13 | TECHNIQUES OF INTEGRATION
13 | TECHNIQUES OF INTEGRATION
This chapter of Single Variable Calculus by Dr JH Klopper is licensed under an Attribution-NonCommercial-NoDerivatives 4.0 International Licence available at http://creativecommons.org/licenses/by-nc-nd/4.0/?ref=chooser-v1 .
13.1 Introduction
13.1 Introduction
Many function allow analytical integration. Here we can use simple techniques such as substitutions and integration by parts to find solutions to integrals.
13.2 u Substitution
13.2 Substitution
u
When calculating the first derivative of the composition of a function, we used the chain rule. Consider the function composition in ().
g(x)h(x)3x+1\[NewLine]f(x)g(h(x))
2
x
2
(3x+1)
(
1
)We can make a substitution with . Now we have that and (u)2u. We also take the derivative of with respect to and have that (x)3. The derivative requires back-substitution, shown in (2), where (x)(u)·(x).
u(x)3x+1
f(u)
2
u
′
f
u
x
′
u
′
f
′
f
′
u
′
f
′
f
′
u
(
2
)This form of the chain rule is named substitution. It is typical to use the symbol and hence the name.
u
u
We have a similar technique that can be used for integration where the integrand is a composition. As example we consider the indefinite integral in (3).
∫dx
2
(x-4)
By expand the integrand, the integrand becomes an expression consisting of the sum of terms. We use the Expand function.
In[]:=
Expand[]
2
(x-4)
Out[]=
16-8x+
2
x
We use the property of the addition of terms to calculate the integral in (3).
-8x+16dxdx+-8xdx+16dx\[NewLine]dx-8xdx+16dx\[NewLine]-+16x+c\[NewLine]-4+16x+c
2
x
2
x
2
x
1
3
3
x
8
2
2
x
1
3
3
x
2
x
(
3
)The is indeed the antiderivative of the integrand . We omit the constant of integration since the derivative of a constant is .
D
function can be used to verify that F(x)-4+16x+c
1
3
3
x
2
x
f(x)-8x+16
2
x
c
0
In[]:=
D-4+16x,x
1
3
3
x
2
x
Out[]=
16-8x+
2
x
It is indeed so that (x)f(x). The Wolfram Language can also be used to calculate the indefinite integral as yet another confirmation that the result in (3) is correct.
′
F
In[]:=
∫x
2
(x-4)
Out[]=
16x-4+
2
x
3
x
3
While the integrand could easily be expanded, we could also use the technique of substitution, where is a function of . For this technique, we let . We take the first derivative in (4).
f(x)
u
u
x
u(x)x-4
du
dx
(
4
)The integral can now be rewritten, as shown in (5), substituting the integrand for and for .
u
dx
du
∫du
2
u
(
5
)This is a very simple integral to solve. We do so as shown in (6), where we use back substitution.
du\[NewLine]\[NewLine](-12+48x-64)\[NewLine]-4+16x++\[NewLine]-4+16x+c
2
u
1
3
3
u
1
3
3
(x-4)
1
3
3
x
2
x
1
3
3
x
2
x
64
3
c
1
1
3
3
x
2
x
(
6
)The initial constant of integration is so as to avoid confusion. We do note, though that plus an arbitrary constant is just a constant (of unknown value) and hence we place a final constant in the final solution.
c
1
64
3
c
Problem 13.2.1 Calculate the indefinite integral in (7).
∫(21-7)dx
2
x
6
(-x-6)
3
x
(
7
)By using the substitution we take the first derivative of with respect to as shown in (8).
u(x)-x-6
3
x
u
x
du
dx
2
x
2
x
(
8
)We make the two substitutions (one for and the other for ) and solve the indefinite integral in (9).
u
du
∫(21-7)dx∫7(3-1)dx\[NewLine]∫(7)du\[NewLine]+c\[NewLine]+c
2
x
6
(-x-6)
3
x
2
x
6
(-x-6)
3
x
6
u
7
u
7
(-x-6)
3
x
(
9
)We verify the solution by calculating the first derivative of the solution with respect to .
x
In[]:=
D[,x]
7
(-x-6)
3
x
Out[]=
7(-1+3)
2
x
6
(-6-x+)
3
x
Recognizing that we are undoing the chain rule, it is a good place to start our substitution by considering the inside function in the composition .
u
h(x)
f(x)g(h(x))
Problem 13.2.2 Solve the indefinite integral in (10).
∫[xsin(-2)]dx
2
x
(
10
)The integrand is the product of two function. We will explore a technique for the integral of the product of functions. Here, though, we recognize the second function as a composition of functions and choose the substitution . The derivative is shown in (11).
u(x)-2
2
x
du
dx
1
2
(
11
)We can now substitute and -2 and solve the indefinite integral in (12).
xdx
2
x
sin(u)dusin(u)du\[NewLine]-cos(u)+c\[NewLine]-cos(-2)
1
2
1
2
1
2
1
2
2
x
(
12
)The solution is verified using the Wolfram Language. Since the cosine function contains a subtraction of two angles, the Wolfram Language will expand the solution using trigonometric expansion. The TrigReduce function returns the reduced form.
In[]:=
TrigReduce∫xSin[-2]x
2
x
Out[]=
-Cos[2-]
1
2
2
x
The solution is as expected because .
cos(-α)cos(α)
What is clear from the problems so far is that the technique of -substitution works well if the derivative of is in the integrand.
u
u(x)
Problem 13.2.3 Solve the indefinite integral in (13).
∫cos(x)sin(x)dx
(
13
)We note that the derivative of with respect to is in the integrand and choose the substitution . The derivative is shown in (14).
sin(x)
x
u(x)sin(x)
du
dx
(
14
)We complete the substitution and find the solution in (15).
udu+c\[NewLine](x)+c
1
2
2
u
1
2
2
sin
(
15
)The Wolfram Language confirms our solution.
In[]:=
∫Cos[x]Sin[x]x
Out[]=
-
1
2
2
Cos[x]
As an exercise, we show that t our solution and that provided by the Wolfram Language is equivalent in (16).
-(x)+(1-(x))++(x)+(x)+-+(x)+c
1
2
2
cos
c
1
(WolframLanguagesolutionwithconstantofintegration)
-c
1
1
2
2
sin
c
1
usetrigonometricidentity(x)=1-(x)
-2
cos
2
sin
1
2
1
2
2
sin
c
1
1
2
2
sin
1
2
c
1
1
2
2
sin
let-+=c
1
2
c
1
(
16
)It should be clear that the Wolfram Language used the substitution . We show the result in (17).
u(x)cos(x)
u(x)=cos(x)=-sin(x)-du=sin(x)dx∫-uu-+c-(x)+c
du
dx
1
2
2
u
1
2
2
cos
(
17
)
13.3 Integration by parts
13.3 Integration by parts
Now we can calculate the solution using integration by parts, shown in (22).
The Wolfram Language confirms the solution.
The rule of thumb hierarchy can be useful. In many problems that are amenable to the use of integration by parts, we may have to use it more than once.
As with many techniques in calculus, integration by parts can only be appreciated through completing many example problems.
Problem 13.3.1 Calculate the indefinite integral in (23).
The Wolfram Language confirms the result.
Problem 13.3.2 Solve the indefinite integral in (25).
The Wolfram Language confirms the result.
In the next problem, we have to apply the technique of integration by parts twice.
Problem 13.3.3 Calculate the indefinite integral in (27).
We verify the solution using the Wolfram Language.
The solution is in fact algebraically similar to the solution in (28).
We repeat Problem 13.3.3 by introduce limits of integration.
Problem 13.3.4 Calculate the definite integral in (29).
We visualize the problem in Figure 13.3.1.
Below we use the definite integral and the Expand function to confirm the result.