9.
Integration
9.1.
Standard integrals
If u is a function of x, then the indeniteZ integral
u dx
is a function whose derivative is u. The standard integrals are:
(1) If a is a constant, then
Z
a dx = ax + c
where c is a constant, called the constant of integration; this means that the derivative
of axZ + c is a. x
(2) x dx = n + 1 + c for n 6= −1.
For example,
Z
Z
n+1
n
√
1
x1/2
√ dx = x−1/2 dx =
+ c = 2 x + c.
1/2
x
Z
Z
sin x dx = − cos x + c
cos x dx = sin x + c
Z
ex dx = ex + c
Z
1
dx = ln x+c
x
x
(3)
;
.
(4)
.
Z
(5)
, or to allow for the possibility of being negative,
(6) The inverse trigonometric functions
also give some integrals, like
Z
x2
9.2.
1
dx = ln |x|+c
x
1
dx = tan−1 x + c .
+1
Basic rules
The rules for dierentiation imply the following basic rules for integration.
1. If a is a constant then
Z
Z
au dx = a u dx ;
for example, 3 cos x dx = 3 cos x dx = 3 sin x + c.
Note that you don't need to write the constant of integration until the nal answer.
2.
Z
Z
Z
R
For example
R
(u ± v) dx =
Z
x
3
(e + x ) dx =
Z
u dx ±
Z
x
e dx +
44
v dx
x3 dx = ex +
x4
+ c.
4
.
Example
9.3.
9.1
.
Z Z 1
1 2
1 2 (x2 )2 + 2(x2 ) +
x +
dx =
dx =
x
x
x
Z
x5
x4 + 2x + x−2 dx =
+ x2 − x−1 + c .
5
2
Denite integrals
The denite integral of f (x) from a toZ b is denoted by
b
f (x) dx,
and it is equal to
[F (x)] = F (b) − F (a),
R
where F (x) = f (x) dx is the indenite integral.
You don't need to include the constant of integration in F (x) if you do it cancels out
so there is no constant of integration in the nal answer.
9.2 Z
a
b
a
Example
.
π/2
Example
9.3
0
h
iπ/2
π
cos x dx = sin x
= sin − sin 0 = 1 − 0 = 1.
2
0
.
Z
2
x2 dx =
h x 3 i2
3
−2
−2
=
23 (−2)3
8 −8
16
−
= −
=
.
3
3
3
3
3
The Fundamental Theorem of Calculus says that the indenite integral R f (x) dx
measures the area under the curve y = f (x) in the region a ≤ x ≤ b.
9.4 What is the area of the triangular region under y = mx between x = 0 and
x=b?
It is
Z
h
i
b
a
Example
.
b
mx2
2
mx dx =
b
=
mb2
.
2
Note that this is just half the base times the height, as expected.
0
9.4.
0
Integration by substitution
To integrate R y dx by substitution, do the following:
(a) Choose a suitable substitution, say u = g(x). Use it to write x in terms of u if necessary.
(b) Find dx/du, either directly, or from the formula dx/du = 1/(du/dx).
(c) Rewrite the integral using the formula
Z
Z
y dx =
45
y
dx
du
du
(this is valid because of the chain rule).
(d) Try to express the integrand y(dx/du)
as a function of u alone, not involving x, so that
the integral becomes of theR form R h(u)du. If you can't do that, the substitution is no use.
(e) Compute the integral h(u) du, then use the substitution u = g(x) to write it in terms
of x.
Z
9.5 Find cos 2x dx using the substitution u = 2x.
We have x = u, so dx/du = . Therefore
Example
.
1
2
1
2
Z
Z
1
1
1
cos 2x dx =
cos u du = sin u + c = sin 2x + c
2
2
2
Z
sin x
dx using the substitution u = cos x + 1.
Example
. Find
cos x + 1
du/dx = − sin x
dx/du = −1/ sin x
Z
Z
Z
sin x 1
1
sin x
dx = −
du = −
du = − ln |u| + c = − ln | cos x + 1| + c .
cos x + 1
u sin x
u
Z
1
√
Example
. Find
dx.
9 − x2
x = 3u
u = x/3
dx/du = 3
Z
Z
1
1
√
√
dx =
3 du
9 − x2
9 − x2
Z
Z
1
1
√
√
3 du =
du = sin−1 u + c
2
2
9 − 9u
1−u
We have
9.6
, so
9.7
, so that
Let
9.8
. Then
. We have
, so
= sin−1 (x/3) + c.
Using the substitution u = x2 + 1, nd the area under the curve y =
x/(x + 1) between x = 0 and x = 3.
Example
2
.
It is
Z
To nd the indenite integral
use the substitution
0
3
x2
x
dx .
+1
Z
x
dx,
x2 + 1
u = x2 + 1
du/dx = 2x
dx/du = 1/(2x)
Z
Z
x
x
1
dx
=
.
du
x2 + 1
x2 + 1 2x
Z
1
1
1
=
du = ln |u| + c = ln |x2 + 1| + c .
2u
2
2
. Then
, so
46
. Therefore
Therefore
Z
0
Special
case.
R
If
3
h1
i3 1
x
1
1
2
dx
=
ln
|x
+
1|
= ln 10 − ln 1 = ln 10.
2
x +1
2
2
2
2
0
f (x) dx = F (x) + c
, andZa and b are constants, then
1
f (ax + b)dx = F (ax + b) + c
a
u = ax + b
Z
1
1
dx = ln(3x + 5) + c
.
3x + 5
3
R
1 (4x + 1)6
+c
.
(4x + 1)5 dx =
4
6
(this follows by putting
9.9
9.10
Example
Example
9.5.
).
.
Integration by parts
To integrate R y dx by parts, do the following:
(a) Write y as a product of two terms, and choose one of them that you can integrate.
(b) Let dv/dx be the term you can integrate, and let v be the integral.
(c) Let u be the other term, and compute du/dx.
(d) Rewrite the integral using
the formula
Z
Z
Z
y dx =
u
dv
dx = uv −
dx
du
v dx
dx
(this follows from the product
formula).
Z
9.11 Find x cos x dx by parts.
The integrand x cos x is the product of x and cos x.
We can integrate both terms, but if you choose x, it won't help in fact, it makes the
integral more complicated. Instead we choose to integrate cos x.
Let dv/dx = cos x, so that v = sin x. (You don't need to include the +c here)
Let u = x. Then du/dx = 1.
Then Z
Z
Z
Z
dv
du
Example
.
x cos x dx =
u
dx = uv −
We know this last integral, soZ we get
dx
dx
v dx = x sin x −
x cos x dx = x sin x + cos x + c .
9.12 Find the denite integral R
First we want the indenite integral Z
Example
.
2
(2x
0
+ 1)e−x dx.
(2x + 1)e−x dx.
47
sin x dx .
Take dv/dx = e andZ u = 2x + 1. ThenZv = −e and du/dx
= 2. Therefore
Z
−x
−x
(2x + 1)e−x dx =
−x
= −(2x + 1)e
Z
+2
u
dv
dx = uv −
dx
du
v dx
dx
e−x dx = −(2x + 1)e−x − 2e−x + c
= −(2x + 3)e−x + c .
Then
Z
2
0
2
(2x + 1)e−x dx = −(2x + 3)e−x 0 = 3 − 7e−2 .
9.13 Find R x sin x dx.
One needs to use integration by parts twice.
Take dv/dx
= sin x, so v = − cos x, and u = x , so du/dx = 2x. Then
Z
Z
Z
Z
Example
2
.
2
x2 sin x dx =
Now we already found
u
dv
dx = uv −
dx
du
v dx = −x2 cos x +
dx
2x cos x dx .
Z
x cos x dx = x sin x + cos x + c
using integration by Zparts, so we get
x2 sin x dx = −x2 cos x + 2x sin x + 2 cos x + c.
9.14 Find R ln x dx.
This doesn't look like a product, but write it as
Example
.
ln x = ln x × 1
with dv/dx = Z1 and u = ln x. Then
v = x and du/dx = 1/x. Therefore
Z
Z
ln x dx = uv −
9.6.
du
v dx = x ln x −
dx
Integration of rational functions
1 dx = x ln x − x + c.
To compute the integral of a rational function, convert it into partial fractions. Then use
the Zfollowing integrals:
dx
1
(1) Zax1+ b dx = a1 lnZ |ax + b| + c. Namely, letting u = ax + b we have du
= a, so
= .
dx
du
a
1
1 1
1
1
Then ax + b dx = u × a du = a ln |u| + c = a ln |ax + b| + c.
48
(2)
Z
. To see this, let
; then we have
1
1 1 +c
u = ax + b
dx
=
−
(ax + b)2
a ax + b
Z
Z
1
1
11
1 1 1
du
=
−
+
c
=
−
+ c.
dx
=
×
(ax + b)2
u2 a
a u
a ax + b
Example
9.15
.
Write
What is
Z
1
dx ?
(x + 1)2 (x + 2)
1
1
1
1
=
+
.
−
2
2
(x + 1) (x + 2)
(x + 1)
x+1 x+2
Then
Z
Z
Z
Z
1
1
1
1
dx =
dx +
dx
dx −
2
2
(x + 1) (x + 2)
(x + 1)
x+1
x+2
1
=−
− ln |x + 1| + ln |x + 2| + c .
x+1
To integrate
rational functions with the quadratic x
Z
(3) x +1 a dx = a1 tan xa + c ;
Z
(4) x +x a dx = 21 ln(x + a ) + c .
9.16
Z
Z
5x + 1
x
2
+ a2
in the denominator, use
−1
2
2
2
2
2
Example
2
.
Z
1
dx = 5
dx +
dx.
2
2
+3
x +3
x +3
√
a= 3
Z
x 5
1
2x + 1
2
−1
√
√ + c.
dx
=
ln(x
+
3)
+
tan
x2 + 3
2
3
3
x2
This ts the formulae with
. Therefore
To integrate rational functions with a more general quadratic in the denominator, complete
the square, and use a substitution.
9.17 Find
Z
x+5
Example
.
We complete the square:
so
x2 + 6x + 18
dx .
x2 + 6x + 25 = (x2 + 6x + 9) + 16 = (x + 3)2 + 16
Z
x+5
dx =
2
x + 6x + 18
Z
49
x+5
dx .
(x + 3)2 + 16
Now use theZ substitution u = x Z+ 3. Then x = uZ− 3, so dx/du = Z1. Therefore
x2
9.7.
x+5
u+2
u
1
dx =
du =
du + 2
du
2
2
2
+ 6x + 25
u + 16
u + 16
u + 16
2
1
= ln(u2 + 16) + tan−1 (u/4) + c
2
4
1
1
2
= ln(x + 6x + 25) + tan−1 ((x + 3)/4) + c.
2
2
Worked examples
Example
9.18 Calculate the indenite integrals:
.
(i) R tan x dx
(ii) R x ln x dx
(iii) R sin (x) dx
(iv) e dx
9.19 Calculate the integralZ
R
2
2x
Example
.
9.20 Calculate the integral
Z
2
1
Example
1
dx .
x(x + 1)2
.
π/4
0
cos x − sin x
dx .
cos x + sin x
50