim is to ensure that the new integral is easier to evaluate, so we must carefully choose our functions and . Once we have chosen these, we will need to differentiate and integrate to form the functions and respectively.

Letβs now see an example of how to use integration by parts to evaluate the integral of .

### Example 1: Integrating the Product of a Polynomial and a Trigonometric Function

Use integration by parts to evaluate .

### Answer

The integration by parts formula tells us that, for differentiable functions and ,

Since we are integrating , we need to establish which factor we will define as and which factor we will define as .

Notice that if we choose , then when we differentiate this function, we will obtain . Since this is a constant, it will make the integrand in the final term of this formula much less complicated than the original.

Letβs set

Next, we find by differentiating , and by integrating :

### Note:

In principle, we should obtain a constant of integration every time we integrate. However, we will eventually combine this with a second constant and so we generally choose not to include one at this stage.

The integration by parts formula then becomes

Using integration by parts, we find that

In our first example, we saw that, by being careful with our choice of , we produced a second integral that was much easier to evaluate. Had we instead chosen , we would have obtained a second integral that was more complex. In this case, the derivative of was a constant term. If, however, it is not clear which function to choose for , the acronym LIATE can help us to decide. Whichever function comes first in the list is the function we should choose to be . It is worth noting that although it is a useful trick, there are exceptions to the LIATE rule.

### How To: Applying the LIATE Rule

In integration by parts, the LIATE rule tells us to choose to be the function that appears first in this list.

L | Logarithmic functions | , , etc. |
---|---|---|

I | Inverse trigonometric functions | , , etc. |

A | Algebraic functions | , , etc. |

T | Trigonometric functions | , , etc. |

E | Exponential functions | , , etc. |

In the next example, we will look at how to use this acronym to integrate the product of an exponential and a polynomial function.

### Example 2: Finding the Integral of an Exponential Function Multiplied by a Polynomial Using Integration by Parts

Determine .

### Answer

The integrand is the product of two functions. This is an indication to us that we might need to use integration by parts to evaluate the integral.

Integration by parts tells us that, for differentiable functions and ,

We begin by choosing the functions and . The LIATE rule tells us to choose to be the function that appears first in the list: logarithmic functions, inverse trigonometric functions, algebraic functions, trigonometric functions, and exponential functions.

Our integrand is the product of a polynomial (algebraic) function and an exponential function. Since A occurs before E in the acronym, we choose the algebraic function to be .

Hence, we set

Next, we find by differentiating , and by integrating . The general power rule tells how to find the derivative of a differentiable function raised to a constant exponent, :

Differentiating gives us

To obtain , we integrate :

Therefore,

Remember, while we should obtain a constant of integration every time we integrate, we will eventually combine this with a second constant and so we generally choose not to include one at this stage.

Using integration by parts,

Notice that we now have a second integrand, which is the product of two functions. We might be worried that we chose the wrong function . However, we notice that the derivative of is a constant, which means we can evaluate this new integral using integration by parts. We will take the constant factor of 6 outside the integral and apply the formula once again to evaluate:

Let so that

Substituting these into the integration by parts formula gives

We can now substitute this expression into the equation for our original integral:

To simplify this result, we take out a factor of :

By applying integration by parts,

In the previous example, we saw that it is sometimes necessary to apply integration by parts multiple times. Each time we applied integration by parts, the power of the algebraic function decreased, eventually becoming constant and thus creating a simple final integral. In our next example, we will see how the LIATE acronym has its exceptions and how we might need to rearrange our result to evaluate an indefinite integral.

### Example 3: Finding the Indefinite Integral of the Product of an Exponential and a Trigonometric Function

By setting and , evaluate by integrating by parts.

### Answer

The integration by parts formula tells us that, for differentiable functions and ,

We are told to set and . In other words,

We will need to evaluate by differentiating , and by integrating :

The integration by parts formula tells us

We cannot evaluate directly, so we apply integration by parts once more.

Let so that

Therefore,

Notice that the second integral we obtain is equal to the original integral, so we do not need to continue integrating. Instead, by substituting this expression into our previous integration by parts, we have

We can now add to both sides of this equation and include a constant of integration:

Finally, dividing by 2, we obtain

In this example, we saw that we do not always need to apply the LIATE acronym and we are able to reverse the order in which we choose and . This will rarely be the case but shows that it is possible to choose our functions in a different manner.

Next, recall that integration by parts lets us find the antiderivative of a function, while the second part of the fundamental theorem of calculus says that we can evaluate definite integrals by using any antiderivative. Thus, we can use the integration by parts formula to evaluate a definite integral.

### Theorem: Integration by Parts for a Definite Integral

For differentiable functions and , where and are continuous on ,

In our next example, we will demonstrate how to evaluate such an integral.

### Example 4: Using Integration by Parts to Evaluate a Definite Integral

Evaluate .

### Answer

The integrand is the product of two functions, and . This is an indication to us that we might need to use integration by parts to evaluate the integral.

Remember that, for differentiable functions and , where and are continuous on ,

To apply integration by parts, we first need to decide which factor is and which factor is . We note that both functions, and , are differentiable and have continuous derivatives.

The acronym LIATE can help us decide. We choose to be the function that appears first in this list: logarithmic functions, inverse trigonometric functions, algebraic functions, trigonometric functions, and exponential functions.

Since is an algebraic function, we will define this to be .

Let

Then, we see

Note that the derivative of is not going to yield a simple integral; however, we will be able to apply the integration by parts formula a second time.

Substituting these expressions into the integration by parts formula gives us

We cannot directly evaluate this new definite integral, so instead we will apply integration by substitution again. Let and on the second integral:

Therefore,

Since this is a definite integral, we do not need to include a constant of integration at any stage:

Finally, we evaluate our antiderivative at the limits of integration:

Therefore,

Notice that it is not uncommon to need to use integration by parts more than once to evaluate an integral. We will now look at a common integral that is evaluated using integration by parts.

### Example 5: Integrating the Natural Logarithm Function

Integrate by parts using and .

### Answer

We are told to use integration by parts to evaluate this integral.

The integration by parts formula tells us that, for differentiable functions and ,

We are given

To apply the integration by parts formula, we differentiate and integrate :

Once we have each of these expressions, we can substitute them into the formula to find

We can simplify by factoring :

This previous example demonstrated that we can use the integration by parts formula to evaluate the integral of the natural logarithm function, writing it as the product of and 1. We choose since if we were to choose , then we would still need to integrate in the second integral.

In our final example, we will look at how the integration by parts formula can help us integrate an inverse trigonometric function.

### Example 6: Integrating an Inverse Trigonometric Function

Calculate

### Answer

While it is not instantly obvious how to evaluate this integral, we do know some information about the derivative of :

If we rewrite our integrand as , we can then use integration by parts to evaluate this integral.

We recall that, for differentiable functions and , where and are continuous on ,

To choose the function , we can use the fact that we know the derivative of . We also know that is continuous and differentiable over the set of real numbers, while its derivative is continuous over the same set. Thus, we can apply the given formula.

Let

Then,

Substituting each expression into the integration by parts formula gives

We might now notice that the numerator of the integrand is a scalar multiple of the derivative of the denominator.

We can, therefore, use a special version of integration by substitution, which says that, for a differentiable function ,

Letting and , we see that our numerator is half of this value of , so we can rewrite our integral as

In fact, we do not need this constant since we are going to be evaluating a definite integral:

Therefore,

Now that we have demonstrated how to use the integration by parts formula to evaluate a number of different types of integrals, let us recap some of the key points.

### Key Points

- We can evaluate integrals of products of functions by using the integration by parts formula: where and are differentiable functions.
- When integrating by parts, we try to choose to be the function that, when differentiated, will create a more easily evaluated second integral.
- The acronym LIATE can help us decide which function to choose for , by choosing the function that appears first in this list.
L Logarithmic functions , , etc. I Inverse trigonometric functions , , etc. A Algebraic functions , , etc. T Trigonometric functions , , etc. E Exponential functions , , etc. - We can use the formula to integrate special functions, such as and , by writing each as or .
- For differentiable functions and , where , , , and are continuous on ,