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

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

Integration by substitution, also known as โ€œ๐‘ข-substitutionโ€ or โ€œchange of variablesโ€, is a method of finding unknown integrals by replacing one variable with another and changing the integrand into something that is known or can be easily integrated using other methods. After performing the integration, we usually change back to our original variable by reversing the substitution to give the final result in terms of that variable.

The ability to carry out integration by substitution is a skill that develops with practice and experience. This is why it is best to look at a lot of examples and practice as much as possible. Sometimes, an apparently sensible substitution does not lead to an integral that is easy to evaluate and you must be prepared to try an alternative change of variable.

We need to be able to write the integrand in a particular way as integration by substitution is usually applied to an integral when it takes the special form ๏„ธ๐‘“(๐‘”(๐‘ฅ))๐‘”โ€ฒ(๐‘ฅ)๐‘ฅ.d

We can use the chain rule and the fundamental theorem of calculus to derive a substitution rule for integrals of this type. We recall the first part of the fundamental theorem of calculus: if ๐‘“(๐‘ฅ) is a continuous real-valued function on some interval and ๐นโ€ฒ(๐‘ฅ)=๐‘“(๐‘ฅ) (i.e., ๐น is an antiderivative of ๐‘“), then we have the indefinite integral ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=๐น(๐‘ฅ)+,dC where C is known as the constant of integration. Also, recall the chain rule for derivatives of composite functions: if ๐‘“ and ๐‘” are differentiable functions, then the chain rule expresses the derivative of their composite ๐‘“(๐‘”(๐‘ฅ)) as dd๐‘ฅ(๐‘“(๐‘”(๐‘ฅ)))=๐‘“โ€ฒ(๐‘”(๐‘ฅ))๐‘”โ€ฒ(๐‘ฅ).

Now, suppose ๐น(๐‘ข) is an antiderivative of ๐‘“(๐‘ข) and ๐‘ข=๐‘”(๐‘ฅ) is a differentiable function. We can apply the chain rule to obtain dddd๐‘ฅ(๐น(๐‘ข))=๐‘ฅ(๐น(๐‘”(๐‘ฅ)))=๐นโ€ฒ(๐‘”(๐‘ฅ))๐‘”โ€ฒ(๐‘ฅ)=๐‘“(๐‘”(๐‘ฅ))๐‘”โ€ฒ(๐‘ฅ).

Therefore, from the fundamental theorem of calculus, ๏„ธ๐‘“(๐‘”(๐‘ฅ))๐‘”โ€ฒ(๐‘ฅ)๐‘ฅ=๐น(๐‘ข)+=๏„ธ๐‘“(๐‘ข)๐‘ข.dCd

This leads us to the following substitution rule, which is similar to the chain rule for differentiation but in reverse.

Definition: Substitution Rule

If ๐‘ข=๐‘”(๐‘ฅ) is a differentiable function whose range is an interval ๐ผโŠ†โ„ and ๐‘“(๐‘ฅ) is continuous on ๐ผ, then ๏„ธ๐‘“(๐‘”(๐‘ฅ))๐‘”โ€ฒ(๐‘ฅ)๐‘ฅ=๏„ธ๐‘“(๐‘ข)๐‘ข,dd where dd๐‘ข=๐‘”โ€ฒ(๐‘ฅ)๐‘ฅ.

The key to finding the right substitution, ๐‘ข=๐‘”(๐‘ฅ), is finding a piece of the integrand whose derivative is also in the integrand. It is usually chosen from the โ€œcomplicatedโ€ part of the integrand that we want to simplify, which can be written as a composite function ๐‘“(๐‘”(๐‘ฅ)) for some ๐‘“(๐‘ฅ) that is continuous in a subset of the range of ๐‘”(๐‘ฅ).

When applying this substitution rule for indefinite integrals, we substitute ๐‘ข=๐‘”(๐‘ฅ), integrate with respect to the variable ๐‘ข, and reverse the substitution in the resulting antiderivative to express the final result back in terms of the variable ๐‘ฅ.

We may also manipulate differentials with derivatives by treating them as a fraction, which is useful mathematical shorthand but may not be mathematically rigorous. The reason we can do this is the chain rule dddd๐‘ข=๏€ฝ๐‘ข๐‘ฅ๏‰๐‘ฅ or dddd๐‘ฅ=๏€ฝ๐‘ฅ๐‘ข๏‰๐‘ข.

However, for the purposes of this explainer and simplicity, we may treat the derivative as a fraction as it is relevant for problems with integration by substitution.

The best way to understand this method is to see it in action. Consider the indefinite integral ๏„ธ(๐‘ฅ+1)๐‘ฅ.๏Šซd

In order to evaluate this, we could expand out the integrand and use the power rule for integration, but here we will use the substitution rule. We can recognize that a part of the integrand contains a composite function: ๐‘“(๐‘”(๐‘ฅ))=(๐‘ฅ+1),๏Šซ with ๐‘“(๐‘ฅ)=๐‘ฅ๏Šซ and ๐‘”(๐‘ฅ)=๐‘ฅ+1. Since ๐‘“(๐‘ฅ) is a polynomial, it is continuous in the range of ๐‘”(๐‘ฅ) and we can use the substitution ๐‘ข=๐‘”(๐‘ฅ)=๐‘ฅ+1.

Taking the derivative, we have dd๐‘ข๐‘ฅ=1 or, equivalently by manipulating the differentials, dd๐‘ฅ=๐‘ข.

Thus, the integral can be written as ๏„ธ(๐‘ฅ+1)๐‘ฅ=๏„ธ๐‘ข๐‘ข=16๐‘ข+.๏Šซ๏Šซ๏ŠฌddC

We can now substitute back ๐‘ข=๐‘ฅ+1 to do a reverse substitution to get the final result in terms of ๐‘ฅ. Finally, we have ๏„ธ(๐‘ฅ+1)๐‘ฅ=16(๐‘ฅ+1)+.๏Šซ๏ŠฌdC

An integral that starts out difficult can sometimes become very easy with an appropriate substitution. Letโ€™s consider the following more complicated-looking integral: ๏„ธ๐‘ฅ๏€น๐‘ฅ+2๏…๐‘ฅ.๏Šจ๏Šฉ๏Šฎd

For this integral, we could expand ๏€น๐‘ฅ+2๏…๏Šฉ๏Šฎ using the binomial theorem then multiply the result by ๐‘ฅ๏Šจ and integrate using the power rule for integration. However, this can be tedious and prone to errors. A better way to find the integral would be by substitution.

Before we talk about the substitution, notice what happens when we differentiate ๏€น๐‘ฅ+2๏…๏Šฉ๏Šฏ using the chain rule (or the general power rule): dd๐‘ฅ๏€ป๏€น๐‘ฅ+2๏…๏‡=9๏€น3๐‘ฅ๏…๏€น๐‘ฅ+2๏…=27๐‘ฅ๏€น๐‘ฅ+2๏….๏Šฉ๏Šฏ๏Šจ๏Šฉ๏Šฎ๏Šจ๏Šฉ๏Šฎ

This is equal to the integrand up to a constant factor. In particular, we can divide by 27 to obtain dd๐‘ฅ๏€ผ127๏€น๐‘ฅ+2๏…๏ˆ=๐‘ฅ๏€น๐‘ฅ+2๏….๏Šฉ๏Šฏ๏Šจ๏Šฉ๏Šฎ

We can integrate both sides of the expression using the fundamental theorem of calculus to obtain ๏„ธ๐‘ฅ๏€น๐‘ฅ+2๏…๐‘ฅ=127๏€น๐‘ฅ+2๏…+.๏Šจ๏Šฉ๏Šฎ๏Šฉ๏ŠฏdC

This technique is known as recognition, but integration by substitution does this automatically without having to find the derivative and compare terms. We can recognize that a part of the integrand contains a composite function: ๐‘“(๐‘”(๐‘ฅ))=๏€น๐‘ฅ+2๏…,๏Šฉ๏Šฎ with ๐‘“(๐‘ฅ)=๐‘ฅ๏Šฎ and ๐‘”(๐‘ฅ)=๐‘ฅ+2๏Šฉ. Since ๐‘“(๐‘ฅ) is a polynomial, it is continuous in the range of ๐‘”(๐‘ฅ) and we can use the substitution ๐‘ข=๐‘”(๐‘ฅ)=๐‘ฅ+2.๏Šฉ

Taking the derivative of this with respect to ๐‘ฅ, we have dd๐‘ข๐‘ฅ=3๐‘ฅ๏Šจ or, equivalently after manipulating the differentials, dd๐‘ฅ=๐‘ข3๐‘ฅ.๏Šจ

We can now rewrite the integral in terms of the variable ๐‘ข using the substitution rule and integrate the result using the power rule for integration: ๏„ธ๐‘ฅ๏€น๐‘ฅ+2๏…๐‘ฅ=๏„ธ๐‘ข๐‘ฅ๏€ฝ๐‘ข3๐‘ฅ๏‰=13๏„ธ๐‘ข๐‘ข=13๏€ผ19๐‘ข๏ˆ+=127๐‘ข+.๏Šจ๏Šฉ๏Šฎ๏Šฎ๏Šจ๏Šจ๏Šฎ๏Šฏ๏ŠฏdddCC

Finally, we substitute ๐‘ข=๐‘ฅ+2๏Šฉ back to get the final result in terms of ๐‘ฅ, ๏„ธ๐‘ฅ๏€น๐‘ฅ+2๏…๐‘ฅ=127๏€น๐‘ฅ+2๏…+,๏Šจ๏Šฉ๏Šฎ๏Šฉ๏ŠฏdC which is the same result we get from recognition.

Most of the time when you make a substitution, you only need to replace one part of the integrand as the other part, involving ๐‘ฅ, would be cancelled by the expression for d๐‘ฅ. However, if this does not happen, then you may need to replace any ๐‘ฅ by making it the subject in the substitution. To see this in action, consider the integral ๏„ธ๐‘ฅ๏€น๐‘ฅ+1๏…๐‘ฅ.๏Šฉ๏Šจ๏Šซd

We can recognize that a part of the integrand contains a composite function: ๐‘“(๐‘”(๐‘ฅ))=๏€น๐‘ฅ+1๏…,๏Šจ๏Šซ with ๐‘“(๐‘ฅ)=๐‘ฅ๏Šซ and ๐‘”(๐‘ฅ)=๐‘ฅ+1๏Šจ. Since ๐‘“(๐‘ฅ) is a polynomial, it is continuous in the range of ๐‘”(๐‘ฅ) and we can use the substitution ๐‘ข=๐‘”(๐‘ฅ)=๐‘ฅ+1.๏Šจ

Taking the derivative, we have dd๐‘ข๐‘ฅ=2๐‘ฅ or, equivalently by manipulating the differentials, dd๐‘ฅ=๐‘ข2๐‘ฅ.

We can now rewrite the integral in terms of the variable ๐‘ข using the substitution rule: ๏„ธ๐‘ฅ๏€น๐‘ฅ+1๏…๐‘ฅ=๏„ธ๐‘ฅ๐‘ข๏€ฝ๐‘ข2๐‘ฅ๏‰=12๏„ธ๐‘ฅ๐‘ข๐‘ข.๏Šฉ๏Šจ๏Šซ๏Šฉ๏Šซ๏Šจ๏Šซddd

Notice that we still have an ๐‘ฅ๏Šจ appearing in the integrand that did not get cancelled, so we have to replace this by ๐‘ข by making ๐‘ฅ๏Šจ the subject in the substitution: ๐‘ฅ=๐‘ขโˆ’1.๏Šจ

Therefore, we can replace everything in terms of ๐‘ข and integrate the result using the power rule for integration: ๏„ธ๐‘ฅ๏€น๐‘ฅ+1๏…๐‘ฅ=12๏„ธ๐‘ฅ๐‘ข๐‘ข=12๏„ธ(๐‘ขโˆ’1)๐‘ข๐‘ข=12๏„ธ๏€น๐‘ขโˆ’๐‘ข๏…๐‘ข=114๐‘ขโˆ’112๐‘ข+.๏Šฉ๏Šจ๏Šซ๏Šจ๏Šซ๏Šซ๏Šฌ๏Šซ๏Šญ๏ŠฌddddC

Finally, we can make the reverse substitution ๐‘ข=๐‘ฅ+1๏Šจ to rewrite the result in terms of ๐‘ฅ: ๏„ธ๐‘ฅ๏€น๐‘ฅ+1๏…๐‘ฅ=114๏€น๐‘ฅ+1๏…โˆ’112๏€น๐‘ฅ+1๏…+.๏Šฉ๏Šจ๏Šซ๏Šจ๏Šญ๏Šจ๏ŠฌdC

Now, letโ€™s look at a few examples in order to practice and deepen our understanding. The first two examples are about finding the indefinite integral of a function involving polynomials in ๐‘ฅ.

Example 1: Finding the Integration of a Function Using Integration by Substitution

Determine ๏„ธ๐‘ฅ๏€น๐‘ฅ+9๏…๐‘ฅ๏Šซ๏Šฌ๏Šญd.

Answer

In this example, we want to find the indefinite integral of a polynomial function using integration by substitution.

We first note that a part of the integrand contains a composite function: ๐‘“(๐‘”(๐‘ฅ))=๏€น๐‘ฅ+9๏…,๏Šฌ๏Šญ with ๐‘“(๐‘ฅ)=๐‘ฅ๏Šญ and ๐‘”(๐‘ฅ)=๐‘ฅ+9๏Šฌ. Since ๐‘“(๐‘ฅ) is a polynomial, it is continuous in the range of ๐‘”(๐‘ฅ) and we can use the substitution ๐‘ข=๐‘”(๐‘ฅ)=๐‘ฅ+9.๏Šฌ

The derivative of this with respect to ๐‘ฅ is dd๐‘ข๐‘ฅ=6๐‘ฅ,๏Šซ or, equivalently by manipulating the differentials, dd๐‘ฅ=๐‘ข6๐‘ฅ.๏Šซ

Next, we apply this substitution to the integral to change the variable from ๐‘ฅ to ๐‘ข and integrate the resulting expression using the power rule for integration: ๏„ธ๐‘ฅ๏€น๐‘ฅ+9๏…๐‘ฅ=๏„ธ๐‘ฅ๐‘ข๏€ฝ๐‘ข6๐‘ฅ๏‰=16๏„ธ๐‘ข๐‘ข=16๏€พ๐‘ข8๏Š+=148๐‘ข+.๏Šซ๏Šฌ๏Šญ๏Šซ๏Šญ๏Šซ๏Šญ๏Šฎ๏ŠฎdddCC

Finally, we apply the reverse substitution ๐‘ข=๐‘ฅ+9๏Šฌ to get the final result in terms of ๐‘ฅ: ๏„ธ๐‘ฅ๏€น๐‘ฅ+9๏…๐‘ฅ=148๏€น๐‘ฅ+9๏…+.๏Šซ๏Šฌ๏Šญ๏Šฌ๏ŠฎdC

Example 2: Finding the Integration of a Function Using Integration by Substitution

Determine ๏„ธ8๐‘ฅ(8๐‘ฅ+9)๐‘ฅ๏Šจd by using the substitution method.

Answer

In this example, we want to find the indefinite integral of a polynomial function using integration by substitution.

We first note that a part of the integrand contains a composite function: ๐‘“(๐‘”(๐‘ฅ))=(8๐‘ฅ+9),๏Šจ with ๐‘“(๐‘ฅ)=๐‘ฅ๏Šจ and ๐‘”(๐‘ฅ)=8๐‘ฅ+9. Since ๐‘“(๐‘ฅ) is a polynomial, it is continuous in the range of ๐‘”(๐‘ฅ) and we can use the substitution ๐‘ข=๐‘”(๐‘ฅ)=8๐‘ฅ+9.

The derivative of this with respect to ๐‘ฅ is dd๐‘ข๐‘ฅ=8 or, equivalently by manipulating the differentials, dd๐‘ฅ=๐‘ข8.

Next, we apply this substitution to the integral to change the variable from ๐‘ฅ to ๐‘ข: ๏„ธ8๐‘ฅ(8๐‘ฅ+9)๐‘ฅ=๏„ธ8๐‘ฅ๐‘ข๏€ฝ๐‘ข8๏‰=๏„ธ๐‘ฅ๐‘ข๐‘ข.๏Šจ๏Šจ๏Šจddd

Notice that we still have an ๐‘ฅ in the integrand that did not get cancelled, so we have to replace ๐‘ฅ with ๐‘ข by making ๐‘ฅ the subject in the substitution: ๐‘ฅ=18(๐‘ขโˆ’9).

We can use this to eliminate any ๐‘ฅ that appears in the integrand and then integrate the resulting expression using the power rule for integration: ๏„ธ8๐‘ฅ(8๐‘ฅ+9)๐‘ฅ=๏„ธ๐‘ฅ๐‘ข๐‘ข=๏„ธ๏€ผ18(๐‘ขโˆ’9)๏ˆ๐‘ข๐‘ข=18๏„ธ๏€น๐‘ขโˆ’9๐‘ข๏…๐‘ข=18๏€พ๐‘ข4โˆ’9๐‘ข3๏Š+=132๐‘ขโˆ’38๐‘ข+.๏Šจ๏Šจ๏Šจ๏Šฉ๏Šจ๏Šช๏Šฉ๏Šช๏ŠฉddddCC

Finally, we apply the reverse substitution ๐‘ข=8๐‘ฅ+9 to get the final result in terms of ๐‘ฅ: ๏„ธ8๐‘ฅ(8๐‘ฅ+9)๐‘ฅ=132(8๐‘ฅ+9)โˆ’38(8๐‘ฅ+9)+.๏Šจ๏Šช๏ŠฉdC

The next example involves finding the indefinite integral of a root function.

Example 3: Finding the Integration of a Function Involving a Root Function Using Integration by Substitution

Determine ๏„ธ48โˆ’6๐‘ฅโˆš16โˆ’2๐‘ฅ๐‘ฅ๏Žคd.

Answer

In this example, we want to find the indefinite integral of a function involving a root using integration by substitution.

We first note that a part of the integrand contains a composite function: ๐‘“(๐‘”(๐‘ฅ))=1โˆš16โˆ’2๐‘ฅ,๏Žค with ๐‘“(๐‘ฅ)=1โˆš๐‘ฅ๏Žค and ๐‘”(๐‘ฅ)=16โˆ’2๐‘ฅ. Since ๐‘“(๐‘ฅ) is continuous in a subset of the range of ๐‘”(๐‘ฅ) (excluding ๐‘ฅ=0 for ๐‘“), we can use the substitution ๐‘ข=๐‘”(๐‘ฅ)=16โˆ’2๐‘ฅ.

The derivative of this with respect to ๐‘ฅ is dd๐‘ข๐‘ฅ=โˆ’2 or, equivalently by manipulating the differentials, dd๐‘ฅ=โˆ’๐‘ข2.

Next, we apply this substitution to the integral to change the variable from ๐‘ฅ to ๐‘ข, noting that 3๐‘ข=48โˆ’6๐‘ฅ appears in the integrand. We then integrate the resulting expression using the power rule for integration: ๏„ธ48โˆ’6๐‘ฅโˆš16โˆ’2๐‘ฅ๐‘ฅ=๏„ธ3๐‘ขโˆš๐‘ข๏€ฝโˆ’๐‘ข2๏‰=โˆ’32๏„ธ๐‘ข๐‘ข=โˆ’32๏5๐‘ข9๏+=โˆ’56๐‘ข+.๏Žค๏Žค๏Žฃ๏Žค๏Žจ๏Žค๏Žจ๏ŽคdddCC

Finally, we apply the reverse substitution ๐‘ข=16โˆ’2๐‘ฅ to get the final result in terms of ๐‘ฅ: ๏„ธ48โˆ’6๐‘ฅโˆš16โˆ’2๐‘ฅ๐‘ฅ=โˆ’56(16โˆ’2๐‘ฅ)+.๏Žค๏Žจ๏ŽคdC

The next example involves finding the indefinite integral of trigonometric functions.

Example 4: Finding the Integration of a Function Involving Trigonometric Functions Using Integration by Substitution

Determine ๏„ธ๏€ป๏€นโˆ’24๐‘ฅ+306๐‘ฅ๏…๏€นโˆ’6๐‘ฅโˆ’56๐‘ฅ๏…๏‡๐‘ฅ๏Šฉ๏Šช๏Šซsincosd.

Answer

In this example, we want to find the indefinite integral of a trigonometric function using integration by substitution.

We first note that a part of the integrand contains a composite function: ๐‘“(๐‘”(๐‘ฅ))=๏€นโˆ’6๐‘ฅโˆ’56๐‘ฅ๏…,๏Šช๏Šซcos with ๐‘“(๐‘ฅ)=๐‘ฅ๏Šซ and ๐‘”(๐‘ฅ)=โˆ’6๐‘ฅโˆ’56๐‘ฅ๏Šชcos. Since ๐‘“(๐‘ฅ) is a polynomial, it is continuous in the range of ๐‘”(๐‘ฅ) and we can use the substitution ๐‘ข=๐‘”(๐‘ฅ)=โˆ’6๐‘ฅโˆ’56๐‘ฅ.๏Šชcos

The derivative of this with respect to ๐‘ฅ is ddsin๐‘ข๐‘ฅ=โˆ’24๐‘ฅ+306๐‘ฅ๏Šฉ or, equivalently by manipulating the differentials, ddsin๐‘ฅ=๐‘ข(โˆ’24๐‘ฅ+306๐‘ฅ).๏Šฉ

Now, we apply this substitution to the integral to change the variable from ๐‘ฅ to ๐‘ข and integrate the resulting expression using the power rule for integration: ๏„ธ๏€ป๏€นโˆ’24๐‘ฅ+306๐‘ฅ๏…๏€นโˆ’6๐‘ฅโˆ’56๐‘ฅ๏…๏‡๐‘ฅ=๏„ธ๏€นโˆ’24๐‘ฅ+306๐‘ฅ๏…๐‘ข๏€ฝ๐‘ข(โˆ’24๐‘ฅ+306๐‘ฅ)๏‰=๏„ธ๐‘ข๐‘ข=16๐‘ข+.๏Šฉ๏Šช๏Šซ๏Šฉ๏Šซ๏Šฉ๏Šซ๏ŠฌsincosdsindsindC

Finally, we apply the reverse substitution ๐‘ข=โˆ’6๐‘ฅโˆ’56๐‘ฅ๏Šชcos to get the final result in terms of ๐‘ฅ: ๏„ธ๏€ป๏€นโˆ’24๐‘ฅ+306๐‘ฅ๏…๏€นโˆ’6๐‘ฅโˆ’56๐‘ฅ๏…๏‡๐‘ฅ=16๏€นโˆ’6๐‘ฅโˆ’56๐‘ฅ๏…+.๏Šฉ๏Šช๏Šซ๏Šช๏ŠฌsincosdcosC

The next example involves finding the indefinite integral of a logarithmic function.

Example 5: Finding the Integration of a Function Involving a Logarithmic Function Using Integration by Substitution

Determine ๏„ธโˆ’116๐‘ฅโˆš๐‘ฅ๐‘ฅ๏Žขlnd.

Answer

In this example, we want to find the indefinite integral of a function involving logarithms using integration by substitution.

We first note that a part of the integrand contains a composite function: ๐‘“(๐‘”(๐‘ฅ))=โˆš๐‘ฅ,๏Žขln with ๐‘“(๐‘ฅ)=โˆš๐‘ฅ๏Žข and ๐‘”(๐‘ฅ)=๐‘ฅln. Since ๐‘“(๐‘ฅ) is continuous in the range of ๐‘”(๐‘ฅ), we can use the substitution ๐‘ข=๐‘”(๐‘ฅ)=๐‘ฅ.ln

The derivative of this with respect to ๐‘ฅ is dd๐‘ข๐‘ฅ=1๐‘ฅ or, equivalently by manipulating the differentials, dd๐‘ฅ=๐‘ฅ๐‘ข.

Now, we apply this substitution to the integral to change the variable from ๐‘ฅ to ๐‘ข and integrate the resulting expression using the power rule for integration: ๏„ธโˆ’116๐‘ฅโˆš๐‘ฅ๐‘ฅ=๏„ธโˆ’116๐‘ฅ๐‘ข(๐‘ฅ๐‘ข)=โˆ’116๏„ธ๐‘ข๐‘ข=โˆ’116๏3๐‘ข4๏+=โˆ’118๐‘ข+.๏Žข๏Ž ๏Žข๏Ž ๏Žข๏Žฃ๏Žข๏Žฃ๏ŽขlndddCC

Finally, we apply the reverse substitution ๐‘ข=๐‘ฅln to get the final result in terms of ๐‘ฅ: ๏„ธโˆ’116๐‘ฅโˆš๐‘ฅ๐‘ฅ=โˆ’118(๐‘ฅ)+.๏Žข๏Žฃ๏ŽขlndlnC

The next example involves finding the indefinite integral of reciprocal trigonometric functions.

Example 6: Integrating Reciprocal Trigonometric Functions

Determine ๏„ธ47๏€ผ7๐‘ฅ5+1๏ˆ๐‘ฅsecd๏Šจ.

Answer

In this example, we want to find the indefinite integral of a reciprocal trigonometric a function using integration by substitution.

We first note that a part of the integrand contains a composite function: ๐‘“(๐‘”(๐‘ฅ))=๏€ผ7๐‘ฅ5+1๏ˆ,sec๏Šจ with ๐‘“(๐‘ฅ)=๐‘ฅsec๏Šจ and ๐‘”(๐‘ฅ)=7๐‘ฅ5+1. Since ๐‘“(๐‘ฅ) is continuous in a subset of the range of ๐‘”(๐‘ฅ), we can use the substitution ๐‘ข=๐‘”(๐‘ฅ)=7๐‘ฅ5+1.

The derivative of this with respect to ๐‘ฅ is dd๐‘ข๐‘ฅ=75 or, equivalently by manipulating the differentials, dd๐‘ฅ=57๐‘ข.

Now, we apply this substitution to the integral to change the variable from ๐‘ฅ to ๐‘ข and integrate the resulting expression using the power rule for integration: ๏„ธ47๏€ผ7๐‘ฅ5+1๏ˆ๐‘ฅ=๏„ธ47๐‘ข๏€ผ57๐‘ข๏ˆ=2049๏„ธ๐‘ข๐‘ข=2049๐‘ข+.secdsecdsecdtanC๏Šจ๏Šจ๏Šจ

Finally, we apply the reverse substitution ๐‘ข=7๐‘ฅ5+1 to get the final result in terms of ๐‘ฅ: ๏„ธ47๏€ผ7๐‘ฅ5+1๏ˆ๐‘ฅ=2049๏€ผ7๐‘ฅ5+1๏ˆ+.secdtanC๏Šจ

The last example is about finding the indefinite integral involving both exponential and trigonometric functions.

Example 7: Finding the Integration of a Function Involving Trigonometric and Exponential Functions Using Integration by Substitution

Determine ๏„ธ(7โˆ’99๐‘ฅ)๐‘’๐‘ฅsind(๏Šญ๏—๏Šฐ๏Šฏ๏—)cos.

Answer

In this example, we want to find the indefinite integral of an integrand involving an exponential and trigonometric function using integration by substitution.

We first note that a part of the integrand contains a composite function: ๐‘“(๐‘”(๐‘ฅ))=๐‘’,(๏Šญ๏—๏Šฐ๏Šฏ๏—)cos with ๐‘“(๐‘ฅ)=๐‘’๏— and ๐‘”(๐‘ฅ)=7๐‘ฅ+9๐‘ฅcos. Since ๐‘“(๐‘ฅ) is an exponential function, it is continuous in the range of ๐‘”(๐‘ฅ) and we can use the substitution ๐‘ข=๐‘”(๐‘ฅ)=7๐‘ฅ+9๐‘ฅ.cos

The derivative of this with respect to ๐‘ฅ is ddsin๐‘ข๐‘ฅ=7โˆ’99๐‘ฅ or, equivalently by manipulating the differentials, ddsin๐‘ฅ=๐‘ข(7โˆ’99๐‘ฅ).

Now, we apply this substitution to the integral to change the variable from ๐‘ฅ to ๐‘ข and integrate the resulting expression using the power rule for integration: ๏„ธ(7โˆ’99๐‘ฅ)๐‘’๐‘ฅ=๏„ธ(7โˆ’99๐‘ฅ)๐‘’๏€ฝ๐‘ข(7โˆ’99๐‘ฅ)๏‰=๏„ธ๐‘’๐‘ข=๐‘’+.sindsindsindC(๏Šญ๏—๏Šฐ๏Šฏ๏—)๏‘๏‘๏‘cos

Finally, we apply the reverse substitution ๐‘ข=7๐‘ฅ+9๐‘ฅcos to get the final result in terms of ๐‘ฅ: ๏„ธ(7โˆ’99๐‘ฅ)๐‘’๐‘ฅ=๐‘’+.sindC(๏Šญ๏—๏Šฐ๏Šฏ๏—)(๏Šญ๏—๏Šฐ๏Šฏ๏—)coscos

Key Points

  • Integration by substitution can be used to find the indefinite integral of complicated functions involving roots, trigonometric functions, logarithmic functions, and many more.
  • The substitution rule we use is similar to the chain rule for differentiation, but in reverse: ๏„ธ๐‘“(๐‘”(๐‘ฅ))๐‘”โ€ฒ(๐‘ฅ)๐‘ฅ=๏„ธ๐‘“(๐‘ข)๐‘ข.dd
  • To choose our substitution, ๐‘ข, we look for a factor of the integrand whose derivative also appears in the integrand or the inner function of the โ€œcomplicatedโ€ part ot the integrand.
    In particular, if part of the integrand contains a derivative ๐‘”โ€ฒ(๐‘ฅ) and/or a composite function of the form ๐‘“(๐‘”(๐‘ฅ)), we use the substitution ๐‘ข=๐‘”(๐‘ฅ).

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