Integration Review
I. Integration Formulas
!
x n dx =
2.
x n +1
+ C (n " #1)
n +1
13.
!
e x dx = e x + C
!
dx
= ln x + C
x
14.
!
a x dx =
3.
!
sin x dx = " cosx + C
15.
"
4.
!
cos x dx = sin x + C
16.
!
dx
= arctan x + C
1+ x 2
5.
! sec
17.
"
dx
= arcsec x + C
x x2 ! 1
6.
!
csc 2 x dx = " cot x + C
18.
"
dx
= ln x ! a + C
x !a
7.
! sec x tan x dx = sec x + C
19.
! sin kx dx = " k cos kx + C
8.
!
csc x cot x dx = " csc x + C
20.
!
coskx dx =
9.
!
tan x dx = ln sec x + C
21.
!
e kx dx =
10.
!
cot x dx = ln sin x + C
22.
"
dx
x
= arcsin + C (a > 0)
2
a
a !x
11.
! sec x dx = lnsec x + tan x + C
23.
!
dx
1
x
arctan + C (a > 0)
2 =
a +x
a
a
12.
!
24.
"
dx
1
x
=
arcsec
+ C (a > 0)
a
x x 2 ! a2 a
1.
2
x dx = tan x + C
csc x dx = ln csc x " cot x + C
25.
26.
ax
ln a
+ C (a > 0, a " 1)
dx
= arcsin x + C
1! x 2
1
1
k
1
k
sin kx + C (k " 0)
e kx + C (k " 0)
2
2
! ln x dx = x ln x " x + C
1
1
! sec x dx = 2 sec x tan x + 2 ln sec x + tan x + C
3
(k # 0)
II. Algebraic Substitutions
If we wish to integrate a function that contains a term of the type [f(x)]n , then often the
substitution u = [f(x)]n or u = f(x) will simplify the integral. For example, if an integral
contains the expression 3 5x + 1 , try letting u = 3 5x + 1 or u = 5x + 1. Of course, any time an
integral contains a power of an expression, say [f(x)]n, where n might be an integer or a
fraction, it is always a good idea to think of the substitution u = f(x), so that [f(x)]n becomes un.
III. Quadratic Expressions
If we have a quadratic expression with a linear term, then it is often convenient to
complete the square and make a substitution. For example, if we wish to integrate a function that
contains the expression x2 – 6x + 10, we write this as
x2 – 6x + 10 = (x2 – 6x + 9) + 1 = (x – 3)2 + 1.
If we let u = x – 3, then (x – 3)2 + 1 = u2 + 1.
As another example, suppose we want to integrate a function containing the expression
3 + 2x ! x 2 . A good thing to do here is factor –1 from the terms containing x; so
3 + 2x ! x 2 = 3 ! (x 2 ! 2x) = 3 ! (x 2 ! 2x + 1) + 1 = 4 ! (x ! 1)2 .
Letting u = x – 1, we obtain
4 ! (x ! 1)2 = 4 ! u 2 .
IV. Integration by Parts
! u dv = uv " ! v du
This method allows us to evaluate an integral by replacing it by a "simpler" integral. If
we use this method, we often let u equal some expression f(x) for which f !(x) is simpler
than f(x). For example, typical choices for u might be x or x2 or x3, or possibly ln x or
tan–1 x or sin–1 x. Often dv involves the most complicated expression that can be easily
integrated.
Some common integrals where the integration by parts method is used are:
1.
! x sin x dx
2.
! x cos x dx
3.
! xe
4.
!x
5.
! x cos 3x dx
6.
"x e
7.
!x
8.
!x e
dx
9.
! sin
x dx
x
2
dx
sin x dx
2 !x
3
dx
sin 2x dx
3 5x
–1
10.
! tan
11.
! ln x dx
12.
!x
13.
!
14.
! x sec
15.
! x sec x tan x dx
16.
! x csc x cot x dx
3
2
–1
x dx
ln x dx
x ln x dx
2
x dx
There are some integrals for which integration by parts is used twice and the original
integral is solved for as in an equation. Integrals of the following type fit into this pattern:
!e
17.
!e
18.
! sec
2x
sin 3x dx
3
x dx
ax
sin bx dx
or
!e
ax
cosbx dx
(a " 0, b " 0)
V. Trigonometric Integrals
Integrals of the type ! sin m x cos n x dx
If m is an odd positive integer, let u = cos x.
If n is an odd positive integer, let u = sin x.
If both m and n are even nonnegative integers, then use the identities:
cos2 x = 1 + cos 2x
2
Integrals of the type
! tan
m
sin2 x = 1 ! cos 2x
2
x sec n x dx
If m is an odd positive integer, then normally let u = sec x.
If n is an even positive integer, then normally let u = tan x.
If m is an even positive integer and n is an odd positive integer, then a formula
and/or integration by parts should be used.
In any case, the identity 1 + tan2x = sec2x is frequently used.
Integrals of the type
! cot
m
x csc n x dx
! tan
are handled in the same manner as
m
x sec n x dx .
The important identity here is 1 + cot2x = csc2x
1.
! sin x cos
4.
! sin
7.
!
3
2
2
x dx
2x cos 3 2x dx
sin x cos x dx
2.
! sin
5.
! cos
8.
! tan x sec
11.
! tan x dx
5
4
x dx
3.
! sin
x dx
6.
! sin
3
10.
! cot
13.
3
! tan x dx
14.
16.
! sec
17. ! sec 3 x dx
3
2
x csc 3 x dx
x dx
! tan
4
x dx
x dx
3
x cos 3 x dx
4
x cos 2 x dx
9.
! tan
2
x sec 2 x dx
12.
! tan
2
x dx
15.
! sec x dx
18. ! sec 4 x dx
VI. Trigonometric Substitution
a 2 ! x 2 , where a > 0, then
x
the substitution x = a sin θ is often helpful. Note that if x = a sin θ, then sin θ = . The
a
x
actual substitution is θ = arcsin .
a
1. If we wish to integrate a function which involves a power of
2. If we wish to integrate a function involving a power of
x
x = a tan θ is often helpful. [ θ = arctan ]
a
a 2 + x 2 , then the substitution
3. If we wish to integrate a function involving a power of
x
x = a sec θ is often helpful. [ θ = arcsec ]
a
x 2 ! a 2 , then the substitution
VII. Partial Fractions
To integrate
f (x)
! g(x) dx , where
f(x) and g(x) are polynomials:
1. If the degree of f(x) is greater than or equal to the degree of g(x), divide f(x) by g(x) to
obtain a proper rational function.
2. Factor g(x) completely as a product of linear factors and irreducible quadratic factors.
(The quadratic ax2 + bx + c is irreducible if b2 – 4ac < 0.)
3. The following example gives a typical partial fraction decomposition:
If f(x) is a polynomial of degree less than 11, then
f (x)
A B C
D
E
Fx + G
Hx + I
Jx + K
= + 2+ 3+
+
+ 2
+ 2
+ 2
2
2
3
2
2
x (x + 2) (x + 16)
x x
x
x + 2 (x + 2)
x + 16 (x + 16) (x + 16)3
3
VIII. A Practice Exam on Integration Techniques (Grade the exam yourself.)
1. (20 pts.)
5x 3 ! 5x 2 + 11x + 19
" (x ! 1)2 (x 2 + 9) dx
2. (10 pts.)
!x
3. (10 pts.)
! ln(x + 1) dx
4. (10 pts.)
! sin
5. (10 pts.)
"
5
sin x 3 dx
5
2x cos 4 2x dx
1
6x ! x 2
x4
6. (5 pts.)
!
7. (5 pts.)
! x tan x
8. (10 pts.)
!
9. (10 pts.)
" (9 ! x
10. (10 pts.)
4
dx
x5 + 1
2
dx
x
dx
x+2
1
"x
dx
2
2 3/2
)
dx
1
dx
! 4x + 20