sum rule: If f (x) and g(x) are functions, then Z
difference rule: If f (x) and g(x) are functions, Z
Activity 5.1 Derive the constant multiple, sum and difference rules from the linear combination rule.
Activity 5.2 Use antiderivatives to show that the linear combination rule works.
Example 5.3 Using these rules we see that:
Z where c is an arbitrary constant.
So, in the case of linear combinations of functions such as these, we see that the integral of the linear combination is given by the linear combination of the integrals.
Activity 5.3 Use the rules above to integrate the following functions with respect to x.
(a) − 3 cos x, (b) ex+ cos x, (c) 3 sin x− 3 x.
We now look at the other rules of integration, i.e. the ones that will allow us to integrate other combinations of functions. But, unlike what we saw with the rules of differentiation in Section 3.2.2, we shall see that these are harder to apply.
5.2.3 Integration by substitution
Integration by substitution is a way of dealing with integrands that involve the composition of two functions and, as such, it is closely related to the chain rule of differentiation. To see how it works, we will start by seeing how integration by substitution is related to the chain rule and then we will describe how to apply this rule. We will then apply this rule in some simple examples and then some harder ones.
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5.2. How to find indefinite integrals
Why integration by substitution works
We start by noting that the chain rule for differentiation tells us that if h(x) = (f ◦ g)(x) = f(g(x)),
then we write h as f (g) so that, on differentiating, we get dh
dx = df dg
dg dx. But, because of this we can see that
h(x) = (f◦ g)(x) = f(g(x)) is an antiderivative of dh dx = df
dg dg dx,
and so, we have Z
df dg
dg
dxdx = f (g(x)) + c,
which is the basis of integration by substitution. However, this is quite hard to apply and so, as a useful way of applying this rule, we think of
dg as dg
dxdx, so that we have
Z df
dg dg = f (g) + c, and this is the key to the method that we shall be using here.
How to integrate by substitution
We can now see how to apply integration by substitution. The basic idea is that, if you are given an integrand that involves a composition of two functions, this rule of
integration sometimes allows you to turn it into an easier integral by making a substitution. That is:
The integral involves the derivative of a composition and has the form Z
f0(g(x))g0(x) dx.
Write f0(g(x)) as f0(g) and g0(x)dx as dg. This should give you the easier integral Z
f0(g) dg.
Find this integral and replace all occurrences of g with g(x) to get your final answer.
Now, to make this clearer, let’s look at some examples.
Some simple applications of integration by substitution
Easy integrations by substitution involve an integrand which is nothing more than a simple composition of two functions and so there can be no doubt about which function should be ‘g’. To see this, let’s consider what happens when we want to integrate a simple composition which involves the function 3x + 1.
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where c is an arbitrary constant.Example 5.5 Find where c is an arbitrary constant.
Example 5.6 Find where c is an arbitrary constant.
In particular, observe what changes in these examples and what stays the same. Indeed, just for comparison, we can see what would happen if we had a composition which is like the one in Example 5.4 but it now involves the function 4x + 7 instead of 3x + 1.
Example 5.7 Find where c is an arbitrary constant.
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5.2. How to find indefinite integrals
Activity 5.4 In a similar manner, find
Z 1
4x + 7dx and Z
e4x+7 dx.
Note that in all of these examples, the substitution works because we have g(x) = ax + b and hence
dg
dx = a =⇒ dg = a dx =⇒ 1
adg = dx,
where a6= 0 and b are constants. Indeed, as we end up with an integrand involving 1a, which is a constant, it can be moved out of the integral using the constant multiple rule of integration. So, if our integrand is a composition, i.e. f (g(x)), and g(x) is a linear function, i.e. it has the form ax + b where a6= 0 and b are constants, this kind of substitution will always work and this leads to the general result that
Z
f (ax + b) dx = 1
aF (ax + b) + c,
where F (x) is an antiderivative of f (x) and c is an arbitrary constant.
Activity 5.5 Suppose that a6= 0 and b are constants. Use this result to find an
expression for Z
(ax + b)ndx, when n is a constant. Also find expressions for
Z
Activity 5.6 Using the expressions you found in Activity 5.5, verify your answers to Activity 5.4.
Some less simple applications of integration by substitution
We will also see slightly harder integrations by substitution where the integrand
involves a composition of two functions multiplied by another function. Although, even in these cases, there can be little doubt about which function should be ‘g’. To see this, let’s consider what happens when we want to integrate a simple composition which involves the function x2+ 1.
Example 5.8 Find
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i.e. the extra ‘x’ in the integrand was actually needed for the substitution g = x2+ 1 to work. i.e. the extra ‘x’ in the integrand was, again, needed for the substitution g = x2+ 1 to work. i.e. the extra ‘x’ in the integrand was, again, needed for the substitution g = x2+ 1 to work.
In particular, observe what changes in these examples and what stays the same. Indeed, just for comparison, we can see what would happen if we had a composition which is like the one in Example 5.8 but it now involves the function 3x2+ 7 instead of x2+ 1. i.e. the extra ‘x’ in the integrand was actually needed for the substitution
g = 3x2+ 7 to work.
Activity 5.7 In a similar manner, find
Z x
3x2+ 7dx and Z
x e3x2+7 dx.
To summarise, it is worth noting that in all of these examples, the substitution works because we have g(x) = ax2+ b and hence
dg
dx = 2ax =⇒ dg = 2ax dx,
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5.2. How to find indefinite integrals
where a6= 0 and b are constants. But, 2ax is not a constant and so we can not deal with this by taking it out of the integral as we did in the last set of examples. However, in these cases, the substitution still works because we have
dg
dx = 2ax =⇒ dg = 2ax dx =⇒ 1
2adg = xdx,
and there is also an ‘x’ in the integrand to facilitate the transition from ‘dx’ to ‘dg’.
Indeed, in the absence of this extra ‘x’, the substitution would produce a more complicated integral and we would not be able to proceed!
Integration by substitution more generally
The general lesson that we should be drawing from the last two sets of examples is that integration by substitution works when we have an integrand which is the product of
the composition of two functions f0(g(x)), and a constant multiple of g0(x).
The first of these enables us to replace f0(g(x)) with f0(g) and the second enables us to replace dx with some constant multiple of dg. Having done this, the substitution has turned a hard integral into an easier one and we can proceed. Let’s now consider some more complicated examples.
Example 5.12 Find Z
(x3+ x2)7(3x2+ 2x) dx.
Here the composition is (x3 + x2)7 and so we take g = x3+ x2. As such, we have dg
dx = 3x2+ 2x,
which is the other part of the product in the integrand, i.e. this substitution will work. Thus, we see that
dg = (3x2+ 2x) dx, and so the substitution gives
Z
(x3+ x2)7(3x2+ 2x) dx = Z
g7dg = g8
8 + c = (x3+ x2)8
8 + c.
Here, the extra ‘3x2+ 2x’ in the integrand was needed for the substitution g = x3+ x2 to work.
Example 5.13 Find
Z 2x + 2 x2+ 2x + 2dx.
Here the composition is (x2+ 2x + 2)−1 and so we take g = x2+ 2x + 2. As such, we
have dg
dx = 2x + 2,
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which is the other part of the product in the integrand, i.e. this substitution will work. Thus, we see that
dg = (2x + 2) dx, and so the substitution gives
Z 2x + 2
x2+ 2x + 2dx = Z 1
g dg = ln|g| + c = ln |x2+ 2x + 2| + c.
Here, the extra ‘2x + 2’ in the integrand was needed for the substitution g = x2+ 2x + 2 to work.
Example 5.14 Find Z
(x2+ 1) ex3+3x+7 dx.
Here the composition is ex3+3x+7 and so we take g = x3+ 3x + 7. As such, we have dg
dx = 3x2+ 3 = 3(x2+ 1),
which is a constant multiple of the other part of the product in the integrand, i.e.
this substitution will work. Thus, we see that dg = 3(x2+ 1) dx =⇒ 1
3dg = (x2+ 1) dx, and so the substitution gives
Z
(x2+ 1) ex3+3x+7 dx = Z
eg
1 3dg
= 1 3
Z
eg dg = 1
3eg+c = 1
3ex3+3x+7+c.
Here, the extra ‘x2+ 1’ in the integrand was needed for the substitution g = x3+ 3x + 7 to work.
Activity 5.8 Find Z
x sin(x2) dx.
Integration by substitution with trigonometric functions
Sometimes we can straightforwardly apply what we have just seen to find integrals that involve compositions of trigonometric functions as the following examples show.
Example 5.15 Find Z
sin2x cos x dx.
Here the composition is sin2x and so we take g = sin x. As such, we have dg
dx = cos x,
which is the other part of the product in the integrand, i.e. this substitution will
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5.2. How to find indefinite integrals
work. Thus, we see that
dg = cos x dx, and so the substitution gives
Z
sin2x cos x dx = Z
g2dg = g3
3 + c = 1
3sin3x + c.
Here, of course, the extra ‘cos x’ in the integrand was needed for the substitution g = sin x to work.
Activity 5.9 Find Z
cos2x sin x dx.
Indeed, as the next example shows, this kind of substitution allows us to find another useful result.
Example 5.16 Find Z
tan x dx.
In (2.1), we saw that
tan x = sin x cos x,
which means that the composition is (cos x)−1 and so we take g = cos x. As such, we
have dg
dx =− sin x,
which, up to a minus, is the other part of the product in the integrand, i.e. this substitution will work. Thus, we see that
dg = − sin x dx, and so the substitution gives
Z
tan x dx =
Z sin x
cos xdx =− Z dg
g =− ln |g| + c = − ln | cos x| + c.
Here, of course, the extra ‘sin x’ in the integrand was needed for the substitution g = cos x to work.
Activity 5.10 Find Z
cot x dx.
However, not every trigonometric substitution is so easy to spot as the next example shows.
Example 5.17 Find
Z dx
(x + a)2+ b2.