Chapter 4
Methods of Integration
§4.1 Integration by Substitution
The chain rule for differentiation allows us to find
chain rule:
d 100
u
dx
=
d 100 d u
u .dx .
du
d
(x
dx
+ 1)100 by letting u = x + 1, and applying the
There is no chain rule for integration, but
a kind of a reversal to the chain rule for differentiation.
R there is100
R
For example, suppose we want to find (x + 2) d x. This is not a Rstandard integral, but u 100 d u is
and this is the hint. We let u = x + 2. This means that the integral is u 100 , d x. However, in this form,
integration is not possible because the variable of integration and the variable of the function don’t
match up. We need to relate d x to d u. For this, we havee u = x + 2 and so dd ux = 1. We can read this as
d u = d x. Therefore,
Z
Z
100
(x + 2) d x = u 100 d u where u = x + 2,
= u 101 /101 + C
=
(x + 2)101
101
+C
Notice that at the end we replaced u by x. Its the polite thing to do.
This form of integration, where the variable is switched to make the integral appear as a standard integral, is called integration by substitution. Unlike the chain rule for differentiation however, the procedure is not straightforward, because it may not be at all obvious what is the correct substitution to
make.
4.1 Example
1.
Z
cos(3t + 2) d t =
25
Z
cos u
du
3
du
dt
where u = 3t + 2 and so
= 3 implies d u = 3d t
=
1
3
1
Z
cos u d u
= [sin u + C ]
3
sin(3t + 2)
=
+ C ′.
3
Notice that there is no point in writing C3 since this is just a different constant function. C ′
indicates that we can still get any constant, but it has changed from the previous line C .
2.
Z
1
3t + 6
dt =
where u = 3t + 6, and so
Z
du
dt
=
1
3
1
u
dt
= 3 implies d u = 3d t
Z
1
u
1
1
= (ln |u| + C ) =
+ C ′.
3
3 ln |3t + 6|
du
3. In this example, it is not so clear beforehand what we should substitute for:
Z
3x 2 (x 3 + 1)5 d x.
Should we let u = 3x 2, or u = 3x 2 (x 3 + 1)5 or u = x 3 + 1 or something else. It happens that the
most fruitful substitution is u = x 3 + 1.
Z
Z
3x 2 (x 3 + 1)5 d x = 3x 2 u 5 d x
Now,
du
dx
= 3x 2 so the integral becomes
=
=
=
Z
u 6d u
u6
+C
6
(x 3 + 1)6
6
+C
Why was this the correct substitution? The reason lies behind the fact that substitution is akin to
a reversal of the chain rule. The chain rule says that if h(x) = f (g (x)) then h ′(x) = f ′ (g (x))g ′ (x).
R
So f ′ (g (x))g ′ (x) d x is easy to integrate– we just substitute u = g (x) because then dd ux = g ′ (x)
and so d u = g ′ (x)d x. In other words,
Z
Z
′
′
f (g (x))g (x)d x = f ′ (u)d u = f (u) + C = f (g (x)) + C .
Therefore, when deciding what substitution to make, keep an eye out for a function and its
derivative appearing somewhere in the integral.
26
4.
x2
Z
where u = x 3 + 1. Since
du
dx
p
x3 + 1
dx =
Z
x2
p dx
u
1
Z
= 3x 2 , x 2 d x = 13 d u
=
3
1
1
u− 2 d u
1
= (2u 2 + C )
3
2Æ 3
x + 1 + C ′.
=
3
5. To integrate
we get
Z
p
x sin2 (x 3/2 + 1)d x we substitute u = x 3/2 + 1. Since
Z
p
x sin2 (x 3/2 + 1)d x =
2
3
Z
du
dx
1
1
= 32 x 2 , d u = 23 x 2 d x and
sin2 u d u.
The substitution has simplified the integral but still has not made it into a standard integral. We
need another trick to integrate sin2 and we use the formula sin2 x = 21 (1 − cos2x) to write
=
1
3
1
Z
(1 − cos2u) d u
1
= (u − sin 2u + C )
3
2
1 3/2
1
= (x + 1) − sin(2x 3/2 + 1) + C ′ .
3
6
6.
Z
5
cos x sin x d x =
Z
u 5 (−d u)
u6
+C
6
cos6 x
+C
=−
6
=−
Definite integration by Substitution When calculating a definite integral, which involves a substitution one has to be careful to substitute for the limits. For example
Z
x2
1
0
p
x3 + 1
dx =
27
Z
x=1 1 d u
3
x=0
p
u
where we have substituted u = x 3 + 1. Now make sure to change the limits of integration to the new
variable. When x = 0, u = 1 and when x = 1, u = 2. So we get
=
1
3
2
Z
2
1
u−2 d u
1
2
= u 1
3
p
2
= ( 2 − 1).
3
1
2
§4.2 Integration by Parts
The product rule for differentiation says that
d
dx
u(x)v(x) = u ′ (x)v(x) + u(x)v ′ (x).
If we integrate both sides with respect to x, we have
Z
Z
d
u(x)v(x)d x =
u ′ (x)v(x) + u(x)v ′(x) d x
dx
and so, forgetting about constants of integration for the moment,
Z
Z
′
u(x)v(x) = u (x)v(x) d x + u(x)v ′ (x d x.
or in other words,
Z
′
u(x)v (x) d x = u(x)v(x) −
Z
u ′ (x)v(x)d x
This is the formula for integration by parts.
R
4.2 Example
(i) x sin 2x d x. Here, the integrand is the product of two functions f (x) = x and
g (x) = sin 2x. The formula for integration by parts does not tell you which function to pick for
u and which to pick for v ′ . The formula changes the integrand by differentiating u and integrating
v ′ . We want this change to make the integrand simpler. The correct choice is to take u(x) = x
and v ′ (x) = sin 2x, the reason being that x simplifies when differentiated while sin 2x does not
become any more complicated when it is integrated. We have u(x) = x =⇒ u ′ (x) = 1 while
v ′ (x) = sin 2x =⇒ v(x) = − 12 cos 2x.
Z
Z
− cos 2x
− cos 2x
− 1.
dx
x sin 2x d x = x
2
2
Z
−x cos 2x 1
cos(2x) d x
=
+
2
2
−x cos 2x 1
=
+ sin 2x + C .
2
4
A rule of thumb when trying to use integration by parts is to let u be the “polynomial part”
which reduces its degree when differentiated, and to let v be the trigonometric or exponential
part, which doesn’t get significantly more complicated when integrated.
28
(ii)
Z
x e −7x d x =
Z
u(x)v ′ (x) d x
where u(x) = x and v ′ (x) = e −7x .
= u(x)v(x) −
Since u(x) = x, we get u ′ (x) = 1. We find v(x) =
w = −7x, to get v(x) = − 71 e −7x .
1
−7x
R
Z
u ′ (x)v(x) d x
v ′ (x) d x =
1
R
e −7x d x by the substitution,
Z
= − xe
+
e −7x d x
7
7
1
1
= − x e −7x − e −7x . + C
7
49
Sometimes, you have to persevere, and use integration by parts more than once.
(iii)
Z
2
x cos x d x =
Z
u(x)v ′ (x)d x
where u(x) = x 2 and v ′ (x) = cos 2x gives u ′ (x) = 2x and v(x) = 12 sin 2x.
Z
u ′ (x)v(x) d x
Z
1 2
= x sin 2x − 2 x sin 2x d x.
2
= u(x)v(x) −
Now we have to do it again! This time take u(x) = x to get u ′ (x) = 1 and v ′ (x) = sin 2x to get
v(x) = − 21 cos 2x.
1
Z
1
1
= x sin 2x − 2 − x cos 2x − (− cos 2x) d x
2
2
2
1
sin 2x
= x 2 sin 2x + x cos 2x −
+C.
2
4
2
In general, if p(x) is a polynomial of degree n then to integrate p(x) cos(a x), p(x) sin(a x) or
p(x)e ax , first split it into a sum of integrals of the form x m cos(a x) where m ≤ n. Each one of
these integrals requires n applications of integration by parts.
To integrate, p(x) ln(a x), letting u(x) = ln(a x) works better...
(iv)
Z
3
x ln x d x =
Z
u(x)v ′ (x) d x
29
where u(x) = ln x and v ′ (x) = x 4 gives u ′ (x) =
1
x
and v(x) = 15 x 5 .
Z
= u(x)v(x) − u ′ (x)v(x) d x
Z
11 5
1 5
x dx
= x ln x −
5
x5
Z
1
1 5
x4 d x
= x ln x −
5
5
1
1
= x 5 ln x − x 5 .
5
25
(v) Parts can be used to
R integrate Rln x, even though we have to artificially treat the integrand as a
product. We write ln x d x = 1. ln x d x and take u = ln x, v ′ (x) = 1 to get
Z
ln x d x =
Z
(ln x)1 d x
Z
1
xdx
= x ln x −
x
Z
= x ln x − 1 d x
= x ln x − x.
(vi) Sometimes, you can get the answer by going around in circles.
Z
Z
e x sin x d x = u(x)v ′ (x) d x
where u(x) = e x and v ′ (x) = sin x. Thus u ′ (x) = e x and v(x = − cos x so the integration by parts
formula gives
x
= −e cos x +
Z
e x cos x d x.
Now integrate by parts again, this time u(x) = e x and v ′ (x) = cos x to get
x
x
Z
x
= −e cos x + e sin x − e sin x d x
Z
x
= e (sin x − cos x) − e x sin x d x.
It appears we are back where we started, but this time we have an equation involving the quantity
we want to find.
Z
2 e x sin x d x = e x (sin x − cos x)
Z
ex
x
e sin x d x = (sin x − cos x) + C .
2
30
§4.3 Integration by Partial Fractions
R 1
To integrate x+4
d x is a simple matter of substituting u = x + 4. However, if the denominator has a
R
1
d x.
higher degree then a substitution is unlikely to work. For example,
2
x +2x−3
1
1
− x+3
) then the integration is straightforward.
= 41 ( x−1
‚Z
Œ
Z
1
dx
dx
1
−
dx =
4
x −1
x +3
x 2 + 2x − 3
1
= (ln |x − 1| − ln |x + 3| + C )
4 1 x − 1
+C.
= ln 4
x + 3
However, if one “notices” that
Z
1
x 2 +2x−3
This is the idea behind integration by partial fractions. Given
factors, g (x) = g1 (x)g2 (x) . . . gn (x), we try to write
4.3 Example
1. Integrate
R
5x−13
d x.
(x−3)(x−2)
f (x)
g (x)
as
A1
g1 (x)
+
f (x)
d x where g (x)
g (x)
a2
An
+ · · · + g (x)
and
g2 (x)
n
R
Let us try to write the integrand as
5x − 13
(x − 3)(x − 2)
=
=
A
+
A
x−3
+
is a product of
then integrate.
B
.
x−2
B
x −3 x −2
A(x − 2) + B(x − 3)
(x − 3)(x − 2)
and so
5x − 13 = A(x − 2) + B(x − 3).
This has to hold for all x so if we take x = 2 we see that −3 = B(−1) so B = 3. Taking x = 3 gives
2 = A. Thus
2
3
5x − 13
=
+
(x − 3)(x − 2) x − 3 x − 2
and the integral becomes
Z
Z
Z
5x − 13
2
3
dx =
dx +
dx
(x − 3)(x − 2)
x −3
x −2
= 2 ln |x − 3X + 3 ln |x − 2| + C .
R x+3
2. Integrate
d x.
3
2x −8x
B
C
= Ax + x−2
2x 3 − 8x factorises as x(x − 2)(x + 2) so we try to write x+3
+ x+2
. Adding fractions
2x 3 −8x
and equating the numerator gives x + 3 = A(x − 2)(x + 2) + B(x)(x + 2) + C (x)(x − 2).
When x = 2 this gives 5 = B(2)(4) and so B = 58 . Taking x = 0 gives 3 = −4A so A = − 34 . Lastly,
x = −2 gives 1 = C (−2)(−4) so C = 81 . Thus
Z
Z
Z
Z
3 1
5
1
1
1
x +3
d
x
=
−
d
x
+
d
x
+
dx
4 x
8 x −2
8 x +2
2x 3 − 8x
5
1
3
= − ln |x| + ln |x − 2| + ln |x + 2| + C .
4
8
8
31
3.
R
1
d x.
x 2 −3x+2
=
For this, we factorise x 2 −3x+2 = (x−2)(x−1). Equating 2 1
x −3x+2
R
1
−1
1
+ x−1 . Thus,
d x = ln |x − 2| − ln |x − 1| + C .
2
x−2
A
+ B
x−2 x−1
gives 1x 2 − 3x + 2 =
x −3x+2
4. There is a slight complication if a linear factor repeats in the denominator, e.g.
case, try writing
f (x)
(x−a)2
=
For example, to calculate
A
x−a
R
+
B
.
(x−a)2
6x+7
d x,
(x+2)2
R
f (x)
d x.
(x−a)2
In this
we write
6x + 7
(x + 2)2
=
A
x +2
+
B
(x + 2)2
to get 6x + 7 = A(x + 2) + B. The substitution x = −2 gives B = −5 while x = 0 gives 7 = 2A − 5
so A = 6. Now the integral becomes
Z
Z
Z
6x + 7
1
dx =6
d x − 5 d x(x + 2)2
x +2
(x + 2)2
(
= 6 ln |x + 2| − 5 + 2)−1 + C .
x
5.
R
dx
(x+2)(x−1)2
§4.4 Logarithm and exponential functions
You can forget everything you know about e x and ln x, we’re gooing to start from scratch.
§4.4.1 Natural logarithm
4.4 Definition The natural log function, ln x is defined by
Zx
1
dt
ln x =
1 t
x > 0.
The graph of the log function looks like Figure 4.1.
We make some remarks on this important function.
4.5 Note (a) Since 1/x is continuous on (0, ∞), the integral exists for all x > 0,
R1
(b) ln 1 = 1 1t d t = 0,
(c) ln x is defined only for x > 0, and is not defined for zero or any negative number,
(d)
d
dx
ln x = x1 ,
(e) ln x is an increasing function and hence. . .
(f) ln x is injective and so has an inverse,
32
1
0
1
2
x
3
4
0
-1
-2
-3
Figure 4.1: Graph of ln x
33
5
6
(g) ln x > 0 when x > 1 and ln x < 0 when 0 < x < 1,
(h) ln x is unbounded. That is ln x → ∞ as x → ∞.
Some properties of ln x which we will prove are the following.
(i)
d
dx
ln x = 1x ,
(ii) ln xy = ln x + ln y,
(iii) ln x1 = − ln x,
(iv) ln yx = ln x − ln y,
(v) ln x n = n ln x,
(vi) ln x is a strictly increasing function on (0, ∞).
(Of course these are all for x, y > 0.
Proofs.
(i) The first part is an easy consequence of the fundamental theorem of calculus ?? which says that
Rx
d
ln x = ddx 1 1t d t = x1 .
dx
(ii) Here we use the result from differential calculus which says that if f ′ (x) = g ′ (x) at each point of
an interval I then there is a constant C such that f (x) = g (x) + C on I . sa-learn –mbox –spam
/mail/SpamLow ; sa-learn –mbox –ham /mail/saved-messages
Consider the functions f (x) = ln xy and g (x) = ln x+ln y for some fixed number y. f ′ (x) =
1
y
xy
=
1
x
and g ′ (x) = x1 + 0 = f ′ (x). So f and g agree up to a constant. That is ln xy = ln x + ln y + C . To
find C , take x = 1. Since ln 1 = 0 this gives ln y = 0 + ln y + C . So C = 0 and
ln xy = ln x + ln y.
(iii) Now we can use part (ii) to write 0 = ln 1 = ln x x1 = ln x + ln x1 and so ln x1 = − ln x.
(iv) ln yx = ln x 1y = ln x + ln 1y = ln x − ln y.
(v) ln x 2 = ln x.x = ln x + ln x = 2 ln x. Finish using proof by induction.
(vi) Let f (x) = ln x. Since f ′ (x) =
1
x
> 0 for all x > 0, f is an increasing function.
§4.4.2 Exponential function(s)
Since ln(x) is increasing for x > 0, it is an injective function on (0, ∞) by Theorem 1.7 it has an inverse.
Moreover, since ln : (0, ∞) → R, we have ln−1 : R → (0, ∞).
4.6 Definition We define the exponential function exp by
exp(x) = ln−1 (x).
34
That is exp(x) is the number y such that ln y = x. In particular, we have
ln(exp x) = x = exp(ln x)
Graph of exp x.
4.7 Note We make some remarks.
1. exp x is defined for all x ∈ R, that is the domain of exp x is (−∞, ∞),
2. exp x is always > 0, that is, Range(exp) = (0, ∞),
3. exp 0 = 1,
4. exp(1) is denoted by e and is approximately equal to 2.718281828459045236. In particular, we
have that ln e = 1.
5. exp is an increasing function.
We will prove some important properties of the exponential function.
4.8 Lemma The exponential function has the following properties.
1.
d
dx
exp x = exp x ,
2. exp(x + y) = exp x exp y ,
3. exp(−x) =
1
,
exp x
4. exp(x − y) =
exp x
,
exp y
5. (exp x)y = exp(xy).
6. exp x = e x ,
Proofs.
1. Differentiate both sides of the equation ln(exp x) = x (using the chain rule) to get
1
d
exp x d x
which of course implies that
d
dx
exp x = 1
exp x = exp x.
2. From the properties of ln, ln(exp(x + y)) = x + y = ln(exp x) + ln(exp y) = ln(exp x exp y) and
since ln is an injective function we must have that exp(x + y) = exp x exp y.
3.
4.
5. (Similar to 2.)
35
6. This follows from 5: e x = (exp 1) x = exp(1.x) = exp x.
EX
4.9 Exercise Show that e x is an increasing function. More generally, show that if f is an increasing function (and so has an inverse) then f −1 is an increasing function. Hint: Two ways–
either use Definition 2.15 of increasing, or use the derivative characterisation of increasing (Corollary 2.16) and calculate the derivative of f −1 in terms of the derivative of f using the chain rule.
Of course, the second method only works if f and f −1 are differentiable functions.
4.10 Note
1. It pays to be
R familiar with the rules for manipulating logs and exponentials. For example, try to calculate ln(31x 7 8) d x.
2. Here, we have just looked at “the” log and exponential functions, that is, the natural log ln x and
the exponential to the base e. There are, of course, other exponential functions such as f (x) = 10 x
or g (x) = b x where b is any positive number. Each of these has an inverse, the log function to
the base b , written as logb .
y = b x ⇐⇒ x = logb y
In this notation, ln y = loge y.
§4.5 Derivatives of Inverse Trigonometric Functions
The derivatives of the six inverse functions are listed in the standard list of formulae. We’ll prove a
couple of them. First, a little result to help calculate the derivative of any inverse function.
d
dx
4.11 Lemma Let f and f −1 be differentiable. Then
f −1 (x) =
1
′
f (f
−1
(x))
.
Proof. Exercise ƒ
4.12 Example We calculate the derivative of x 7→ sin−1 x for x ∈ (−π/2, π/2). Let f (x) = sin x then
we want ddx f −1 (x). Using Lemma 4.11, this is
d
dx
f −1 (x) =
=
=
4.13 Example We can calculate the
dy
dx
|x|
x 2 −1
f (f
−1
(x))
1
cos( f −1 (x))
1
cos(sin−1 x)
1
.
=p
1 − x2
for y = tan−1 x in the same way. Letting f (x) = tan x we have
Other inverse trigonometric derivatives are
p−1
1
′
d
dx
cos−1 x = p−1 2 ,
1−x
.
In particular we have the following integrals,
36
d
dx
sec−1 (x) =
|x|
p1
x 2 −1
, and
d
dx
cosec−1 (x) =
1.
R
2.
R
3.
R
p1
1−x 2
d x = sin−1 x + C
1
dx
1+x 2
|x|
p1
= tan−1 x + C
x 2 −1
d x = sec−1 x + C
These give us a new bunch of functions which we can integrate.
4.14 Example Find
R
1
d x.
4x 2 +9
Notice that if the denominator were 4x 2 − 9 then we would integrate using parts. But 4x 2 + 9 doesn’t
factorise and so the integral takes quite a different form.
Z
Z
1
1
1
dx
dx =
2
2
9 ( 3 x)2 + 1
4x + 9
Z
1
1 3
=
dx
9 u2 + 1 2
1
= tan−1 u + C
6
1
2x
= tan−1 ( ) + C .
6
3
§4.5.1 Integrals of Inverse Trigonometric functions
Most of the inverse trigonometric functions can be integrated quite easily using integration by parts.
4.15 Example Find
R
sin−1 x d x.
Z
sin
−1
xdx =
Z
1. sin−1 x d x
(let u = sin−1 x and v ′ = 1)
= x sin−1 x −
= x sin
−1
x−
Z
Z
(make the substitution u = 1 − x 2)
1
p
1 − x2
x
p
1 − x2
xdx
dx
1 1
− u− 2 d u
2
Æ
= x sin−1 x + 1 − x 2 + C .
= x sin
−1
37
x−
Z
4.16 Example Find
R
tan−1 x d x.
Z
−1
tan
xdx =
Z
1. tan−1 x d x
(let u = tan−1 x and v ′ = 1)
x−
Z
= x tan−1 x −
Z
= x tan−1 x −
Z
−1
= x tan
1
1 + x2
x
1 + x2
x dx
dx
(make the substitution u = 1 + x 2)
1
1
2
u −1 d u
= x tan−1 x − ln(u) + C
2
1
= x tan−1 x − ln(1 + x 2) + C .
2
38