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Lesson Explainer: Integration by Substitution: Definite Integrals Mathematics • Higher Education

In this explainer, we will learn how to use integration by substitution for definite integrals.

Let us first recall how to integrate by substitution by going through a quick example. Let ๐ผ=๏„ธ๐‘ฅ๏€น2๐‘ฅ+5๏…๐‘ฅ.๏Šจ๏Šฉ๏Šชd

We compute this integral by making the substitution ๐‘ข=2๐‘ฅ+5๏Šฉ. We have that dd๐‘ข๐‘ฅ=6๐‘ฅ,๏Šจ and so ๐ผ=16๏„ธ๏€น2๐‘ฅ+5๏…6๐‘ฅ๐‘ฅ=16๏„ธ๐‘ข๐‘ข๐‘ฅ๐‘ฅ=16๏„ธ๐‘ข๐‘ข.๏Šฉ๏Šช๏Šจ๏Šช๏Šชddddd

In this form, the integral is straightforward to compute: 16๏„ธ๐‘ข๐‘ข=16ร—15๐‘ข+=130๐‘ข+.๏Šช๏Šซ๏ŠซdCC

Finally, we substitute ๐‘ข=2๐‘ฅ+5๏Šฉ back in: ๐ผ=130๐‘ข+=130๏€น2๐‘ฅ+5๏…+.๏Šซ๏Šฉ๏ŠซCC

In general, if we have a function of a function ๐‘“(๐‘”(๐‘ฅ)) and an integral of the form ๏„ธ๐‘“(๐‘”(๐‘ฅ))๐‘”๐‘ฅ๐‘ฅddd, then the substitution ๐‘ข=๐‘”(๐‘ฅ) yields dddd๐‘ข๐‘ฅ=๐‘”๐‘ฅ, and so ๏„ธ๐‘“(๐‘”(๐‘ฅ))๐‘”๐‘ฅ๐‘ฅ=๏„ธ๐‘“(๐‘ข)๐‘ข.dddd

This technique works for definite integrals just as it does for indefinite ones; however, when we make a substitution, the limits of integration need to change.

Formula: Integration by Substitution for Definite Integrals

If ๐‘“ is a continuous function and ๐‘” is differentiable with continuous derivative, then ๏„ธ๐‘“(๐‘”(๐‘ฅ))๐‘”๐‘ฅ๐‘ฅ=๏„ธ๐‘“(๐‘ข)๐‘ข,๏Œป๏Œบ๏€(๏Œป)๏€(๏Œบ)dddd where ๐‘ข=๐‘”(๐‘ฅ).

Let us look at an example of using substitution to evaluate a definite integral.

Example 1: Finding the Value of a Definite Integral Using Substitution

Use the substitution ๐‘ข=๐‘ฅโˆ’4 to evaluate the definite integral ๏„ธ๐‘ฅ(๐‘ฅโˆ’4)๐‘ฅ๏Šฉ๏Šง๏Šจd.

Answer

Recall that if we are integrating a function of the form ๐‘“(๐‘”(๐‘ฅ))๐‘”๐‘ฅdd, then we can make the substitution ๐‘ข=๐‘”(๐‘ฅ): ๏„ธ๐‘“(๐‘”(๐‘ฅ))๐‘”๐‘ฅ๐‘ฅ=๏„ธ๐‘“(๐‘ข)๐‘ข.๏Œป๏Œบ๏€(๏Œป)๏€(๏Œบ)dddd

Observe that after substituting the limits of integration change from ๐‘Ž and ๐‘ to ๐‘”(๐‘Ž) and ๐‘”(๐‘).

We want to evaluate ๏„ธ๐‘ฅ(๐‘ฅโˆ’4)๐‘ฅ๏Šฉ๏Šง๏Šจd. Taking ๐‘ข=๐‘ฅโˆ’4, we have the following:

  • dd๐‘ข๐‘ฅ=1
  • ๐‘ฅ=๐‘ข+4
  • the limits of integration ๐‘ข(1)=1โˆ’4=โˆ’3 and ๐‘ข(3)=3โˆ’4=โˆ’1

Thus, ๏„ธ๐‘ฅ(๐‘ฅโˆ’4)๐‘ฅ=๏„ธ(๐‘ข+4)๐‘ข๐‘ข=๏„ธ๐‘ข+4๐‘ข๐‘ข=๏”14๐‘ข+43๐‘ข๏ =14(โˆ’1)+43(โˆ’1)โˆ’๏€ผ14(โˆ’3)+43(โˆ’3)๏ˆ=443.๏Šฉ๏Šง๏Šจ๏Šฑ๏Šง๏Šฑ๏Šฉ๏Šจ๏Šฑ๏Šง๏Šฑ๏Šฉ๏Šฉ๏Šจ๏Šช๏Šฉ๏Šฑ๏Šง๏Šฑ๏Šฉ๏Šช๏Šฉ๏Šช๏Šฉddd

If we are not told which substitution to use, we will have to choose a suitable substitution ourselves.

Example 2: Finding the Value of a Definite Integral Using Substitution

Find ๏„ธ๐‘ฅโˆš๐‘ฅ+5๐‘ฅ๏Šช๏Šฑ๏Šงd to the nearest thousandth.

Answer

Recall that if we have a function of the form ๐‘“(๐‘”(๐‘ฅ))๐‘”๐‘ฅdd, then we can make the substitution ๐‘ข=๐‘”(๐‘ฅ): ๏„ธ๐‘“(๐‘”(๐‘ฅ))๐‘”๐‘ฅ๐‘ฅ=๏„ธ๐‘“(๐‘ข)๐‘ข.๏Œป๏Œบ๏€(๏Œป)๏€(๏Œบ)dddd

Observe that after substituting the limits of integration change from ๐‘Ž and ๐‘ to ๐‘”(๐‘Ž) and ๐‘”(๐‘).

In making a substitution, we are looking for two things. The first thing is an expression, call it ๐‘”(๐‘ฅ), that appears in the integrand and whose derivative, dd๐‘”๐‘ฅ, also appears in the integrand. It is worth remembering that if ๐‘”(๐‘ฅ) is linear in ๐‘ฅ (i.e., ๐‘”(๐‘ฅ) is of the form ๐‘Ž๐‘ฅ+๐‘), then its derivative is a constant; dd๐‘”๐‘ฅ=๐‘Ž. This means that a linear substitution can always be made.

The second thing we are looking for is a substitution that will make the integral easier to evaluate. When the integrand contains a function of a function, the inner function is often a good candidate for substitution.

These two considerations suggest that ๐‘ข=๐‘ฅ+5 would be a good substitution to try. Indeed, taking ๐‘ข=๐‘ฅ+5, we have the following:

  • dd๐‘ข๐‘ฅ=1
  • ๐‘ฅ=๐‘ขโˆ’5
  • the limits of integration ๐‘ข(โˆ’1)=โˆ’1+5=4 and ๐‘ข(4)=4+5=9

Thus, ๏„ธ๐‘ฅโˆš๐‘ฅ+5๐‘ฅ=๏„ธ(๐‘ขโˆ’5)โˆš๐‘ข๐‘ข=๏„ธ(๐‘ขโˆ’5)๐‘ข๐‘ข=๏„ธ๐‘ขโˆ’5๐‘ข๐‘ข=๏•25๐‘ขโˆ’103๐‘ข๏ก=259โˆ’1039โˆ’๏€ฝ254โˆ’1034๏‰=25ร—243โˆ’103ร—27โˆ’25ร—32+103ร—8=4225โˆ’1903=31615.๏Šช๏Šฑ๏Šง๏Šฏ๏Šช๏Šฏ๏Šช๏Šฏ๏Šช๏Šฏ๏Šชdddd๏Ž ๏Žก๏Žข๏Žก๏Ž ๏Žก๏Žค๏Žก๏Žข๏Žก๏Žค๏Žก๏Žข๏Žก๏Žค๏Žก๏Žข๏Žก

Therefore, the value of our definite integral to the nearest thousandth is 21.067.

The method of substitution is often an effective way of tackling trigonometric integrals, as we will see in the next example.

Example 3: Finding the Value of a Definite Integral Using Substitution

Find ๏„ธโˆ’9(๐‘ง)(๐‘ง)๐‘ง๏‘ฝ๏Žฃ๏Šฆ๏Šจtansecd.

Answer

Recall that if we have a function of the form ๐‘“(๐‘”(๐‘ฅ))๐‘”๐‘ฅdd, then we can make the substitution ๐‘ข=๐‘”(๐‘ฅ): ๏„ธ๐‘“(๐‘”(๐‘ฅ))๐‘”๐‘ฅ๐‘ฅ=๏„ธ๐‘“(๐‘ข)๐‘ข.๏Œป๏Œบ๏€(๏Œป)๏€(๏Œบ)dddd

Observe that after substituting the limits of integration change from ๐‘Ž and ๐‘ to ๐‘”(๐‘Ž) and ๐‘”(๐‘).

In making a substitution, we are looking for an expression, call it ๐‘”(๐‘ฅ), that appears in the integrand and whose derivative, dd๐‘”๐‘ฅ, also appears in the integrand. Now, trigonometric functions differentiate to other trigonometric functions. For this reason, when presented with an integral involving multiple trigonometric functions, it is worth checking to see if any of them are the derivatives of other ones.

We want to evaluate ๏„ธโˆ’9(๐‘ง)(๐‘ง)๐‘ง๏‘ฝ๏Žฃ๏Šฆ๏Šจtansecd. Taking ๐‘ข=(๐‘ง)tan, we have the following:

  • ddsec๐‘ข๐‘ง=(๐‘ง)๏Šจ
  • the limits of integration ๐‘ข(0)=(0)=0tan and ๐‘ข๏€ป๐œ‹4๏‡=๏€ป๐œ‹4๏‡=1tan

Thus, ๏„ธโˆ’9(๐‘ง)(๐‘ง)๐‘ง=๏„ธโˆ’9๐‘ข๐‘ข=๏”โˆ’92๐‘ข๏ =โˆ’921โˆ’920=โˆ’92.๏‘ฝ๏Žฃ๏Šฆ๏Šจ๏Šง๏Šฆ๏Šจ๏Šง๏Šฆ๏Šจ๏Šจtansecdd

Recall that the derivative of ln(๐‘ฅ) is 1๐‘ฅ. This means that quotients with ln(๐‘ฅ) in the numerator are often good candidates for integration by substitution, as in the next example.

Example 4: Finding the Value of a Definite Integral Using Substitution

Determine ๏„ธ๐‘ฅ๐‘ฅ๐‘ฅ๏Œพ๏Šงlnd.

Answer

Recall that if we have a function of the form ๐‘“(๐‘”(๐‘ฅ))๐‘”๐‘ฅdd, then we can make the substitution ๐‘ข=๐‘”(๐‘ฅ): ๏„ธ๐‘“(๐‘”(๐‘ฅ))๐‘”๐‘ฅ๐‘ฅ=๏„ธ๐‘“(๐‘ข)๐‘ข.๏Œป๏Œบ๏€(๏Œป)๏€(๏Œบ)dddd

Observe that after substituting the limits of integration change from ๐‘Ž and ๐‘ to ๐‘”(๐‘Ž) and ๐‘”(๐‘).

Since our integrand in this case consists of a fraction with ln(๐‘ฅ) in the numerator, ๐‘ข=(๐‘ฅ)ln is an obvious candidate for substitution. Indeed, taking ๐‘ข=(๐‘ฅ)ln, we have the following:

  • dd๐‘ข๐‘ฅ=1๐‘ฅ
  • the limits of integration ๐‘ข(1)=(1)=0ln and ๐‘ข(๐‘’)=1

Thus, ๏„ธ๐‘ฅ๐‘ฅ๐‘ฅ=๏„ธ๐‘ข๐‘ข=๏”12๐‘ข๏ =12ร—1โˆ’12ร—0=12.๏Œพ๏Šง๏Šง๏Šฆ๏Šจ๏Šง๏Šฆ๏Šจ๏Šจlndd

Let us look at a more complicated example of using substitution to evaluate a definite integral involving the natural logarithm.

Example 5: Finding the Value of a Definite Integral Using Substitution

Evaluate ๏„ธ5๐‘ฅโˆ’7๏€บ๏€บ(๐‘ฅโˆ’7)๏†+1๏†๐‘ฅ๏Šญ๏Šฐ๏Œพ๏Šฎ๏Šจ๏Šฉlnd.

Answer

Recall that if we have a function of the form ๐‘“(๐‘”(๐‘ฅ))๐‘”๐‘ฅdd, then we can make the substitution ๐‘ข=๐‘”(๐‘ฅ): ๏„ธ๐‘“(๐‘”(๐‘ฅ))๐‘”๐‘ฅ๐‘ฅ=๏„ธ๐‘“(๐‘ข)๐‘ข.๏Œป๏Œบ๏€(๏Œป)๏€(๏Œบ)dddd

Observe that after substituting the limits of integration change from ๐‘Ž and ๐‘ to ๐‘”(๐‘Ž) and ๐‘”(๐‘).

Although the integrand in this example looks rather complicated, we should immediately notice that it contains a natural logarithm and a fraction whose denominator is closely related to the argument of that logarithm. By using the chain rule, we get ddln๐‘ฅ๏€บ(๐‘ฅโˆ’7)๏†=2(๐‘ฅโˆ’7)(๐‘ฅโˆ’7)=2๐‘ฅโˆ’7.๏Šจ๏Šจ

This suggests that ๐‘ข=๏€บ(๐‘ฅโˆ’7)๏†ln๏Šจ might be an extremely good candidate for substitution. However, observe that the constant term in ln๏€บ(๐‘ฅโˆ’7)๏†+1๏Šจ differentiates to 0, so we may as well take ๐‘ข=๏€บ(๐‘ฅโˆ’7)๏†+1ln๏Šจ. With ๐‘ข=๏€บ(๐‘ฅโˆ’7)๏†+1ln๏Šจ, we have the following:

  • dd๐‘ข๐‘ฅ=2๐‘ฅโˆ’7
  • the limits of integration ๐‘ข(8)=(1)+1=1ln and ๐‘ข(๐‘’+7)=๏€น๐‘’๏…+1=2+1=3ln๏Šจ

Thus, ๏„ธ5๐‘ฅโˆ’7๏€บ๏€บ(๐‘ฅโˆ’7)๏†+1๏†๐‘ฅ=๏„ธ52๐‘ข๐‘ข,=๏”58๐‘ข๏ =58ร—3โˆ’58ร—1=5ร—(81โˆ’1)8=4008=50.๏Šญ๏Šฐ๏Œพ๏Šฎ๏Šจ๏Šฉ๏Šฉ๏Šง๏Šฉ๏Šช๏Šฉ๏Šง๏Šช๏Šชlndd

Finally, let us look at an example combining an exponential function and trigonometric functions.

Example 6: Finding the Value of a Definite Integral Using Substitution

Evaluate ๏„ธ๐‘’๐œƒ๐œƒ๏Ž„๏Šจ๏ผ๏‘ฝ๏Žขsincosd.

Answer

Recall that if we have a function of the form ๐‘“(๐‘”(๐‘ฅ))๐‘”๐‘ฅdd, then we can make the substitution ๐‘ข=๐‘”(๐‘ฅ): ๏„ธ๐‘“(๐‘”(๐‘ฅ))๐‘”๐‘ฅ๐‘ฅ=๏„ธ๐‘“(๐‘ข)๐‘ข.๏Œป๏Œบ๏€(๏Œป)๏€(๏Œบ)dddd

Observe that after substituting the limits of integration change from ๐‘Ž and ๐‘ to ๐‘”(๐‘Ž) and ๐‘”(๐‘). Consider the expression ๏„ธ๐‘’๐œƒ๐œƒ.๏Ž„๏Šจ๏ผ๏‘ฝ๏Žขsincosd

Notice the presence of a sine function in the exponent and right next to it its derivative, a cosine. Indeed, with ๐‘ข=(๐œƒ)sin, we have ddcos๐‘ข๐œƒ=๐œƒ, and so ๏„ธ๐‘’๐œƒ๐œƒ=๏„ธ๐‘’๐‘ข.๏Ž„๏Šจ๏ผ๏‘(๏Ž„)๏‘()๏Šจ๏‘๏‘ฝ๏Žข๏‘ฝ๏Žขsincosdd

We have ๐‘ข๏€ป๐œ‹3๏‡=๏€ป๐œ‹3๏‡=โˆš32sin and ๐‘ข(๐œ‹)=(๐œ‹)=0sin. Therefore, ๏„ธ๐‘’๐œƒ๐œƒ=๏„ธ๐‘’๐‘ข=๏”12๐‘’๏ =12โˆ’12๐‘’=12๏€ป1โˆ’๐‘’๏‡.๏Ž„๏Šจ๏ผ๏‘(๏Ž„)๏‘()๏Šจ๏‘๏Šจ๏‘๏Šฆโˆš๏Šฉโˆš๏Šฉ๏‘ฝ๏Žข๏‘ฝ๏Žขโˆš๏Žข๏Žกsincosdd

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • We can identify situations where a substitution can be used to simplify a definite integral.
  • In particular, we can recognize expressions of the form ๐‘“(๐‘”(๐‘ฅ))๐‘”๐‘ฅ,dd which are the product of a function of a function ๐‘“(๐‘”(๐‘ฅ)) and the derivative dd๐‘”๐‘ฅ of the inner function.
  • In this situation, we can make the substitution ๐‘ข=๐‘”(๐‘ฅ); that is, ๏„ธ๐‘“(๐‘”(๐‘ฅ))๐‘”๐‘ฅ๐‘ฅ=๏„ธ๐‘“(๐‘ข)๐‘ข.๏Œป๏Œบ๏€(๏Œป)๏€(๏Œบ)dddd
  • We can evaluate such definite integrals, taking care that when we make a substitution ๐‘ข=๐‘”(๐‘ฅ), the limits of integration change: ๐‘Žโ†ฆ๐‘”(๐‘Ž)๐‘โ†ฆ๐‘”(๐‘).

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