Lesson Explainer: Integration by Substitution: Indefinite Integrals | Nagwa Lesson Explainer: Integration by Substitution: Indefinite Integrals | Nagwa

Lesson Explainer: Integration by Substitution: Indefinite Integrals Mathematics • Third Year of Secondary School

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

Integration by substitution, also known as “𝑢-substitution” or “change of variables”, 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 𝑓(𝑔(𝑥))𝑔(𝑥)𝑥.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 𝑓(𝑥) is a continuous real-valued function on some interval and 𝐹(𝑥)=𝑓(𝑥) (i.e., 𝐹 is an antiderivative of 𝑓), then we have the indefinite integral 𝑓(𝑥)𝑥=𝐹(𝑥)+,dC where C is known as the constant of integration. Also, recall the chain rule for derivatives of composite functions: if 𝑓 and 𝑔 are differentiable functions, then the chain rule expresses the derivative of their composite 𝑓(𝑔(𝑥)) as dd𝑥(𝑓(𝑔(𝑥)))=𝑓(𝑔(𝑥))𝑔(𝑥).

Now, suppose 𝐹(𝑢) is an antiderivative of 𝑓(𝑢) and 𝑢=𝑔(𝑥) is a differentiable function. We can apply the chain rule to obtain dddd𝑥(𝐹(𝑢))=𝑥(𝐹(𝑔(𝑥)))=𝐹(𝑔(𝑥))𝑔(𝑥)=𝑓(𝑔(𝑥))𝑔(𝑥).

Therefore, from the fundamental theorem of calculus, 𝑓(𝑔(𝑥))𝑔(𝑥)𝑥=𝐹(𝑢)+=𝑓(𝑢)𝑢.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 𝑢=𝑔(𝑥) is a differentiable function whose range is an interval 𝐼 and 𝑓(𝑥) is continuous on 𝐼, then 𝑓(𝑔(𝑥))𝑔(𝑥)𝑥=𝑓(𝑢)𝑢,dd where dd𝑢=𝑔(𝑥)𝑥.

The key to finding the right substitution, 𝑢=𝑔(𝑥), is finding a piece of the integrand whose derivative is also in the integrand. It is usually chosen from the “complicated” part of the integrand that we want to simplify, which can be written as a composite function 𝑓(𝑔(𝑥)) for some 𝑓(𝑥) that is continuous in a subset of the range of 𝑔(𝑥).

When applying this substitution rule for indefinite integrals, we substitute 𝑢=𝑔(𝑥), integrate with respect to the variable 𝑢, and reverse the substitution in the resulting antiderivative to express the final result back in terms of the variable 𝑥.

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 dddd𝑢=𝑢𝑥𝑥 or dddd𝑥=𝑥𝑢𝑢.

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 (𝑥+1)𝑥.d

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: 𝑓(𝑔(𝑥))=(𝑥+1), with 𝑓(𝑥)=𝑥 and 𝑔(𝑥)=𝑥+1. Since 𝑓(𝑥) is a polynomial, it is continuous in the range of 𝑔(𝑥) and we can use the substitution 𝑢=𝑔(𝑥)=𝑥+1.

Taking the derivative, we have dd𝑢𝑥=1 or, equivalently by manipulating the differentials, dd𝑥=𝑢.

Thus, the integral can be written as (𝑥+1)𝑥=𝑢𝑢=16𝑢+.ddC

We can now substitute back 𝑢=𝑥+1 to do a reverse substitution to get the final result in terms of 𝑥. Finally, we have (𝑥+1)𝑥=16(𝑥+1)+.dC

An integral that starts out difficult can sometimes become very easy with an appropriate substitution. Let’s consider the following more complicated-looking integral: 𝑥𝑥+2𝑥.d

For this integral, we could expand 𝑥+2 using the binomial theorem then multiply the result by 𝑥 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 𝑥+2 using the chain rule (or the general power rule): dd𝑥𝑥+2=93𝑥𝑥+2=27𝑥𝑥+2.

This is equal to the integrand up to a constant factor. In particular, we can divide by 27 to obtain dd𝑥127𝑥+2=𝑥𝑥+2.

We can integrate both sides of the expression using the fundamental theorem of calculus to obtain 𝑥𝑥+2𝑥=127𝑥+2+.dC

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: 𝑓(𝑔(𝑥))=𝑥+2, with 𝑓(𝑥)=𝑥 and 𝑔(𝑥)=𝑥+2. Since 𝑓(𝑥) is a polynomial, it is continuous in the range of 𝑔(𝑥) and we can use the substitution 𝑢=𝑔(𝑥)=𝑥+2.

Taking the derivative of this with respect to 𝑥, we have dd𝑢𝑥=3𝑥 or, equivalently after manipulating the differentials, dd𝑥=𝑢3𝑥.

We can now rewrite the integral in terms of the variable 𝑢 using the substitution rule and integrate the result using the power rule for integration: 𝑥𝑥+2𝑥=𝑢𝑥𝑢3𝑥=13𝑢𝑢=1319𝑢+=127𝑢+.dddCC

Finally, we substitute 𝑢=𝑥+2 back to get the final result in terms of 𝑥, 𝑥𝑥+2𝑥=127𝑥+2+,dC 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 𝑥, would be cancelled by the expression for d𝑥. However, if this does not happen, then you may need to replace any 𝑥 by making it the subject in the substitution. To see this in action, consider the integral 𝑥𝑥+1𝑥.d

We can recognize that a part of the integrand contains a composite function: 𝑓(𝑔(𝑥))=𝑥+1, with 𝑓(𝑥)=𝑥 and 𝑔(𝑥)=𝑥+1. Since 𝑓(𝑥) is a polynomial, it is continuous in the range of 𝑔(𝑥) and we can use the substitution 𝑢=𝑔(𝑥)=𝑥+1.

Taking the derivative, we have dd𝑢𝑥=2𝑥 or, equivalently by manipulating the differentials, dd𝑥=𝑢2𝑥.

We can now rewrite the integral in terms of the variable 𝑢 using the substitution rule: 𝑥𝑥+1𝑥=𝑥𝑢𝑢2𝑥=12𝑥𝑢𝑢.ddd

Notice that we still have an 𝑥 appearing in the integrand that did not get cancelled, so we have to replace this by 𝑢 by making 𝑥 the subject in the substitution: 𝑥=𝑢1.

Therefore, we can replace everything in terms of 𝑢 and integrate the result using the power rule for integration: 𝑥𝑥+1𝑥=12𝑥𝑢𝑢=12(𝑢1)𝑢𝑢=12𝑢𝑢𝑢=114𝑢112𝑢+.ddddC

Finally, we can make the reverse substitution 𝑢=𝑥+1 to rewrite the result in terms of 𝑥: 𝑥𝑥+1𝑥=114𝑥+1112𝑥+1+.dC

Now, let’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 𝑥.

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

Determine 𝑥𝑥+9𝑥d.

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: 𝑓(𝑔(𝑥))=𝑥+9, with 𝑓(𝑥)=𝑥 and 𝑔(𝑥)=𝑥+9. Since 𝑓(𝑥) is a polynomial, it is continuous in the range of 𝑔(𝑥) and we can use the substitution 𝑢=𝑔(𝑥)=𝑥+9.

The derivative of this with respect to 𝑥 is dd𝑢𝑥=6𝑥, or, equivalently by manipulating the differentials, dd𝑥=𝑢6𝑥.

Next, we apply this substitution to the integral to change the variable from 𝑥 to 𝑢 and integrate the resulting expression using the power rule for integration: 𝑥𝑥+9𝑥=𝑥𝑢𝑢6𝑥=16𝑢𝑢=16𝑢8+=148𝑢+.dddCC

Finally, we apply the reverse substitution 𝑢=𝑥+9 to get the final result in terms of 𝑥: 𝑥𝑥+9𝑥=148𝑥+9+.dC

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

Determine 8𝑥(8𝑥+9)𝑥d 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: 𝑓(𝑔(𝑥))=(8𝑥+9), with 𝑓(𝑥)=𝑥 and 𝑔(𝑥)=8𝑥+9. Since 𝑓(𝑥) is a polynomial, it is continuous in the range of 𝑔(𝑥) and we can use the substitution 𝑢=𝑔(𝑥)=8𝑥+9.

The derivative of this with respect to 𝑥 is dd𝑢𝑥=8 or, equivalently by manipulating the differentials, dd𝑥=𝑢8.

Next, we apply this substitution to the integral to change the variable from 𝑥 to 𝑢: 8𝑥(8𝑥+9)𝑥=8𝑥𝑢𝑢8=𝑥𝑢𝑢.ddd

Notice that we still have an 𝑥 in the integrand that did not get cancelled, so we have to replace 𝑥 with 𝑢 by making 𝑥 the subject in the substitution: 𝑥=18(𝑢9).

We can use this to eliminate any 𝑥 that appears in the integrand and then integrate the resulting expression using the power rule for integration: 8𝑥(8𝑥+9)𝑥=𝑥𝑢𝑢=18(𝑢9)𝑢𝑢=18𝑢9𝑢𝑢=18𝑢49𝑢3+=132𝑢38𝑢+.ddddCC

Finally, we apply the reverse substitution 𝑢=8𝑥+9 to get the final result in terms of 𝑥: 8𝑥(8𝑥+9)𝑥=132(8𝑥+9)38(8𝑥+9)+.dC

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 486𝑥162𝑥𝑥d.

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: 𝑓(𝑔(𝑥))=1162𝑥, with 𝑓(𝑥)=1𝑥 and 𝑔(𝑥)=162𝑥. Since 𝑓(𝑥) is continuous in a subset of the range of 𝑔(𝑥) (excluding 𝑥=0 for 𝑓), we can use the substitution 𝑢=𝑔(𝑥)=162𝑥.

The derivative of this with respect to 𝑥 is dd𝑢𝑥=2 or, equivalently by manipulating the differentials, dd𝑥=𝑢2.

Next, we apply this substitution to the integral to change the variable from 𝑥 to 𝑢, noting that 3𝑢=486𝑥 appears in the integrand. We then integrate the resulting expression using the power rule for integration: 486𝑥162𝑥𝑥=3𝑢𝑢𝑢2=32𝑢𝑢=325𝑢9+=56𝑢+.dddCC

Finally, we apply the reverse substitution 𝑢=162𝑥 to get the final result in terms of 𝑥: 486𝑥162𝑥𝑥=56(162𝑥)+.dC

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 24𝑥+306𝑥6𝑥56𝑥𝑥sincosd.

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: 𝑓(𝑔(𝑥))=6𝑥56𝑥,cos with 𝑓(𝑥)=𝑥 and 𝑔(𝑥)=6𝑥56𝑥cos. Since 𝑓(𝑥) is a polynomial, it is continuous in the range of 𝑔(𝑥) and we can use the substitution 𝑢=𝑔(𝑥)=6𝑥56𝑥.cos

The derivative of this with respect to 𝑥 is ddsin𝑢𝑥=24𝑥+306𝑥 or, equivalently by manipulating the differentials, ddsin𝑥=𝑢(24𝑥+306𝑥).

Now, we apply this substitution to the integral to change the variable from 𝑥 to 𝑢 and integrate the resulting expression using the power rule for integration: 24𝑥+306𝑥6𝑥56𝑥𝑥=24𝑥+306𝑥𝑢𝑢(24𝑥+306𝑥)=𝑢𝑢=16𝑢+.sincosdsindsindC

Finally, we apply the reverse substitution 𝑢=6𝑥56𝑥cos to get the final result in terms of 𝑥: 24𝑥+306𝑥6𝑥56𝑥𝑥=166𝑥56𝑥+.sincosdcosC

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 116𝑥𝑥𝑥lnd.

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: 𝑓(𝑔(𝑥))=𝑥,ln with 𝑓(𝑥)=𝑥 and 𝑔(𝑥)=𝑥ln. Since 𝑓(𝑥) is continuous in the range of 𝑔(𝑥), we can use the substitution 𝑢=𝑔(𝑥)=𝑥.ln

The derivative of this with respect to 𝑥 is dd𝑢𝑥=1𝑥 or, equivalently by manipulating the differentials, dd𝑥=𝑥𝑢.

Now, we apply this substitution to the integral to change the variable from 𝑥 to 𝑢 and integrate the resulting expression using the power rule for integration: 116𝑥𝑥𝑥=116𝑥𝑢(𝑥𝑢)=116𝑢𝑢=1163𝑢4+=118𝑢+.lndddCC

Finally, we apply the reverse substitution 𝑢=𝑥ln to get the final result in terms of 𝑥: 116𝑥𝑥𝑥=118(𝑥)+.lndlnC

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

Example 6: Integrating Reciprocal Trigonometric Functions

Determine 477𝑥5+1𝑥secd.

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: 𝑓(𝑔(𝑥))=7𝑥5+1,sec with 𝑓(𝑥)=𝑥sec and 𝑔(𝑥)=7𝑥5+1. Since 𝑓(𝑥) is continuous in a subset of the range of 𝑔(𝑥), we can use the substitution 𝑢=𝑔(𝑥)=7𝑥5+1.

The derivative of this with respect to 𝑥 is dd𝑢𝑥=75 or, equivalently by manipulating the differentials, dd𝑥=57𝑢.

Now, we apply this substitution to the integral to change the variable from 𝑥 to 𝑢 and integrate the resulting expression using the power rule for integration: 477𝑥5+1𝑥=47𝑢57𝑢=2049𝑢𝑢=2049𝑢+.secdsecdsecdtanC

Finally, we apply the reverse substitution 𝑢=7𝑥5+1 to get the final result in terms of 𝑥: 477𝑥5+1𝑥=20497𝑥5+1+.secdtanC

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 (799𝑥)𝑒𝑥sind()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: 𝑓(𝑔(𝑥))=𝑒,()cos with 𝑓(𝑥)=𝑒 and 𝑔(𝑥)=7𝑥+9𝑥cos. Since 𝑓(𝑥) is an exponential function, it is continuous in the range of 𝑔(𝑥) and we can use the substitution 𝑢=𝑔(𝑥)=7𝑥+9𝑥.cos

The derivative of this with respect to 𝑥 is ddsin𝑢𝑥=799𝑥 or, equivalently by manipulating the differentials, ddsin𝑥=𝑢(799𝑥).

Now, we apply this substitution to the integral to change the variable from 𝑥 to 𝑢 and integrate the resulting expression using the power rule for integration: (799𝑥)𝑒𝑥=(799𝑥)𝑒𝑢(799𝑥)=𝑒𝑢=𝑒+.sindsindsindC()cos

Finally, we apply the reverse substitution 𝑢=7𝑥+9𝑥cos to get the final result in terms of 𝑥: (799𝑥)𝑒𝑥=𝑒+.sindC()()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: 𝑓(𝑔(𝑥))𝑔(𝑥)𝑥=𝑓(𝑢)𝑢.dd
  • To choose our substitution, 𝑢, we look for a factor of the integrand whose derivative also appears in the integrand or the inner function of the “complicated” part ot the integrand.
    In particular, if part of the integrand contains a derivative 𝑔(𝑥) and/or a composite function of the form 𝑓(𝑔(𝑥)), we use the substitution 𝑢=𝑔(𝑥).

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