Lesson Explainer: Integration by Parts Mathematics • Higher Education

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 it shows that it is possible to choose our functions in a different manner.

In the next example, we will see how picking which part of the function we integrate and which part we differentiate is important and not always immediately obvious, and how we cannot always rely on the acronym to decide which part to integrate and which part to differentiate.

Example 4: Using Integration by Parts to Integrate a Function

Determine ο„Έ2𝑒π‘₯3(π‘₯+1)π‘₯ο—οŠ¨d.

Answer

The expression we have been asked to integrate here is 2𝑒π‘₯3(π‘₯+1)ο—οŠ¨, which is an algebraic fraction multiplied by an exponential function. Since this is a product of two functions, we will have to use integration by parts. The formula for doing this is 𝑒𝑣π‘₯π‘₯=π‘’π‘£βˆ’ο„Έπ‘£π‘’π‘₯π‘₯.dddddd

It is important to chose the correct parts of the function to be 𝑒 and dd𝑣π‘₯. If we consider the acronym LIATE, we can see that we should pick the algebraic fraction to be 𝑒. This would give us 𝑒=2π‘₯3(π‘₯+1)𝑣π‘₯=𝑒.οŠ¨ο—anddd

We will need to differentiate 𝑒 and integrate dd𝑣π‘₯. We should apply the quotient rule to differentiate 𝑒. Using the quotient rule, we will get dd𝑒π‘₯=2Γ—3(π‘₯+1)βˆ’2π‘₯Γ—6(π‘₯+1)9(π‘₯+1)=2Γ—3(π‘₯+1)βˆ’2π‘₯Γ—69(π‘₯+1)=βˆ’2π‘₯3(π‘₯+1).οŠͺ

Integrating the exponential term will leave it unchanged, so we have 𝑣=𝑒.

Let us now consider what we will have for the integral term in the formula for integration by parts: ο„Έβˆ’2𝑒π‘₯3(π‘₯+1)π‘₯.ο—οŠ©d

Looking at this integral, we can see it is very similar to the original integral we were trying to find, except this time the power of the denominator is one higher. If we tried to integrate this term using integration by parts like we did originally (choosing to differentiate the algebraic fraction and integrate the exponential function), then we would need to perform another integration, but the denominator would have an even higher power. This means if we keep trying to integrate in this fashion, we will keep increasing the power of the denominator and never reach the solution.

In order to integrate, we need to change the values of 𝑒 and dd𝑣π‘₯. One thing we can notice here is that if we consider part of the function we are integrating and rewrite it, we will obtain 23(π‘₯+1)=23(π‘₯+1).

Typically, when using integration by parts, we are looking to reduce the power of a term by differentiating it. However, this only works when the power is positive. Since this power is negative, we can get the power closer to zero by instead integrating. So, let us pick this part of the function to be dd𝑣π‘₯, and the remaining part to be 𝑒. We have 𝑒=π‘₯𝑒𝑣π‘₯=23(π‘₯+1).ο—οŠ±οŠ¨anddd

Using the product rule, we can differentiate 𝑒 to get dd𝑒π‘₯=𝑒+π‘₯𝑒=(1+π‘₯)𝑒.

Next, we integrate dd𝑣π‘₯ to get 𝑣=βˆ’23(π‘₯+1)=βˆ’23(π‘₯+1).

Substituting these into the formula for integration by parts, we have ο„Έ2𝑒π‘₯3(π‘₯+1)π‘₯=βˆ’2π‘₯𝑒3(π‘₯+1)βˆ’ο„Έβˆ’23(π‘₯+1)Γ—(1+π‘₯)𝑒π‘₯=βˆ’2π‘₯𝑒3(π‘₯+1)+ο„Έ23𝑒π‘₯.ο—οŠ¨ο—ο—ο—ο—ddd

We can see that the integration by parts formula has worked out really nicely for us here, since the (π‘₯+1) term cancelled out in the integral. In order to find our solution, we just need to integrate that last term. This gives us the result that ο„Έ2𝑒π‘₯3(π‘₯+1)π‘₯=βˆ’2π‘₯𝑒3(π‘₯+1)+23𝑒+=βˆ’2π‘₯𝑒3(π‘₯+1)+2𝑒(π‘₯+1)3(π‘₯+1)+=2𝑒3(π‘₯+1)+.ο—οŠ¨ο—ο—ο—ο—ο—dCCC

We have seen how important it can be when it comes to picking which part of the function we differentiate and which part we integrate in the previous example. In the next example, we will see how we can use integration by parts to integrate a logarithmic function.

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 the final example, we will see how we can integrate the product of a logarithm and an algebraic function.

Example 6: Using Integration by Parts to Integrate a Function

Determine ο„Έπ‘₯π‘₯π‘₯lnd.

Answer

We can see that, in our integral, we have a logarithmic function multiplied by an algebraic function. In order to integrate this, we will need to use integration by parts. The formula required to do this is 𝑒𝑣π‘₯π‘₯=π‘’π‘£βˆ’ο„Έπ‘£π‘’π‘₯π‘₯.dddddd

Typically, we would choose to differentiate the algebraic part, but if we did that it would mean we would have to integrate the logarithmic part, which is very tricky to do, so instead we shall integrate the algebraic part. Hence, we will have 𝑒=π‘₯𝑣π‘₯=π‘₯.lnanddd

Now we can differentiate and integrate these, respectively, to get ddand𝑒π‘₯=1π‘₯𝑣=13π‘₯.

Substituting these into the integration by parts formula, we obtain ο„Έπ‘₯π‘₯π‘₯=13π‘₯π‘₯βˆ’ο„Έ13π‘₯Γ—1π‘₯π‘₯=13π‘₯π‘₯βˆ’ο„Έ13π‘₯π‘₯.lndlndlnd

All that is left to do is to complete the integration of the last term and we will reach our solution: ο„Έπ‘₯π‘₯π‘₯=13π‘₯π‘₯βˆ’19π‘₯+=19π‘₯(3π‘₯βˆ’1)+.lndlnClnC

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.

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