Lesson Explainer: Integration by Parts

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 𝑦=𝑒𝑣, where 𝑒 and 𝑣 are differentiable functions: dddddd𝑦π‘₯=𝑒𝑣π‘₯+𝑣𝑒π‘₯.

Rearranging this equation gives 𝑒𝑣π‘₯=𝑦π‘₯βˆ’π‘£π‘’π‘₯𝑒𝑣π‘₯=π‘₯(𝑒𝑣)βˆ’π‘£π‘’π‘₯.dddddddddddd

Next, we integrate both sides of this equation with respect to π‘₯: 𝑒𝑣π‘₯π‘₯=ο„Έο€½π‘₯(𝑒𝑣)βˆ’π‘£π‘’π‘₯π‘₯𝑒𝑣π‘₯π‘₯=ο„Έπ‘₯(𝑒𝑣)π‘₯βˆ’ο„Έπ‘£π‘’π‘₯π‘₯.ddddddddddddddddd

Then, by the first part of the fundamental theorem of calculus, the first term on the right-hand side simplifies to 𝑒𝑣+C, though the constant of integration will merge with the constant from the other indefinite integral. This gives us the formula for integration by parts: 𝑒𝑣π‘₯π‘₯=π‘’π‘£βˆ’ο„Έπ‘£π‘’π‘₯π‘₯.dddddd

Theorem: Integration by Parts

For two differentiable functions 𝑒 and 𝑣, 𝑒𝑣π‘₯π‘₯=π‘’π‘£βˆ’ο„Έπ‘£π‘’π‘₯π‘₯.dddddd

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 𝑒 and dd𝑣π‘₯. Once we have chosen these, we will need to differentiate 𝑒 and integrate dd𝑣π‘₯ to form the functions dd𝑒π‘₯ and 𝑣 respectively.

Let’s now see an example of how to use integration by parts to evaluate the integral of π‘₯π‘₯sin.

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

Use integration by parts to evaluate ο„Έπ‘₯π‘₯π‘₯sind.

Answer

The integration by parts formula tells us that, for differentiable functions 𝑒 and 𝑣, 𝑒𝑣π‘₯π‘₯=π‘’π‘£βˆ’ο„Έπ‘£π‘’π‘₯π‘₯.dddddd

Since we are integrating π‘₯π‘₯sin, we need to establish which factor we will define as 𝑒 and which factor we will define as dd𝑣π‘₯.

Notice that if we choose 𝑒=π‘₯, then when we differentiate this function, we will obtain dd𝑒π‘₯=1. Since this is a constant, it will make the integrand in the final term of this formula much less complicated than the original.

Let’s set 𝑒=π‘₯𝑣π‘₯=π‘₯.andddsin

Next, we find dd𝑒π‘₯ by differentiating 𝑒, and 𝑣 by integrating dd𝑣π‘₯: ddandcos𝑒π‘₯=1𝑣=βˆ’π‘₯.

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 ο„Έπ‘₯π‘₯π‘₯=π‘₯Γ—(βˆ’π‘₯)βˆ’ο„Έ(βˆ’π‘₯)Γ—1π‘₯=βˆ’π‘₯π‘₯βˆ’ο„Έβˆ’π‘₯π‘₯=βˆ’π‘₯π‘₯βˆ’(βˆ’π‘₯)+=π‘₯βˆ’π‘₯π‘₯+.sindcoscosdcoscosdcossinCsincosC

Using integration by parts, we find that ο„Έπ‘₯π‘₯π‘₯=π‘₯βˆ’π‘₯π‘₯+.sindsincosC

In our first example, we saw that, by being careful with our choice of 𝑒, we produced a second integral that was much easier to evaluate. Had we instead chosen 𝑒=π‘₯sin, we would have obtained a second integral that was more complex. In this case, the derivative of 𝑒 was a constant term. If, however, it is not clear which function to choose for 𝑒, the acronym LIATE can help us to decide. Whichever function comes first in the list is the function we should choose to be 𝑒. 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 𝑒 to be the function that appears first in this list.

LLogarithmic functionslog(π‘₯), ln(π‘₯), etc.
IInverse trigonometric functionssin(π‘₯), arctan(π‘₯), etc.
AAlgebraic functionsπ‘₯, 5π‘₯, etc.
TTrigonometric functionssin(π‘₯), cos(π‘₯), etc.
EExponential functions2, 𝑒, 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 ο„Έ(3π‘₯+4)𝑒π‘₯οŠ¨ο—d.

Answer

The integrand (3π‘₯+4)π‘’οŠ¨ο— 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 𝑒 and 𝑣, 𝑒𝑣π‘₯π‘₯=π‘’π‘£βˆ’ο„Έπ‘£π‘’π‘₯π‘₯.dddddd

We begin by choosing the functions 𝑒 and dd𝑣π‘₯. The LIATE rule tells us to choose 𝑒 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 𝑒.

Hence, we set 𝑒=(3π‘₯+4)𝑣π‘₯=𝑒.οŠ¨ο—anddd

Next, we find dd𝑒π‘₯ by differentiating 𝑒, and 𝑣 by integrating dd𝑣π‘₯. The general power rule tells how to find the derivative of a differentiable function raised to a constant exponent, 𝑛: ddπ‘₯(𝑓(π‘₯))=𝑛(𝑓(π‘₯))𝑓′(π‘₯).

Differentiating (3π‘₯+4) gives us dd𝑒π‘₯=2(3π‘₯+4)Γ—3=6(3π‘₯+4).

To obtain 𝑣, we integrate dd𝑣π‘₯=𝑒: 𝑒π‘₯=𝑒+.dC

Therefore, ddand𝑒π‘₯=6(3π‘₯+4)𝑣=𝑒.

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, ο„Έ(3π‘₯+4)𝑒π‘₯=(3π‘₯+4)Γ—π‘’βˆ’ο„Έπ‘’Γ—6(3π‘₯+4)π‘₯=𝑒(3π‘₯+4)βˆ’ο„Έ6𝑒(3π‘₯+4)π‘₯.οŠ¨ο—οŠ¨ο—ο—ο—οŠ¨ο—ddd

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 𝑒. However, we notice that the derivative of 3π‘₯+4 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: 6𝑒(3π‘₯+4)π‘₯.d

Let 𝑒=3π‘₯+4𝑣π‘₯=𝑒anddd so that ddand𝑒π‘₯=3𝑣=𝑒.

Substituting these into the integration by parts formula gives 𝑒(3π‘₯+4)π‘₯=(3π‘₯+4)Γ—π‘’βˆ’ο„Έπ‘’Γ—3π‘₯=𝑒(3π‘₯+4)βˆ’ο„Έ3𝑒π‘₯=𝑒(3π‘₯+4)βˆ’3𝑒+.ο—ο—ο—ο—ο—ο—ο—οŠ§dddC

We can now substitute this expression into the equation for our original integral: ο„Έ(3π‘₯+4)𝑒π‘₯=𝑒(3π‘₯+4)βˆ’ο„Έ6𝑒(3π‘₯+4)π‘₯=𝑒(3π‘₯+4)βˆ’6[𝑒(3π‘₯+4)βˆ’3𝑒+]=𝑒(3π‘₯+4)βˆ’6𝑒(3π‘₯+4)+18𝑒+.οŠ¨ο—ο—οŠ¨ο—ο—οŠ¨ο—ο—οŠ§ο—οŠ¨ο—ο—ddCC

To simplify this result, we take out a factor of 𝑒: ο„Έ(3π‘₯+4)𝑒π‘₯=𝑒(3π‘₯+4)βˆ’6(3π‘₯+4)+18+=𝑒9π‘₯+6π‘₯+10+.οŠ¨ο—ο—οŠ¨ο—οŠ¨dCC

By applying integration by parts, ο„Έ(3π‘₯+4)𝑒π‘₯=𝑒9π‘₯+6π‘₯+10+.οŠ¨ο—ο—οŠ¨dC

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 𝑒=𝑒 and dcosd𝑣=π‘₯π‘₯, evaluate 𝑒π‘₯π‘₯cosd by integrating by parts.

Answer

The integration by parts formula tells us that, for differentiable functions 𝑒 and 𝑣, 𝑒𝑣π‘₯π‘₯=π‘’π‘£βˆ’ο„Έπ‘£π‘’π‘₯π‘₯.dddddd

We are told to set 𝑒=𝑒 and dcosd𝑣=π‘₯π‘₯. In other words, 𝑒=𝑒𝑣π‘₯=π‘₯.andddcos

We will need to evaluate dd𝑒π‘₯ by differentiating 𝑒, and 𝑣 by integrating dd𝑣π‘₯: ddandsin𝑒π‘₯=𝑒𝑣=π‘₯.

The integration by parts formula tells us 𝑒π‘₯π‘₯=𝑒×π‘₯βˆ’ο„Έπ‘₯×𝑒π‘₯=𝑒π‘₯βˆ’ο„Έπ‘’π‘₯π‘₯.cosdsinsindsinsind

We cannot evaluate 𝑒π‘₯π‘₯sind directly, so we apply integration by parts once more.

Let 𝑒=𝑒𝑣π‘₯=π‘₯.andddsin so that ddandcos𝑒π‘₯=𝑒𝑣=βˆ’π‘₯.

Therefore, 𝑒π‘₯π‘₯=𝑒×(βˆ’π‘₯)βˆ’ο„Έ(βˆ’π‘₯)×𝑒π‘₯=βˆ’π‘’π‘₯+𝑒π‘₯π‘₯.sindcoscosdcoscosd

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 𝑒π‘₯π‘₯=𝑒π‘₯βˆ’ο•βˆ’π‘’π‘₯+𝑒π‘₯π‘₯=𝑒π‘₯+𝑒π‘₯βˆ’ο„Έπ‘’π‘₯π‘₯.cosdsincoscosdsincoscosd

We can now add 𝑒π‘₯π‘₯cosd to both sides of this equation and include a constant of integration: 2𝑒π‘₯π‘₯=𝑒π‘₯+𝑒π‘₯+.ο—ο—ο—οŠ§cosdsincosC

Finally, dividing by 2, we obtain 𝑒π‘₯π‘₯=12(𝑒π‘₯+𝑒π‘₯)+.cosdsincosC

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 𝑒 and dd𝑣π‘₯. 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 𝑒 and 𝑣, where 𝑒′ and 𝑣′ are continuous on [π‘Ž,𝑏], 𝑒𝑣π‘₯π‘₯=[𝑒𝑣]βˆ’ο„Έπ‘£π‘’π‘₯π‘₯.dddddd

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 ο„Έπ‘₯𝑒π‘₯οŠ§οŠ¦οŠ¨ο—d.

Answer

The integrand is the product of two functions, π‘₯ and 𝑒. This is an indication to us that we might need to use integration by parts to evaluate the integral.

Remember that, for differentiable functions 𝑒 and 𝑣, where 𝑒′ and 𝑣′ are continuous on [π‘Ž,𝑏], 𝑒𝑣π‘₯π‘₯=[𝑒𝑣]βˆ’ο„Έπ‘£π‘’π‘₯π‘₯.dddddd

To apply integration by parts, we first need to decide which factor is 𝑒 and which factor is dd𝑣π‘₯. We note that both functions, π‘₯ and 𝑒, are differentiable and have continuous derivatives.

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

Since π‘₯ is an algebraic function, we will define this to be 𝑒.

Let 𝑒=π‘₯𝑣π‘₯=𝑒.οŠ¨ο—anddd

Then, we see ddand𝑒π‘₯=2π‘₯𝑣=𝑒.

Note that the derivative of 𝑒 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 ο„Έπ‘₯𝑒π‘₯=π‘₯𝑒||βˆ’ο„Έ2π‘₯𝑒π‘₯.οŠ§οŠ¦οŠ¨ο—οŠ¨ο—οŠ§οŠ¦οŠ§οŠ¦ο—dd

We cannot directly evaluate this new definite integral, so instead we will apply integration by substitution again. Let 𝑒=2π‘₯ and dd𝑣π‘₯=𝑒 on the second integral: ddand𝑒π‘₯=2𝑣=𝑒.

Therefore, ο„Έ2π‘₯𝑒π‘₯=2π‘₯𝑒|βˆ’ο„Έ2𝑒π‘₯=[2π‘₯π‘’βˆ’2𝑒].οŠ§οŠ¦ο—ο—οŠ§οŠ¦οŠ§οŠ¦ο—ο—ο—οŠ§οŠ¦dd

Since this is a definite integral, we do not need to include a constant of integration at any stage: ο„Έπ‘₯𝑒π‘₯=π‘₯π‘’βˆ’(2π‘₯π‘’βˆ’2𝑒)=π‘₯π‘’βˆ’2π‘₯𝑒+2𝑒.οŠ§οŠ¦οŠ¨ο—οŠ¨ο—ο—ο—οŠ§οŠ¦οŠ¨ο—ο—ο—οŠ§οŠ¦d

Finally, we evaluate our antiderivative at the limits of integration: ο„Έπ‘₯𝑒π‘₯=1Γ—π‘’βˆ’2Γ—1×𝑒+2π‘’οβˆ’ο‘0Γ—π‘’βˆ’2Γ—0×𝑒+2𝑒=(π‘’βˆ’2𝑒+2𝑒)βˆ’(2)=π‘’βˆ’2.οŠ§οŠ¦οŠ¨ο—οŠ¨οŠ§οŠ§οŠ§οŠ¨οŠ¦οŠ¦οŠ¦d

Therefore, ο„Έπ‘₯𝑒π‘₯=π‘’βˆ’2.οŠ§οŠ¦οŠ¨ο—d

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 ο„Έπ‘₯π‘₯lnd by parts using 𝑒=π‘₯ln and dd𝑣=π‘₯.

Answer

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

The integration by parts formula tells us that, for differentiable functions 𝑒 and 𝑣, 𝑒𝑣π‘₯π‘₯=π‘’π‘£βˆ’ο„Έπ‘£π‘’π‘₯π‘₯.dddddd

We are given 𝑒=π‘₯𝑣=π‘₯,𝑣π‘₯=1.lnandddordd

To apply the integration by parts formula, we differentiate 𝑒 and integrate dd𝑣π‘₯: ddand𝑒π‘₯=1π‘₯𝑣=π‘₯.

Once we have each of these expressions, we can substitute them into the formula to find ο„Έπ‘₯π‘₯=π‘₯Γ—π‘₯βˆ’ο„Έπ‘₯Γ—1π‘₯π‘₯=π‘₯π‘₯βˆ’ο„Έ1π‘₯=π‘₯π‘₯βˆ’π‘₯+.lndlndlndlnC

We can simplify by factoring π‘₯: ο„Έπ‘₯π‘₯=π‘₯(π‘₯βˆ’1)+.lndlnC

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 lnπ‘₯ and 1. We choose 𝑒=π‘₯ln since if we were to choose 𝑒=1, then we would still need to integrate lnπ‘₯ 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 ο„Έ(π‘₯)π‘₯.tand

Answer

While it is not instantly obvious how to evaluate this integral, we do know some information about the derivative of tan(π‘₯): ddtanπ‘₯ο€Ή(π‘₯)=11+π‘₯.

If we rewrite our integrand as 1Γ—(π‘₯)tan, we can then use integration by parts to evaluate this integral.

We recall that, for differentiable functions 𝑒 and 𝑣, where 𝑒′ and 𝑣′ are continuous on [π‘Ž,𝑏], 𝑒𝑣π‘₯π‘₯=[𝑒𝑣]βˆ’ο„Έπ‘£π‘’π‘₯π‘₯.dddddd

To choose the function 𝑒, we can use the fact that we know the derivative of tan(π‘₯). We also know that tan(π‘₯) 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 𝑒=(π‘₯)𝑣π‘₯=1.tananddd

Then, ddand𝑒π‘₯=11+π‘₯𝑣=π‘₯.

Substituting each expression into the integration by parts formula gives ο„Έ(π‘₯)π‘₯=π‘₯(π‘₯)||βˆ’ο„Έπ‘₯1+π‘₯π‘₯.tandtand

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 𝑓, 𝑓′(π‘₯)𝑓(π‘₯)π‘₯=|𝑓(π‘₯)|+.dlnC

Letting 𝑓(π‘₯)=1+π‘₯ and 𝑓′(π‘₯)=2π‘₯, we see that our numerator is half of this value of 𝑓′(π‘₯), so we can rewrite our integral as 12ο„Έ2π‘₯1+π‘₯π‘₯=12||1+π‘₯||+.dlnC

In fact, we do not need this constant since we are going to be evaluating a definite integral: ο„Έ(π‘₯)π‘₯=π‘₯(π‘₯)||βˆ’ο„Έπ‘₯1+π‘₯π‘₯=π‘₯(π‘₯)βˆ’12||1+π‘₯||=ο€Ό1Γ—(1)βˆ’12||1+1||οˆβˆ’ο€Ό0Γ—(0)βˆ’12||1+0||=πœ‹4βˆ’12(2).tandtandtanlntanlntanlnln

Therefore, ο„Έ(π‘₯)π‘₯=πœ‹4βˆ’(2)2.tandln

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: 𝑒𝑣π‘₯π‘₯=π‘’π‘£βˆ’ο„Έπ‘£π‘’π‘₯π‘₯,dddddd where 𝑒 and 𝑣 are differentiable functions.
  • When integrating by parts, we try to choose 𝑒 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 𝑒, by choosing the function that appears first in this list.
    LLogarithmic functionslog(π‘₯), ln(π‘₯), etc.
    IInverse trigonometric functionssin(π‘₯), arctan(π‘₯), etc.
    AAlgebraic functionsπ‘₯, 5π‘₯, etc.
    TTrigonometric functionssin(π‘₯), cos(π‘₯), etc.
    EExponential functions2, 𝑒, etc.
  • We can use the formula to integrate special functions, such as lnπ‘₯ and tan(π‘₯), by writing each as 1Γ—π‘₯ln or 1Γ—(π‘₯)tan.
  • For differentiable functions 𝑒 and 𝑣, where 𝑒, 𝑣, 𝑒′, and 𝑣′ are continuous on [π‘Ž,𝑏], 𝑒𝑣π‘₯π‘₯=[𝑒𝑣]βˆ’ο„Έπ‘£π‘’π‘₯π‘₯.dddddd

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.