Techniques of Integration
Math 203
Lecture 1
Recall on Integration
Techniques of Integration
Lecture Outline
1
Recall on Integration
Indefinite Integrals
Evaluation of Indefinite Integrals
Evaluation of Definite Integrals
2
Techniques of Integration
Substitution Method
Integration by Parts
Techniques of Integration
Recall on Integration
Techniques of Integration
Recommended Reading
Thomas’ Calculus (13th ed.): sections 5.5, 8.1 and 8.2. Or
equally,
Stewart Calculus (7th ed.): sections 4.4, 4.5, and 7.1
Techniques of Integration
Recall on Integration
Techniques of Integration
Indefinite Integrals
Evaluation of Indefinite Integrals
Evaluation of Definite Integrals
Antiderivatives
A function F is an antiderivative of f on an interval I if F 0 (x) = f (x) for all x in I .
Examples
i.
ii.
iii.
1 5
1 5
4
x
= x , for all x in R =⇒ F (x) = x is an antiderivative of f (x) = x 4 on R.
dx 5
5
d
1
1
sin 2x = cos 2x, for all x in R =⇒ F (x) = sin 2x is an antiderivative of f (x) = cos 2x on R.
dx 2
2
1
1
d
[ln x] = , for all x in (0, +∞) =⇒ F (x) = ln x is an antiderivative of f (x) =
on (0, +∞).
dx
x
x
d
Properties of Antiderivatives
(i)
If F (x) is an antiderivative of f (x) on an interval I then so is F (x) + C for any constant C .
(ii)
If F and G are two antiderivatives of f over an interval I then G (x) = F (x) + C for some constant C .
Proof of the claim
d
0
0
0
0
F (x) = G (x) =⇒ G (x)−F (x) = 0 =⇒
[G (x) − F (x)] = 0
=⇒
G (x)−F (x) = C .
|{z}
dx
true only over intervals
Remarks
i.
The premise of property (ii) is that two different anitderivatives of a function on an interval will differ only
by a constant.
ii.
Property (ii) fails if I is not an interval. Can you give a counter example?
Techniques of Integration
Recall on Integration
Techniques of Integration
Indefinite Integrals
Evaluation of Indefinite Integrals
Evaluation of Definite Integrals
Indefinite Integrals
Z
The indefinite integral of a function f (x) on an interval I , denoted
f (x)dx, stands for the most general
antiderivative of f on I .
Remark
Assume that F (x) is an antiderivative of f (x) on an interval I . By property (ii) on the preceding slide, any other
antiderivative of f (x) on I is given by F (x) + C , where C is a constant. Thus,
Z
f (x)dx = F (x) + C
(C an arbitrary constatnt).
Examples
F (x) =
F (x) = 15 x 5 is an antiderivative of f (x) = x 4
Z
1 5
4
on R. Therefore,
x dx = x + C .
5
F (x) = ln x is an antiderivative of f (x) = x1
Z
1
on (0, +∞). Therefore,
dx = ln x + C .
x
1
2
−1
sin
2x is an antiderivative of
1
f (x) = p
on the interval (−1/2, 1/2),
1 − 4x 2
!
d
2
−1
since
(sin
2x) = p
.
2
dx
1 − 4x
Z
dx
1
−1
Therefore,
= sin
2x + C .
p
2
1 − 4x 2
Techniques of Integration
Recall on Integration
Techniques of Integration
Indefinite Integrals
Evaluation of Indefinite Integrals
Evaluation of Definite Integrals
Basic Indefinite Integrals
The following basic integrals should be learned by heart. All
the integrals follow from the simple fact that the derivative of
the right hand side is exactly the integrand on the left hand
side.
For a > 0:
Z
dx
−1
= sin
p
a2 − x 2
x
+ C , on |x| < a
a
1
x
dx
−1
= tan
+ C , on R
a2 + x 2
a
a
Z
dx
x
−1
= sinh
+ C , on R
p
a
x 2 + a2
Z
dx
x
−1
= cosh
+ C , on x > a.
p
a
x 2 − a2
Z
Z
n
x dx =
Z
α
x dx =
x n+1
n+1
x
+ C , on R
(where n ∈ N)
α+1
α+1
+ C , on (0, +∞)
(where
α 6= −1)
Z
1
dx = ln |x| + C , on (0, +∞) or (−∞, 0)
x
Z
1 kx
kx
e dx = e + C , on R
k
Z
1
kx
kx
a dx =
a + C , on R
(a > 0 and a 6= 1)
k ln a
Z
1
cos kxdx =
sin kx + C , on R
k
Z
1
sin kxdx = − cos kx + C , on R
k
Z
1
cosh kxdx =
sinh kx + C , on R
k
Z
1
sinh kxdx =
cosh kx + C , on R
k
Remarks
What about evaluating non basic integrals, that is,
Z
f (x)dx where f (x) is not the derivative of a function
in frequent use, like the ones in the above list?
In fact, evaluating integrals is a difficult problem and
there is no general recipe to find explicit expressions for
antiderivatives.
There are, however, certain types of integrals which
one can tackle with known techniques that we will
explore in the first couple of weeks this semester.
Techniques of Integration
Indefinite Integrals
Evaluation of Indefinite Integrals
Evaluation of Definite Integrals
Recall on Integration
Techniques of Integration
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus
Z x
1
If f is continuous over [a, b] then F (x) =
f (t)dt is an antiderivative of f on (a, b). That is, F
a
is differentiable and
Z x
d
f (t)dt = f (x).
dx a
If f is continuous over [a, b] and F is any antiderivative of f on [a, b], then
Z b
f (x)dx = F (b) − F (a).
0
F (x) =
2
a
Remark
Part (2) of the fundamental theorem of calculus justifies the interest we have in evaluating indefinite integrals.
They help us compute definite integrals which are abundant in applications (more about this later).
Example
Z 1
dx
0
x2 + 1
−1
= tan
x
i1
0
−1
= tan
−1
1 − tan
0=
π
4
−0=
Techniques of Integration
π
4
.
Recall on Integration
Techniques of Integration
Substitution Method
Integration by Parts
Substitution Method
Z
u0
u
Z
Substitution (or Change of Variable) Method
Z
0
e
Z
Z
f (u(x))u (x)dx =
Z
Proof
Notice that it is enough to show that the derivative of the
right hand side is the integrand
on the left hand side. If
Z
u = u(x) and F (u) =
d
dx
F (u)
=
|{z}
chain rule
f (u)du, then
dF du
du dx
0
Z
Z
0
= f (u)u = f (u(x))u (x).
Z
Z
Z
The substitution method implies all the following integrals:
Z
u n+1
n 0
u u dx =
+C
(where n ∈ N)
n+1
Z
α+1
u
α 0
u u dx =
+C
(where α 6= −1))
α+1
Z
Z
u dx =
1 ku
e +C
k
1
ku
a +C
(a > 0 and a 6= 1)
k ln a
1
0
sin ku + C
cos (ku)u dx =
k
1
0
sin (ku)u dx = − cos ku + C
k
1
0
cosh (ku)u dx =
sinh ku + C
k
1
0
sinh (ku)u dx =
cosh ku + C
k
0
u dx
u
−1
= sin
+ C,
(a > 0)
p
2
2
a
a −u
1
u 0 dx
u
−1
= tan
(a > 0)
+ C,
a2 + u 2
a
a
u 0 dx
u
−1
= sinh
(a > 0)
+ C,
p
a
u 2 + a2
0
u dx
u
−1
= cosh
+ C,
(a > 0).
p
a
u 2 − a2
a
f (u)du
dx = ln |u| + C
ku 0
ku 0
u dx =
Techniques of Integration
Recall on Integration
Techniques of Integration
Substitution Method
Integration by Parts
Integration by Parts Formula
Integration by parts is a simple yet very useful technique of integration that follows from the product rule for
derivatives.
Integration by Parts Formula
Z
Z
0
0
uv dx = uv −
u v dx
Z
Z
Proof
0
0
0
0
0
0
(uv ) = u v + uv =⇒ uv = (uv ) − u v =⇒
0
0
(uv ) dx −
uv dx =
Z
Z
0
u v dx = uv −
0
u v dx.
Example
Z
x
xe dx =
Z
x
x
Z
x |{z}
e dx = |{z}
x |{z}
e −
|{z}
u
v0
u
v
x
x
x
1 |{z}
e dx = xe − e + C
|{z}
u0
v
Idea
Integration by parts is applied to integrands which are products f (x)g (x). One thinks of one factor as
being u and of the other as being v 0 .
The integration by parts formula transfers the derivative from the v factor to the u factor. The result is
that the integral in the right hand side of the formula is usually simpler than that on the left hand side.
This point should guide your choice of u and v .
Techniques of Integration
Recall on Integration
Techniques of Integration
Substitution Method
Integration by Parts
Integration by Parts Examples
Examples
Z
i.
2 x
2 x
x
x
x e dx = x e − 2xe + 2e + C .
Z
Z
Z
x sin
x−
x cos
|{z}
|{z}
| {z x} dx = |{z}
x cos x dx =
v0
u
u
x dx
1 sin
|{z}
|{z}
u0
v
v
Z
= x sin x −
sin x dx
| {z }
= x sin x + cos x + C .
a simpler integral
Remark
If one uses the differential notation that dy = y 0 dx
then the integration by parts may be presented in the
differential form as follows:
Z
reduce the integral to
Z
0
u |v {z
dx} = uv −
Notice that the presence of x is what makes evaluating the
integral difficult. Transfering
a derivative to this term will
Z
Z
sin x dx thus simplifying
0
v |u {z
dx}
dv
du
Z
⇐⇒
udv = uv −
v du
integration.
ii.
Z
2 x
Z
x e dx =
2
x
2
x
u
v0
u
v
x |{z}
e dx = |{z}
x |{z}
e −
|{z}
2 x
Z
=x e −2
|
Z
x
2x |{z}
e dx
|{z}
u0
v
Example
x
xe dx
{z }
a simpler integral
Z
Z
ln x dx =
Z
ln x |{z}
dx = |{z}
ln x |{z}
x −
|{z}
u
dv
u
v
1
x
dx
|{z}
x
v |{z}
du
A
integration by parts allows us to evaluate
R second
xe x dx (as computed on the preceding slide) to obtain
Z
= x ln x −
Techniques of Integration
dx = x ln x − x + C .
Recall on Integration
Techniques of Integration
Substitution Method
Integration by Parts
Integration by Parts Examples
Example
Z
Evaluate I =
x
e cos x dx.
Solution
u
Z
v 0 dx
z }| { z x}| {
cos x e dx =
I =
uv
dv
Z
vdu
u
z }| { Z z }| {
z }| { z }|x {
x
x
cos x d(e ) = e cos x −
e d(cos x)
Z
x
= e cos x +
|
x
e sin x dx
{z
}
not any simpler than I .Integrate by parts once more!
Z
x
= e cos x +
x
x
sin x d(e )
x
Z
x
Z
= e cos x + e sin x −
x
x
e d(sin x)
= e cos x + e sin x −
|
x
e cos x dx
{z
}
I +const
Therefore,
x
x
x
x
I = e cos x + e sin x − I + const =⇒ 2I = e cos x + e sin x + const =⇒ I =
1 x
1 x
e cos x + e sin x + C .
2
2
Techniques of Integration
Recall on Integration
Techniques of Integration
Substitution Method
Integration by Parts
Integration by Parts of Definite Integrals
If we combine the formula for integration by parts with Part (2) of the Fundamental Theorem of Calculus, we
obtain integration by parts forumla for definite integrals:
Integration by Parts formula for Definite Integrals
Z b
Z b
b
0
uv dx = uv
−
a
a
0
u v dx
a
Example
u
Z 1
−1
tan
x dx =
0
v0
Z 1 z }| { z}|{
−1
tan
x . 1 dx
0
−1
= x tan
x
−1
= 1. tan
i1
0
Z 1
−
x
0
−1
tan
x dx
0
−1
1 − 0. tan
Z 1
x
0
1 + x2
0−
dx
w0
=
π
4
−
1
Z 1
z}|{
2x
2
0
1+x
| {z }
2
dx
w
=
π
4
−
1
i
π
1
π
ln 2
2 1
ln(1 + x ) =
− (ln 2 − ln 1) =
−
0
2
4
2
4
2
Techniques of Integration
Recall on Integration
Techniques of Integration
Substitution Method
Integration by Parts
Application of Integration by Parts
Example
Thus,
Z
n
cos x dx, where n ≥ 0 is a fixed integer.
Let Fn (x) =
1
Fn (x) =
2
3
4
1
n−1
n−1
cos
x sin x +
Fn−2 (x)
n
n
Use the reduction formula to evaluate F4 (x).
Use the reduction formula to show that for n ≥ 2,
Z π/2
Z π/2
n−1
n−2
n
cos
x dx.
cos x dx =
n
0
0
Deduce that for odd powers n = 2k + 1 of cosine, the
following integration formula is obtained:
Z π/2
2.4.6. . . . .2k
2k+1
cos
x dx =
.
3.5.7. . . . .(2k + 1)
0
Solution
1
Using integration by parts,
Z
Fn (x) =
n−1
cos
n−1
= cos
n−1
= cos
n−1
x sin x + (n − 1)Fn−2 (x) − (n − 1)Fn (x)
n−1
x sin x + (n − 1)Fn−2 (x)
Fn (x) = cos
Prove the so called ”reduction formula” for n ≥ 2:
=⇒ nFn (x) = cos
=⇒ Fn (x) =
2
1
n
n−1
cos
x sin x +
n−1
n
Fn−2 (x).
By the reduction formula,
3
F2 (x)
4
1
and F2 (x) = cos x sin x + F0 (x)
2
2
F4 (x) =
1
4
1
3
cos x sin x +
Z
However, since F0 (x) =
Z
x cos x dx =
x sin x + (n − 1)
x d(sin x)
n−2
x sin x dx
n−2
x(1 − cos x)dx
cos
Z
x sin x + (n − 1)
n−1
cos
Z
dx = x + const,it is
concluded that
cos
2
2
F4 (x) =
=
1
4
1
4
3
3
3
4
3
cos x sin x +
cos x sin x +
Techniques of Integration
8
1
2
cos x sin x +
cos x sin x +
3x
8
x
2
+ const
+ const.
Recall on Integration
Techniques of Integration
Substitution Method
Integration by Parts
Application of Integration by Parts
3
Applying equation (∗) repeatedly,
Z π/2
h
iπ/2
cos x dx = Fn (x)
n
0
0
=
h1
|n
=
2k
I2k+1 =
iπ/2 n − 1 h
iπ/2
+
x sin x
Fn−2 (x)
0
0
n
{z
}
2k + 1
n−1
cos
2k
=
Z π/2
n−2
cos
n
4
If we let In =
I2k−3
2k − 1
2k − 2 2k − 4
2k − 3
I2k−5
.
.
.
x dx.
0
2k − 2 2k − 4
2k
=
Z π/2
2k − 2
2k + 1 2k − 1
=0
n−1
2k + 1 2k − 1 2k − 3
.....
n
cos x dx, then by question
4 2
I1 .
5 |{z}
3
I3
0
(3), for all integer n ≥ 2,
In =
n−1
Z π/2
In−2
(∗)
n
For n = 2k + 1, an odd integer, applying equation
(∗) gives:
Z π/2
I2k+1 =
=
cos
2k+1
0
2k
2k + 1
x dx
However, I1 =
cos x dx = 1.Therefore,
0
I2k+1 =
2.4.6. . . . .2k
3.5.7. . . . .(2k + 1)
,
which proves the integration formula.
Remark
I2k−1
Z π/2
For example,
7
cos x dx =
0
Techniques of Integration
2.4.6
3.5.7
=
16
35
.