# Strategies for Integration Contributed by: In this section, we will learn about:
The techniques to evaluate miscellaneous integrals.
1. 7
TECHNIQUES OF INTEGRATION
2. TECHNIQUES OF INTEGRATION
As we have seen, integration is more
challenging than differentiation.
 In finding the derivative of a function, it is obvious
which differentiation formula we should apply.
 However, it may not be obvious which technique
we should use to integrate a given function.
3. TECHNIQUES OF INTEGRATION
Until now, individual techniques have been
applied in each section.
For instance, we usually used:
 Substitution in Exercises 5.5
 Integration by parts in Exercises 7.1
 Partial fractions in Exercises 7.4
4. TECHNIQUES OF INTEGRATION
7.5
Strategy for Integration
In this section, we will learn about:
The techniques to evaluate miscellaneous integrals.
5. STRATEGY FOR INTEGRATION
In this section, we present a collection
of miscellaneous integrals in random order.
 The main challenge is to recognize which
technique or formula to use.
6. STRATEGY FOR INTEGRATION
No hard and fast rules can be given as to
which method applies in a given situation.
 However, we give some advice on strategy
that you may find useful.
7. STRATEGY FOR INTEGRATION
A prerequisite for strategy selection
is a knowledge of the basic integration
8. STRATEGY FOR INTEGRATION
In the upcoming table, we have
 The integrals from our previous list
 Several additional formulas we have learned
in this chapter
9. STRATEGY FOR INTEGRATION
Most should be memorized.
 It is useful to know them all.
 However, the ones marked with an asterisk
need not be memorized—they are easily derived.
10. TABLE OF INTEGRATION FORMULAS
n 1
n x 1
1. x dx  (n 1) 2.  dx ln | x |
n 1 x
x
x x x a
3. e dx e 4. a dx 
ln a
11. TABLE OF INTEGRATION FORMULAS
5. sin x dx  cos x 6.cos x dx sin x
7. sec 2 x dx tan x 8.csc 2 x dx  cot x
9.sec x tan x dx sec x 10.csc x cot x dx  csc x
11. sec x dx ln sec x  tan x 12. csc x dx ln csc x  cot x
12. TABLE OF INTEGRATION FORMULAS
13. tan x dx ln sec x 14. cot x dx ln sin x
15.sinh x dx cosh x 16.cosh x dx sinh x
dx 1 1  x  dx  x
1
17. 2 2
 tan   18. sin  
x a a a 2
a  x 2
a
13. TABLE OF INTEGRATION FORMULAS
dx 1 x a dx
*19. 2 2
 ln *20. ln x  x 2 a 2
x a 2a x  a x 2 a 2
 Formula 19 can be avoided by using partial fractions.
 Trigonometric substitutions can be used instead of
Formula 20.
14. STRATEGY FOR INTEGRATION
Once armed with these basic integration
formulas, you might try this strategy:
1. Simplify the integrand if possible.
2. Look for an obvious substitution.
3. Classify the integrand according to its form.
4. Try again.
15. 1. SIMPLIFY THE INTEGRAND
Sometimes, the use of algebraic
manipulation or trigonometric identities
will simplify the integrand and make
the method of integration obvious.
16. 1. SIMPLIFY THE INTEGRAND
Here are some examples:
 x  1  x  dx  
x  x dx
tan  sin  2
 sec2  d cos cos  d sin  cos  d  1
2sin 2 d
2 2 2
(sin x  cos x ) dx  (sin x  2sin x cos x  cos x) dx
(1  2sin x cos x) dx
17. 2. LOOK FOR OBVIOUS SUBSTITUTION
Try to find some function u = g(x) in
the integrand whose differential du = g’(x) dx
also occurs, apart from a constant factor.
x
 For instance, in the integral x 2  1 dx ,
notice that, if u = x2 – 1, then du = 2x dx.
 So, we use the substitution u = x2 – 1
instead of the method of partial fractions.
18. 3. CLASSIFY THE FORM
If Steps 1 and 2 have not led to
the solution, we take a look at the form
of the integrand f(x).
19. 3 a. TRIGONOMETRIC FUNCTIONS
We use the substitutions recommended
in Section 7.2 if f(x) is a product of:
 sin x and cos x
 tan x and sec x
 cot x and csc x
20. 3 b. RATIONAL FUNCTIONS
If f is a rational function, we
use the procedure of Section 7.4
involving partial fractions.
21. 3 c. INTEGRATION BY PARTS
If f(x) is a product of a power of x (or
a polynomial) and a transcendental function
(a trigonometric, exponential, or logarithmic
function), we try integration by parts.
 We choose u and dv as per the advice in Section 7.1
 If you look at the functions in Exercises 7.1, you
will see most of them are the type just described.
Particular kinds of substitutions are
2 2
i. If x a occurs, we use a trigonometric
substitution according to the table in section 7.3
ii. If n ax  b occurs, we use the rationalizing
substitution u  n ax  b .
More generally, this sometimes works for n g ( x) .
23. 4. TRY AGAIN
If the first three steps have not produced
the answer, remember there are basically
only two methods of integration:
1. Substitution
2. Parts
24. 4 a. TRY SUBSTITUTION
Even if no substitution is obvious (Step 2),
some inspiration or ingenuity (or even
desperation) may suggest an appropriate
25. 4 b. TRY PARTS
Though integration by parts is used most
of the time on products of the form described
in Step 3 c, it is sometimes effective on single
 Looking at Section 7.1, we see that it works on tan-1x,
sin-1x, and ln x, and these are all inverse functions.
26. 4 c. MANIPULATE THE INTEGRAND
Algebraic manipulations (rationalizing the
denominator, using trigonometric identities)
may be useful in transforming the integral
into an easier form.
 These manipulations may be more substantial
than in Step 1 and may involve some ingenuity.
27. 4 c. MANIPULATE THE INTEGRAND
Here is an example:
dx 1 1  cos x
1  cos x 1  cos x 1  cos x dx
1  cos x
 2
dx
1  cos x
1  cos x
 2 dx
sin x
 2 cos x 
 csc x  2  dx
 sin x 
28. 4 d. RELATE TO PREVIOUS PROBLEMS
When you have built up some experience in
integration, you may be able to use a method
on a given integral that is similar to a method
you have already used on a previous integral.
 You may even be able to express the given
integral in terms of a previous one.
29. 4 d. RELATE TO PREVIOUS PROBLEMS
For instance, ∫ tan2x sec x dx is
a challenging integral.
 If we make use of the identity tan2x = sec2x – 1,
we can write:
2 3
tan x sec x dx sec x dx  sec x dx
 Then, if ∫ sec3x dx has previously been evaluated,
that calculation can be used in the present problem.
30. 4 e. USE SEVERAL METHODS
Sometimes, two or three methods
are required to evaluate an integral.
 The evaluation could involve several successive
substitutions of different types.
 It might even combine integration by parts
with one or more substitutions.
31. STRATEGY FOR INTEGRATION
In the following examples, we
indicate a method of attack, but
do not fully work out the integral.
32. SIMPLIFY INTEGRAND Example 1
3
tan x
cos3 x dx
 In Step 1, we rewrite the integral:
tan 3 x 3 3
cos3 x dx tan x sec x dx
 It is now of the form ∫tanmx secnx dx with m odd.
 So, we can use the advice in Section 7.2
33. TRY SUBSTITUTION Example 1
Suppose, in Step 1, we had written:
tan 3 x sin 3 x 1
cos3 x dx cos3 x cos3 x dx
3
sin x
 6 dx
cos x
Then, we could have continued as follows.
34. TRY SUBSTITUTION Example 1
Substitute u = cos x:
sin 3 x 1  cos 2 x
cos6 x dx  cos6 x sin x dx
2
1 u
 6 (  du )
u
2
u 1
 6 du
u
4 6
(u  u ) du
35. TRY SUBSTITUTION Example 2
x
e dx
 According to (ii) in Step 3 d, we substitute u = √x.
 Then, x = u2, so dx = 2u du and e x dx 2 ueu du
 
 The integrand is now a product of u and
the transcendental function eu.
 So, it can be integrated by parts.
36. RATIONAL FUNCTIONS Example 3
5
x 1
x3  3x 2  10 x dx
 No algebraic simplification or substitution is obvious.
So, Steps 1 and 2 don’t apply here.
 The integrand is a rational function. So, we apply
the procedure of Section 7.4, remembering that
the first step is to divide.
37. TRY SUBSTITUTION Example 4
dx
x ln x
 Here, Step 2 is all that is needed.
 We substitute u = ln x, because its differential is
du = dx/x, which occurs in the integral.
38. MANIPULATE INTEGRAND Example 5
1 x
 dx
1 x
 Although the rationalizing substitution 1 x
u  dx
works here [(ii) in Step 3 d], it leads to 1 x
a very complicated rational function.
 An easier method is to do some algebraic manipulation
(either as Step 1 or as Step 4 c).
39. MANIPULATE INTEGRAND Example 5
Multiplying numerator and denominator
by 1  x , we have:
1 x 1 x
 dx  dx
1 x 1  x2
1 x
 dx   1 x dx
2 2
1 x
1 2
sin x  1  x  C
40. CAN WE INTEGRATE ALL CONTINUOUS FUNCTIONS?
The question arises:
Will our strategy for integration enable us to
find the integral of every continuous function?
 For example, can we use it to evaluate ∫ ex2dx?
41. CAN WE INTEGRATE ALL CONTINUOUS FUNCTIONS?
 At least, we cannot do it in terms
of the functions we are familiar with.
42. ELEMENTARY FUNCTIONS
The functions we have been dealing
with in this book are called elementary
 For instance, the function
x2  1
f ( x)  3  ln(cosh x)  xesin 2 x
x  2x  1
is an elementary function.
43. ELEMENTARY FUNCTIONS
If f is an elementary function, then f’
is an elementary function.
However, ∫ f(x) dx need not be
an elementary function.
44. ELEMENTARY FUNCTIONS
Consider f(x) = ex2.
 Since f is continuous, its integral exists.
x
t2
 If we define the function F by F ( x ) 
e
0
dt
then we know from Part 1 of the Fundamental2
Theorem of Calculus (FTC1) that F '( x) e x
 Thus, f(x) = ex2 has an antiderivative F.
45. ELEMENTARY FUNCTIONS
However, it has been proved that F is not
an elementary function.
 This means that, however hard we try, we will never
succeed in evaluating ∫ ex2dx in terms of the functions
we know.
 In Chapter 11, however, we will see how to express
∫ ex2dx as an infinite series.
46. ELEMENTARY FUNCTIONS
The same can be said of the following
x
e 2 x
x dx sin( x ) dx cos(e ) dx
3 1 sin x
x  1 dx ln x dx  x dx
47. ELEMENTARY FUNCTIONS
In fact, the majority of elementary
functions don’t have elementary
 You may be assured, though, that the integrals
in the exercises are all elementary functions.