5 Integration

5.5 Substitution

We motivate this section with an example. Let f(x)=(x2+3x-5)10. We can compute f(x) using the Chain Rule. It is:


Now consider this: What is (20x+30)(x2+3x-5)9dx? We have the answer in front of us;


How would we have evaluated this indefinite integral without starting with f(x) as we did?

This section explores integration by substitution. It allows us to “undo the Chain Rule.” Substitution allows us to evaluate the above integral without knowing the original function first.

The underlying principle is to rewrite a “complicated” integral of the form f(x)dx as a not-so-complicated integral h(u)du. We’ll formally establish later how this is done. First, consider again our introductory indefinite integral, (20x+30)(x2+3x-5)9dx. Arguably the most “complicated” part of the integrand is (x2+3x-5)9. We wish to make this simpler; we do so through a substitution. Let u=x2+3x-5. Thus


We have established u as a function of x, so now consider the differential of u:


Keep in mind that (2x+3) and dx are multiplied; the dx is not “just sitting there.” margin: Note: Recall from Section 4.3 that the differential of x, denoted dx, is any nonzero real number. If u is a function of x, then the differential of u, denoted du, is defined by du=u(x)dx. Λ

Return to the original integral and do some substitutions through algebra:

(20x+30)(x2+3x-5)9dx =10(2x+3)(x2+3x-5)9dx
=u10+C(replace u with x2+3x-5)

One might well look at this and think “I (sort of) followed how that worked, but I could never come up with that on my own,” but the process is learnable. This section contains numerous examples through which the reader will gain understanding and mathematical maturity enabling them to regard substitution as a natural tool when evaluating integrals.

We stated before that integration by substitution “undoes” the Chain Rule. Specifically, let F(x) and g(x) be differentiable functions and consider the derivative of their composition:




Integration by substitution works by recognizing the “inside” function g(x) and replacing it with a variable. By setting u=g(x), we can rewrite the derivative as


Since du=g(x)dx, we can rewrite the above integral as


This concept is important so we restate it in the context of a theorem.

Theorem 5.5.1 Integration by Substitution

Let F and g be differentiable functions, where the range of g is an interval I contained in the domain of F. Then


If u=g(x), then du=g(x)dx and


The point of substitution is to make the integration step easy. Indeed, the step F(u)du=F(u)+C looks easy, as the antiderivative of the derivative of F is just F, plus a constant. The “work” involved is making the proper substitution. There is not a step-by-step process to memorize; rather, experience will be your guide. To gain experience, we now embark on many examples.

Example 5.5.1 Integrating by substitution

Evaluate xsin(x2+5)dx.

SolutionKnowing that substitution is related to the Chain Rule, we choose to let u be the “inside” function of sin(x2+5). (This is not always a good choice, but it is often the best place to start.)

Let u=x2+5, hence du=2xdx. The integrand has an xdx term, but not a 2xdx term. (Recall that multiplication is commutative, so the x does not physically have to be next to dx for there to be an xdx term.) We can divide both sides of the du expression by 2:


We can now substitute.

xsin(x2+5)dx =sin(x2+5u)xdx12du
=-12cosu+C(now replace u with x2+5)

Thus xsin(x2+5)dx=-12cos(x2+5)+C. We can check our work by evaluating the derivative of the right hand side.

Example 5.5.2 Integrating by substitution

Evaluate cos(5x)dx.

SolutionAgain let u replace the “inside” function. Letting u=5x, we have du=5dx. Since our integrand does not have a 5dx term, we can divide the previous equation by 5 to obtain 15du=dx. We can now substitute.

cos(5x)dx =cos(5xu)dx15du

We can again check our work through differentiation.

The previous example exhibited a common, and simple, type of substitution. The “inside” function was a linear function (in this case, y=5x). When the inside function is linear, the resulting integration is very predictable, so that we can say


For example, sin(7x-4)dx=-17cos(7x-4)+C. Our next example can use this idea, but we will only employ it after going through all of the steps.

Example 5.5.3 Integrating by substituting a linear function

Evaluate 7-3x+1dx.

SolutionWe can view this as a composition of the functions f(g(x)), where f(x)=7/x and g(x)=-3x+1. Employing our understanding of substitution, we let u=-3x+1, the inside function. Thus du=-3dx. The integrand lacks a -3; hence divide the previous equation by -3 to obtain -du/3=dx. We can now evaluate the integral through substitution.

7-3x+1dx =(7u)(du-3)

Not all integrals that benefit from substitution have a clear “inside” function. Several of the following examples will demonstrate ways in which this occurs.

Example 5.5.4 Integrating by substitution

Evaluate sinxcosxdx.

SolutionThere is not a composition of function here to exploit; rather, just a product of functions. Do not be afraid to experiment; when given an integral to evaluate, it is often beneficial to think “If I let u be this, then du must be that …” and see if this helps simplify the integral at all.

In this example, let’s set u=sinx. Then du=cosxdx, which we have as part of the integrand. The substitution becomes very straightforward:

sinxcosxdx =udu

One would do well to ask “What would happen if we let u=cosx?” The result is just as easy to find, yet looks very different. The challenge to the reader is to evaluate the integral letting u=cosx and discover why the answer is the same, yet looks different.

Our examples so far have required “basic substitution.” The next example demonstrates how substitutions can be made that often strike the new learner as being “nonstandard.”

Example 5.5.5 Integrating by substitution

Evaluate xx+3dx.

SolutionRecognizing the composition of functions, set u=x+3. Then du=dx, giving what seems initially to be a simple substitution. But at this stage, we have:


We cannot evaluate an integral that has both an x and an u in it. We need to convert the x to an expression involving just u.

Since we set u=x+3, we can also state that u-3=x. Thus we can replace x in the integrand with u-3. It will also be helpful to rewrite u as u12.

xx+3dx =(u-3)u12du

Checking your work is always a good idea. In this particular case, some algebra will be needed to make one’s answer match the integrand in the original problem.

Example 5.5.6 Integrating by substitution

Evaluate 1xlnxdx.

SolutionThis is another example where there does not seem to be an obvious composition of functions. The line of thinking used in Example 5.5.5 is useful here: choose something for u and consider what this implies du must be. If u can be chosen such that du also appears in the integrand, then we have chosen well.

Choosing u=1/x makes du=-1/x2dx; that does not seem helpful. However, setting u=lnx makes du=1/xdx, which is part of the integrand. Thus:

1xlnxdx =1lnx1/u1xdxdu

The final answer is interesting; the natural log of the natural log. Take the derivative to confirm this answer is indeed correct.

Integrals Involving Trigonometric Functions

Section 8.2 delves deeper into integrals of a variety of trigonometric functions; here we use substitution to establish a foundation that we will build upon.

The next three examples will help fill in some missing pieces of our antiderivative knowledge. We know the antiderivatives of the sine and cosine functions; what about the other standard functions tangent, cotangent, secant and cosecant? We discover these next.

Example 5.5.7 Integration by substitution: antiderivatives of tanx

Evaluate tanxdx.

SolutionThe previous paragraph established that we did not know the antiderivatives of tangent, hence we must assume that we have learned something in this section that can help us evaluate this indefinite integral.

Rewrite tanx as sinx/cosx. While the presence of a composition of functions may not be immediately obvious, recognize that cosx is “inside” the 1/x function. Therefore, we see if setting u=cosx returns usable results. We have that du=-sinxdx, hence -du=sinxdx. We can integrate:

tanxdx =sinxcosxdx

Some texts prefer to bring the -1 inside the logarithm as a power of cosx, as in:

-ln|cosx|+C =ln|(cosx)-1|+C

Thus the result they give is tanxdx=ln|secx|+C. These two answers are equivalent.

Example 5.5.8 Integrating by substitution: antiderivatives of secx

Evaluate secxdx.

SolutionThis example employs a wonderful trick: multiply the integrand by “1” so that we see how to integrate more clearly. In this case, we write “1” as


This may seem like it came out of left field, but it works beautifully. Consider:

secxdx =secxsecx+tanxsecx+tanxdx
Now let u=secx+tanx; this means du=(secxtanx+sec2x)dx, which is our numerator. Thus:
secxdx =duu

We can use similar techniques to those used in Examples 5.5.7 and 5.5.8 to find antiderivatives of cotx and cscx (which the reader can explore in the exercises.) We summarize our results here.

Theorem 5.5.2 Antiderivatives of Trigonometric Functions

(a) sinxdx=-cosx+C (b) cosxdx=sinx+C (c) tanxdx=ln|secx|+C (d) cscxdx=-ln|cscx+cotx|+C (e) secxdx=ln|secx+tanx|+C (f) cotxdx=ln|sinx|+C

Simplifying the Integrand

It is common to be reluctant to manipulate the integrand of an integral; at first, our grasp of integration is tenuous and one may think that working with the integrand will improperly change the results. Integration by substitution works using a different logic: as long as equality is maintained, the integrand can be manipulated so that its form is easier to deal with. The next example demonstrates a common way in which using algebra first makes the integration easier to perform.

Example 5.5.9 Integration by alternate methods

Evaluate x2+2x+3xdx with, and without, substitution.

SolutionWe already know how to integrate this particular example. Rewrite x as x12 and simplify the fraction:


We can now integrate using the Power Rule:

x2+2x+3x1/2dx =(x32+2x12+3x-12)dx

This is a perfectly fine approach. We demonstrate how this can also be solved using substitution as its implementation is rather clever.

Let u=x=x12; therefore


This gives us x2+2x+3xdx=(x2+2x+3)2du. What are we to do with the other x terms? Since u=x12, we have u2=x and u4=x2. We can then replace x2 and x with appropriate powers of u. We thus have

x2+2x+3xdx =(x2+2x+3)2du

which is obviously the same answer we obtained before. In this situation, substitution is arguably more work than our other method. The fantastic thing is that it works. It demonstrates how flexible integration is.

Substitution and Definite Integration

So far this section has focused on learning a new technique for finding antiderivatives. In practice, we will frequently be interested in finding definite integrals. We can use this antiderivative to evaluate the definite integral, but there is a more efficient method.

At its heart, (using the notation of Theorem 5.5.1) substitution converts integrals of the form F(g(x))g(x)dx into an integral of the form F(u)du with the substitution of u=g(x). The following theorem states how the bounds of a definite integral can be changed as the substitution is performed.

Theorem 5.5.3 Substitution with Definite Integrals

Let F and g be differentiable functions, where the range of g is an interval I that is contained in the domain of F. Then


In effect, Theorem 5.5.3 states that once you convert to integrating with respect to u, you do not need to switch back to evaluating with respect to x. A few examples will help one understand.

Example 5.5.10 Definite integrals and substitution: changing the bounds

Evaluate 02cos(3x-1)dx using Theorem 5.5.3.

SolutionObserving the composition of functions, let u=3x-1, hence du=3dx. As 3dx does not appear in the integrand, divide the latter equation by 3 to get du/3=dx.

By setting u=3x-1, we are implicitly stating that g(x)=3x-1. Theorem 5.5.3 states that the new lower bound is g(0)=-1; the new upper bound is g(2)=5. We now evaluate the definite integral:

margin: y=cos(3x-1)-112345-1-0.50.51xy (a)y=13cos(u)-112345-1-0.50.51uy (b) Figure 5.5.1: Graphing the areas defined by the definite integrals of Example 5.5.10. Λ
12cos(3x-1)dx =-15cosudu3

Notice how once we converted the integral to be in terms of u, we never went back to using x.

The graphs in Figure 5.5.1 tell more of the story. In (a) the area defined by the original integrand is shaded, whereas in (b) the area defined by the new integrand is shaded. In this particular situation, the areas look very similar; the new region is “shorter” but “wider,” giving the same area.

Example 5.5.11 Definite integrals and substitution: changing the bounds

Evaluate 0π/2sinxcosxdx using Theorem 5.5.3.

SolutionWe saw the corresponding indefinite integral back in Example 5.5.4. In that example we set u=sinx but stated that we could have let u=cosx. For variety, we do the latter here.

Let u=g(x)=cosx, giving du=-sinxdx and hence sinxdx=-du. The new upper bound is g(π/2)=0; the new lower bound is g(0)=1. Note how the lower bound is actually larger than the upper bound now. We have margin: y=sinxcosx1-0.50.51π2xy (a)y=u1-0.50.51π2uy (b) Figure 5.5.2: Graphing the areas defined by the definite integrals of Example 5.5.11. Λ

0π/2sinxcosxdx =10-udu (switch bounds & change sign)

In Figure 5.5.2 we have again graphed the two regions defined by our definite integrals. Unlike the previous example, they bear no resemblance to each other. However, Theorem 5.5.3 guarantees that they have the same area.

Example 5.5.12 Definite integrals and substitution: changing the bounds

Evaluate 02xex2+1dx using Theorem 5.5.3.

SolutionWe note the composition of functions and let u=x2+1, hence du=2xdx. We divide the differential by 2 to get du2=xdx.

Setting g(x)=u=x2+1, we find that the new lower bound is g(0)=1; the new upper bound is g(2)=5. We now evaluate:

02xex2+1dx =15eudu2

Exercises 5.5


Terms and Concepts

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    Substitution “undoes” what derivative rule?

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    T/F: One can sometimes use algebra to rewrite the integrand of an integral to make it easier to evaluate.


In Exercises 3–56., evaluate the indefinite integral.

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    cotxdx. Do not just refer to Theorem 5.5.2 for the answer; justify it through Substitution.

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    cscxdx. Do not just refer to Theorem 5.5.2 for the answer; justify it through Substitution.

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In Exercises 57–66., evaluate the definite integral.

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