Lesson Explainer: Integration by Parts | Nagwa Lesson Explainer: Integration by Parts | Nagwa

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=𝑢𝑣, where 𝑢 and 𝑣 are differentiable functions: dddddd𝑦𝑥=𝑢𝑣𝑥+𝑣𝑢𝑥.

Rearranging this equation gives 𝑢𝑣𝑥=𝑦𝑥𝑣𝑢𝑥𝑢𝑣𝑥=𝑥(𝑢𝑣)𝑣𝑢𝑥.dddddddddddd

Next, we integrate both sides of this equation with respect to 𝑥: 𝑢𝑣𝑥𝑥=𝑥(𝑢𝑣)𝑣𝑢𝑥𝑥𝑢𝑣𝑥𝑥=𝑥(𝑢𝑣)𝑥𝑣𝑢𝑥𝑥.ddddddddddddddddd

Then, by the first part of the fundamental theorem of calculus, the first term on the right-hand side simplifies to 𝑢𝑣+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: 𝑢𝑣𝑥𝑥=𝑢𝑣𝑣𝑢𝑥𝑥.dddddd

Theorem: Integration by Parts

For two differentiable functions 𝑢 and 𝑣, 𝑢𝑣𝑥𝑥=𝑢𝑣𝑣𝑢𝑥𝑥.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 𝑢 and dd𝑣𝑥. Once we have chosen these, we will need to differentiate 𝑢 and integrate dd𝑣𝑥 to form the functions dd𝑢𝑥 and 𝑣 respectively.

Let’s now see an example of how to use integration by parts to evaluate the integral of 𝑥𝑥sin.

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

Use integration by parts to evaluate 𝑥𝑥𝑥sind.

Answer

The integration by parts formula tells us that, for differentiable functions 𝑢 and 𝑣, 𝑢𝑣𝑥𝑥=𝑢𝑣𝑣𝑢𝑥𝑥.dddddd

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

Notice that if we choose 𝑢=𝑥, then when we differentiate this function, we will obtain dd𝑢𝑥=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 𝑢=𝑥𝑣𝑥=𝑥.andddsin

Next, we find dd𝑢𝑥 by differentiating 𝑢, and 𝑣 by integrating dd𝑣𝑥: ddandcos𝑢𝑥=1𝑣=𝑥.

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 𝑥𝑥𝑥=𝑥×(𝑥)(𝑥)×1𝑥=𝑥𝑥𝑥𝑥=𝑥𝑥(𝑥)+=𝑥𝑥𝑥+.sindcoscosdcoscosdcossinCsincosC

Using integration by parts, we find that 𝑥𝑥𝑥=𝑥𝑥𝑥+.sindsincosC

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

LLogarithmic functionslog(𝑥), ln(𝑥), etc.
IInverse trigonometric functionssin(𝑥), arctan(𝑥), etc.
AAlgebraic functions𝑥, 5𝑥, etc.
TTrigonometric functionssin(𝑥), cos(𝑥), 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 (3𝑥+4)𝑒𝑥d.

Answer

The integrand (3𝑥+4)𝑒 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 𝑢 and 𝑣, 𝑢𝑣𝑥𝑥=𝑢𝑣𝑣𝑢𝑥𝑥.dddddd

We begin by choosing the functions 𝑢 and dd𝑣𝑥. The LIATE rule tells us to choose 𝑢 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 𝑢.

Hence, we set 𝑢=(3𝑥+4)𝑣𝑥=𝑒.anddd

Next, we find dd𝑢𝑥 by differentiating 𝑢, and 𝑣 by integrating dd𝑣𝑥. The general power rule tells how to find the derivative of a differentiable function raised to a constant exponent, 𝑛: dd𝑥(𝑓(𝑥))=𝑛(𝑓(𝑥))𝑓(𝑥).

Differentiating (3𝑥+4) gives us dd𝑢𝑥=2(3𝑥+4)×3=6(3𝑥+4).

To obtain 𝑣, we integrate dd𝑣𝑥=𝑒: 𝑒𝑥=𝑒+.dC

Therefore, ddand𝑢𝑥=6(3𝑥+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, (3𝑥+4)𝑒𝑥=(3𝑥+4)×𝑒𝑒×6(3𝑥+4)𝑥=𝑒(3𝑥+4)6𝑒(3𝑥+4)𝑥.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 𝑢. However, we notice that the derivative of 3𝑥+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: 6𝑒(3𝑥+4)𝑥.d

Let 𝑢=3𝑥+4𝑣𝑥=𝑒anddd so that ddand𝑢𝑥=3𝑣=𝑒.

Substituting these into the integration by parts formula gives 𝑒(3𝑥+4)𝑥=(3𝑥+4)×𝑒𝑒×3𝑥=𝑒(3𝑥+4)3𝑒𝑥=𝑒(3𝑥+4)3𝑒+.dddC

We can now substitute this expression into the equation for our original integral: (3𝑥+4)𝑒𝑥=𝑒(3𝑥+4)6𝑒(3𝑥+4)𝑥=𝑒(3𝑥+4)6[𝑒(3𝑥+4)3𝑒+]=𝑒(3𝑥+4)6𝑒(3𝑥+4)+18𝑒+.ddCC

To simplify this result, we take out a factor of 𝑒: (3𝑥+4)𝑒𝑥=𝑒(3𝑥+4)6(3𝑥+4)+18+=𝑒9𝑥+6𝑥+10+.dCC

By applying integration by parts, (3𝑥+4)𝑒𝑥=𝑒9𝑥+6𝑥+10+.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 𝑢=𝑒 and dcosd𝑣=𝑥𝑥, evaluate 𝑒𝑥𝑥cosd by integrating by parts.

Answer

The integration by parts formula tells us that, for differentiable functions 𝑢 and 𝑣, 𝑢𝑣𝑥𝑥=𝑢𝑣𝑣𝑢𝑥𝑥.dddddd

We are told to set 𝑢=𝑒 and dcosd𝑣=𝑥𝑥. In other words, 𝑢=𝑒𝑣𝑥=𝑥.andddcos

We will need to evaluate dd𝑢𝑥 by differentiating 𝑢, and 𝑣 by integrating dd𝑣𝑥: ddandsin𝑢𝑥=𝑒𝑣=𝑥.

The integration by parts formula tells us 𝑒𝑥𝑥=𝑒×𝑥𝑥×𝑒𝑥=𝑒𝑥𝑒𝑥𝑥.cosdsinsindsinsind

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

Let 𝑢=𝑒𝑣𝑥=𝑥.andddsin so that ddandcos𝑢𝑥=𝑒𝑣=𝑥.

Therefore, 𝑒𝑥𝑥=𝑒×(𝑥)(𝑥)×𝑒𝑥=𝑒𝑥+𝑒𝑥𝑥.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 𝑒𝑥𝑥=𝑒𝑥𝑒𝑥+𝑒𝑥𝑥=𝑒𝑥+𝑒𝑥𝑒𝑥𝑥.cosdsincoscosdsincoscosd

We can now add 𝑒𝑥𝑥cosd to both sides of this equation and include a constant of integration: 2𝑒𝑥𝑥=𝑒𝑥+𝑒𝑥+.cosdsincosC

Finally, dividing by 2, we obtain 𝑒𝑥𝑥=12(𝑒𝑥+𝑒𝑥)+.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 𝑢 and dd𝑣𝑥. 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 𝑢 and 𝑣, where 𝑢 and 𝑣 are continuous on [𝑎,𝑏], 𝑢𝑣𝑥𝑥=[𝑢𝑣]𝑣𝑢𝑥𝑥.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 𝑥𝑒𝑥d.

Answer

The integrand is the product of two functions, 𝑥 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 𝑢 and 𝑣, where 𝑢 and 𝑣 are continuous on [𝑎,𝑏], 𝑢𝑣𝑥𝑥=[𝑢𝑣]𝑣𝑢𝑥𝑥.dddddd

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

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

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

Let 𝑢=𝑥𝑣𝑥=𝑒.anddd

Then, we see ddand𝑢𝑥=2𝑥𝑣=𝑒.

Note that the derivative of 𝑢 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 𝑥𝑒𝑥=𝑥𝑒||2𝑥𝑒𝑥.dd

We cannot directly evaluate this new definite integral, so instead we will apply integration by substitution again. Let 𝑢=2𝑥 and dd𝑣𝑥=𝑒 on the second integral: ddand𝑢𝑥=2𝑣=𝑒.

Therefore, 2𝑥𝑒𝑥=2𝑥𝑒|2𝑒𝑥=[2𝑥𝑒2𝑒].dd

Since this is a definite integral, we do not need to include a constant of integration at any stage: 𝑥𝑒𝑥=𝑥𝑒(2𝑥𝑒2𝑒)=𝑥𝑒2𝑥𝑒+2𝑒.d

Finally, we evaluate our antiderivative at the limits of integration: 𝑥𝑒𝑥=1×𝑒2×1×𝑒+2𝑒0×𝑒2×0×𝑒+2𝑒=(𝑒2𝑒+2𝑒)(2)=𝑒2.d

Therefore, 𝑥𝑒𝑥=𝑒2.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 𝑥𝑥lnd by parts using 𝑢=𝑥ln and dd𝑣=𝑥.

Answer

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

The integration by parts formula tells us that, for differentiable functions 𝑢 and 𝑣, 𝑢𝑣𝑥𝑥=𝑢𝑣𝑣𝑢𝑥𝑥.dddddd

We are given 𝑢=𝑥𝑣=𝑥,𝑣𝑥=1.lnandddordd

To apply the integration by parts formula, we differentiate 𝑢 and integrate dd𝑣𝑥: ddand𝑢𝑥=1𝑥𝑣=𝑥.

Once we have each of these expressions, we can substitute them into the formula to find 𝑥𝑥=𝑥×𝑥𝑥×1𝑥𝑥=𝑥𝑥1𝑥=𝑥𝑥𝑥+.lndlndlndlnC

We can simplify by factoring 𝑥: 𝑥𝑥=𝑥(𝑥1)+.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 ln𝑥 and 1. We choose 𝑢=𝑥ln since if we were to choose 𝑢=1, then we would still need to integrate ln𝑥 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 (𝑥)𝑥.tand

Answer

While it is not instantly obvious how to evaluate this integral, we do know some information about the derivative of tan(𝑥): ddtan𝑥(𝑥)=11+𝑥.

If we rewrite our integrand as 1×(𝑥)tan, we can then use integration by parts to evaluate this integral.

We recall that, for differentiable functions 𝑢 and 𝑣, where 𝑢 and 𝑣 are continuous on [𝑎,𝑏], 𝑢𝑣𝑥𝑥=[𝑢𝑣]𝑣𝑢𝑥𝑥.dddddd

To choose the function 𝑢, we can use the fact that we know the derivative of tan(𝑥). We also know that tan(𝑥) 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 𝑢=(𝑥)𝑣𝑥=1.tananddd

Then, ddand𝑢𝑥=11+𝑥𝑣=𝑥.

Substituting each expression into the integration by parts formula gives (𝑥)𝑥=𝑥(𝑥)||𝑥1+𝑥𝑥.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(𝑥)𝑓(𝑥)𝑥=|𝑓(𝑥)|+.dlnC

Letting 𝑓(𝑥)=1+𝑥 and 𝑓(𝑥)=2𝑥, we see that our numerator is half of this value of 𝑓(𝑥), so we can rewrite our integral as 122𝑥1+𝑥𝑥=12||1+𝑥||+.dlnC

In fact, we do not need this constant since we are going to be evaluating a definite integral: (𝑥)𝑥=𝑥(𝑥)||𝑥1+𝑥𝑥=𝑥(𝑥)12||1+𝑥||=1×(1)12||1+1||0×(0)12||1+0||=𝜋412(2).tandtandtanlntanlntanlnln

Therefore, (𝑥)𝑥=𝜋4(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: 𝑢𝑣𝑥𝑥=𝑢𝑣𝑣𝑢𝑥𝑥,dddddd where 𝑢 and 𝑣 are differentiable functions.
  • When integrating by parts, we try to choose 𝑢 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 𝑢, by choosing the function that appears first in this list.
    LLogarithmic functionslog(𝑥), ln(𝑥), etc.
    IInverse trigonometric functionssin(𝑥), arctan(𝑥), etc.
    AAlgebraic functions𝑥, 5𝑥, etc.
    TTrigonometric functionssin(𝑥), cos(𝑥), etc.
    EExponential functions2, 𝑒, etc.
  • We can use the formula to integrate special functions, such as ln𝑥 and tan(𝑥), by writing each as 1×𝑥ln or 1×(𝑥)tan.
  • For differentiable functions 𝑢 and 𝑣, where 𝑢, 𝑣, 𝑢, and 𝑣 are continuous on [𝑎,𝑏], 𝑢𝑣𝑥𝑥=[𝑢𝑣]𝑣𝑢𝑥𝑥.dddddd

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