Outline
MS121: IT Mathematics
1
Integration by Substitution: The Technique
Integration
2
Integration by Substitution: Worked Examples
3
Integration by Parts: The Technique
4
Integration by Parts: Worked Examples
5
Two Definite Integral Examples
Two Techniques of Integration
John Carroll
School of Mathematical Sciences
Dublin City University
Integration by Substitution: The Technique
Integration by Substitution: The Technique
Outline
1
Learning from Differentiation
Integration by Substitution
Learn from Differentiation
To differentiate
Integration by Substitution: The Technique
y = ex
2
Integration by Substitution: Worked Examples
3
Integration by Parts: The Technique
4
Integration by Parts: Worked Examples
5
Two Definite Integral Examples
Integration (3/4)
MS121: IT Mathematics
3
we substitute u = x 3 and differentiate using the Chain Rule. We have
John Carroll
3 / 37
y
= eu
u = x3
⇒
dy
du
= eu
du
= 3x 2
dx
⇒
dy
dx
=
Integration (3/4)
dy du
3
·
= e u · 3x 2 = 3x 2 e x
du dx
MS121: IT Mathematics
John Carroll
4 / 37
Integration by Substitution: The Technique
Learning from Differentiation
Integration by Substitution: The Technique
Integration by Substitution
Integration by Substitution
Z
y =e
x3
Learning from Differentiation
Find
dy
3
⇒
= 3x 2 e x
dx
3
3x 2 e x dx
We try the same substitution again, so
u = x3
Applying this to Integration
From the above, we therefore know that
Z
3
3
3x 2 e x dx = e x
How could we calculate the integral directly? The reason that it is difficult
to calculate
Z
3
3x 2 e x dx
⇒
du
= 3x 2
dx
du = 3x 2 dx
⇒
Now replace the x’s under the integral (including dx) by the u’s (and du):
Z
Z
3
2 x3
3x e dx =
e x ? 3x 2 dx
Z
=
e u du
= eu = ex
3
3
is similar to that which makes it difficult to calculate the derivative of e x .
Integration (3/4)
MS121: IT Mathematics
Integration by Substitution: The Technique
John Carroll
5 / 37
John Carroll
6 / 37
John Carroll
8 / 37
Outline
Let u = the “most complicated expression”, particularly if that
expression is the argument of another function. This is the same
choice of u that one would use if differentiating using the chain rule
(e.g., choosing u = x 3 above).
1
Integration by Substitution: The Technique
2
Integration by Substitution: Worked Examples
3
Integration by Parts: The Technique
4
Integration by Parts: Worked Examples
5
Two Definite Integral Examples
is also present under the integral.
If the integrand (function to be differentiated) is a quotient, let u =
the denominator.
These hints may be contradictory and they are not foolproof. However,
they are sufficient to tackle a large class of integrals and, with enough
practice, one can develop an intuition for the correct strategy.
Integration (3/4)
MS121: IT Mathematics
Integration by Substitution: Worked Examples
Technique: What is the best choice for u?
The following are some useful tips:
Let u = f (x) if
Integration (3/4)
Learning from Differentiation
Integration by Substitution
f 0 (x) dx
which gives the same answer as before.
MS121: IT Mathematics
John Carroll
7 / 37
Integration (3/4)
MS121: IT Mathematics
Integration by Substitution: Worked Examples
Finding the “Right” Substutution
Integration by Substitution: Worked Examples
Integration by Substitution: Examples
Z
Calculate
Integration by Substitution: Examples
Z
x4
dx
x5 + 2
Calculate
x3
p
16 + x 4 dx
Note that the expression which is the argument of the square root is a
good candidate for substitution. Accordingly, Let
We let
u = x5 + 2
du
= 5x 4
dx
⇒
⇒
du = 5x 4 dx
u = 16 + x 4
du
= 4x 3
dx
⇒
Replacing the x’s by u’s:
Z
x4
x5 + 2
x 4 dx
Z
dx =
x5 + 2
=
1
5
Z
Note: Always write the final answer in terms of the original variable (in
this case x).
Integration (3/4)
MS121: IT Mathematics
Integration by Substitution: Worked Examples
John Carroll
9 / 37
Integration (3/4)
Finding the “Right” Substutution
Calculate
log x
dx
x
10 / 37
Finding the “Right” Substutution
5
x 3 (x 4 + 4) dx
Let u = x 4 + 4, the argument of the fifth power. Then
f 0 (x) dx
=
1
x
dx is present in the integrand.
du
= 4x 3
dx
du
1
dx
=
⇒ du =
dx
x
x
Z
Z
Z
log x
dx
u2
1
dx = log x ?
= u du =
= (log x)2
x
x
2
2
u = log x
Integration (3/4)
John Carroll
Integration by Substitution: Examples
Z
If f (x) = log x, then
Therefore, put
du = 4x 3 dx
MS121: IT Mathematics
Integration by Substitution: Worked Examples
Integration by Substitution: Examples
Z
⇒
Look to arrange 4x 3 dx(= du) in one group in the integral:
Z
Z p
p
3
4
x
16 + x dx =
16 + x 4 ? x 3 dx
Z
Z
√
1
1
=
u ? du =
u 1/2 du
4
4
1 u 1+1/2
u 3/2
(16 + x 4 )3/2
=
=
=
4 1 + 1/2
6
6
du
1
=
u −1 du
u
5
1
1
log u = log(x 5 + 2)
5
5
Z
=
Calculate
Finding the “Right” Substutution
John Carroll
du = 4x 3 dx
Group 4x 3 dx(= du) together and substitute:
Z
Z
5
5
3 4
x (x + 4) dx = (x 4 + 4) .x 3 dx
Z
Z
1
(x 4 + 4)6
1
1 u6
5
=
u ? du =
u 5 du =
=
4
4
4 6
24
⇒
MS121: IT Mathematics
⇒
11 / 37
Integration (3/4)
MS121: IT Mathematics
John Carroll
12 / 37
Integration by Substitution: Worked Examples
Finding the “Right” Substutution
Integration by Substitution: Worked Examples
Integration by Substitution: Examples
Integration by Substitution: Examples
Z
Z
x
dx
2x + 2
Let u = 2x + 2, the argument of the square root. Thus
Calculate
√
du
=2
dx
⇒
Z
x 1
1
√
du =
2
u 2
Z
x
√ du
u
We now replace x by expressing it in terms of u: x = 12 (u − 2), so
du = 2 dx
Z
Z 1
− 2)
x
1
2 (u
√
√
dx =
du
2
u
2x + 2
!
Z √
1
2
1 u 3/2
u 1/2
=
u−√
du =
−2
4
4 3/2
1/2
u
=
MS121: IT Mathematics
Integration by Substitution: Worked Examples
x
√
dx =
2x + 2
Solution (Cont’d)
Now, we replace x (and dx) by u (and du):
Z
Z
Z
x
x 1
x
1
√
√
√ du
dx =
du =
2
u 2
u
2x + 2
Integration (3/4)
Finding the “Right” Substutution
John Carroll
13 / 37
Integration (3/4)
Finding the “Right” Substutution
1 3/2
1
u − u 1/2 = (2x + 2)3/2 − (2x + 2)1/2
6
6
MS121: IT Mathematics
Integration by Substitution: Worked Examples
Integration by Substitution: Trogonometry Examples
John Carroll
14 / 37
Finding the “Right” Substutution
Integration by Substitution: Trogonometry Examples
Z
Calculate
sin 6x dx
Z
sin x
dx
2 + cos x
Since the integrand is a quotient, we put u = 2 + cos x, so
du = − sin x dx. Therefore, we have
Z
Z
Z
sin x dx
−du
sin x
dx =
=
2 + cos x
2 + cos x
u
Calculate
We let u = 6x, the argument of the “sin” function:
du
=6
dx
⇒
du = 6 dx
sin u ?
1
1
du =
6
6
Therefore:
Z
⇒
Z
sin 6x dx
=
=
Integration (3/4)
Z
sin u du
= − log u = − log(2 + cos x)
1
1
· (− cos u) = − cos 6x
6
6
MS121: IT Mathematics
John Carroll
15 / 37
Integration (3/4)
MS121: IT Mathematics
John Carroll
16 / 37
Integration by Substitution: Worked Examples
Finding the “Right” Substutution
Integration by Substitution: Worked Examples
Integration by Substitution: Trogonometry Examples
Z
Calculate
Integration by Substitution: Trogonometry Examples
sin x cos3 x dx
tan5 x
dx
cos2 x
We solve as follows:
Z
Calculate
Let u = cos x, the argument of the cube. Then
du
= − sin x
dx
⇒
d
tan x = sec2 x,
dx
du = − sin x dx
Group − sin x dx(= du) together and substitute:
Z
Z
Z
3
3
sin x cos x dx =
cos x ? sin x dx = u 3 ? (−du)
Z
u4
cos4 x
= − u 3 du = − = −
4
4
Integration (3/4)
Finding the “Right” Substutution
MS121: IT Mathematics
John Carroll
so put u = tan x to obtain
du = sec2 x dx =
Z
⇒
17 / 37
Integration by Parts: The Technique
tan5 x
dx =
cos2 x
Integration (3/4)
Z
u 5 du =
u6
tan6 x
=
6
6
MS121: IT Mathematics
Integration by Parts: The Technique
Outline
dx
cos2 x
John Carroll
18 / 37
Learning from Differentiation
Integration by Parts
The Link to Differentiaton
We have seen how integration by substitution corresponds to the
chain rule for differentiation.
1
Integration by Substitution: The Technique
2
Integration by Substitution: Worked Examples
It is reasonable to ask if there is a rule of integration corresponding to
the product rule?
3
Integration by Parts: The Technique
The answer to this question is “yes” and the rule is that of
integration by parts.
This is a rule which is useful for calculating the integrals of products:
4
5
Integration by Parts: Worked Examples
Z
Two Definite Integral Examples
Integration (3/4)
MS121: IT Mathematics
Z
u dv = uv −
John Carroll
19 / 37
Integration (3/4)
MS121: IT Mathematics
v du
John Carroll
20 / 37
Integration by Parts: Worked Examples
Integration by Parts: Worked Examples
Outline
Integration by Parts:
Z
Calculate
R
Making the “Right” Substitution
u dv = uv −
R
v du
x e 2x dx
We will let u = x and dv = e 2x dx so that
Z
Z
xe 2x dx = u dv
1
Integration by Substitution: The Technique
2
Integration by Substitution: Worked Examples
3
Integration by Parts: The Technique
We also need du and v :
4
5
Integration by Parts: Worked Examples
du
=1
dx
⇒
u=x
⇒
du = dx
i.e. we differentiate u to get du. Also
Two Definite Integral Examples
dv = e
2x
Z
⇒
dx
v=
1
e 2x dx = e 2x
2
i.e. we integrate dv to get v .
Integration (3/4)
MS121: IT Mathematics
Integration by Parts: Worked Examples
Integration by Parts:
Z
Calculate
R
John Carroll
21 / 37
Making the “Right” Substitution
u dv = uv −
R
v du
Integration by Parts:
R
John Carroll
22 / 37
Making the “Right” Substitution
u dv = uv −
R
v du
Z
Calculate
Let
u=x
and
Z
and
v=
1
e 2x dx = e 2x
2
u=x
Now, substitute u, v , du and dv into the integration by parts formula to
obtain:
Z
Z
Z
Z
1 2x
1 2x
2x
x e dx = u dv = uv − v du = x · e −
e dx
2
2
1 2x 1 2x
=
xe − e
2
4
MS121: IT Mathematics
x cos x dx
Choose u = x and dv = cos x dx, so that Inspecting the right-hand side of
the integration by parts formula, we also need du and v :
dv = e 2x dx
so that
Integration (3/4)
MS121: IT Mathematics
Integration by Parts: Worked Examples
x e 2x dx (Cont’d)
du = dx
Integration (3/4)
John Carroll
23 / 37
⇒
dv = cos x dx
du
= 1 ⇒ du = dx
dx
Z
⇒ v = cos x dx = sin x
Substitute u, v , du and dv into the formula to obtain:
Z
Z
Z
x cos x dx = u dv = uv − v du
Z
= x sin x − sin x dx = x sin x + cos x
Integration (3/4)
MS121: IT Mathematics
John Carroll
24 / 37
Integration by Parts: Worked Examples
Integration by Parts:
R
Making the “Right” Substitution
u dv = uv −
R
Integration by Parts: Worked Examples
v du
Integration by Parts:
Z
Calculate
R
Making the “Right” Substitution
u dv = uv −
R
v du
x 2 log x dx
We could choose either
Important Question
How do we know in practice what to choose for u and dv ?
u = x2
To answer the question, we first consider another example where we
make our choice by trial and error.
or
u = log x
Which should one try?
(a) u = x 2 : Then dv = log x dx, so
We then present a set of guidelines to assist in the decision-making
process.
Z
v=
log x dx.
This is not listed in the formulae and tables. We will therefore
abandon this approach.
Integration (3/4)
MS121: IT Mathematics
Integration by Parts: Worked Examples
Integration by Parts:
Z
Calculate
R
John Carroll
25 / 37
Making the “Right” Substitution
u dv = uv −
R
Integration (3/4)
MS121: IT Mathematics
Integration by Parts: Worked Examples
v du
Integration by Parts:
R
John Carroll
26 / 37
“LATE” Guidelines
u dv = uv −
R
v du
x 2 log x dx (Cont’d)
(b) u = log x: Then dv = x 2 dx, so
Z
1
v = x 2 dx = x 3
and
3
du
1
=
dx
x
⇒
du =
Answer to the Important Question
To decide how u should be chosen in general, we advocate that the choice
should be made in the following order:
1
dx
x
Substitute these expressions into the integration by parts formula:
Z
Z
Z
2
x log x dx = u dv = uv − v du
Z
1 3
1 3 1
= log x · x −
x · dx
3
3
x
Z
1 3
1
1
1
=
x log x −
x 2 dx = x 3 log x − x 3
3
3
3
9
Integration (3/4)
MS121: IT Mathematics
John Carroll
L − Logarithm
A − Algebraic: powers of x, e.g. x, x 3
T − Trigonometric
E − Exponential
27 / 37
Integration (3/4)
MS121: IT Mathematics
John Carroll
28 / 37
Integration by Parts: Worked Examples
Integration by Parts:
R
“LATE” Guidelines
u dv = uv −
R
Integration by Parts: Worked Examples
v du
Integration by Parts:
“LATE” Guidelines
The guideline has been used in all the examples:
R 2x
1
xe dx:
x → A, e 2x → E
LATE ⇒ u = x (A), so dv = e 2x dx.
2
3
Z
Calculate
x cos x dx:
x → A, cos x → T
LATE ⇒ u = x (A), so dv = cos x dx.
x 2 log x dx
x 2 → A, log x → L
LATE ⇒ u = log x (L), so dv = x 2 dx.
R
This rule gives very reliable guidance, but occasionally it does not work.
Sometimes, we must integrate by parts successively to get an answer as
the following example will illustrate.
MS121: IT Mathematics
Integration by Parts: Worked Examples
Integration by Parts:
Z
R
x 2 cos x dx = x 2 sin x − 2
John Carroll
u dv = uv −
u dv = uv −
R
v du
x 2 cos x dx
Substitute u, v , du, dv into the formula:
Z
Z
Z
x 2 cos x dx = u dv = uv − v du
Z
Z
2
2
= x sin x − sin x ? 2x dx = x sin x − 2 x sin x dx
We see that we cannot calculate the last integral directly.
Integration (3/4)
29 / 37
Using the “LATE” Guidelines
R
Using the “LATE” Guidelines
According to our notation, this is of type LATE ⇒ u = x 2 (A), so
dv = cos x dx (T). Therefore, we differentiate u and integrate dv to
obtain:
Z
du
= 2x,
v = cos x dx = sin x
dx
R
Integration (3/4)
R
MS121: IT Mathematics
Integration by Parts: Worked Examples
v du
Integration by Parts:
R
John Carroll
30 / 37
Using the “LATE” Guidelines
u dv = uv −
R
v du
Z
x sin x dx
Z
2
x cos x dx
Z
x sin x dx
We must use the integration by parts formula again for
Z
x sin x dx
Z
which is again of type LATE ⇒ u = x (A), so dv = sin x dx (T). We
differentiate u and integrate dv to obtain:
Z
du
= 1,
v = sin x dx = − cos x
dx
2
= x sin x − 2
Z
x sin x dx
= −x cos x + sin x
x 2 cos x dx (Cont’d)
Substituting the second result into the first yields:
Z
Z
2
2
x cos x dx = x sin x − 2 x sin x dx
Substituting into the formula yields:
Z
Z
x sin x dx = −x cos x − (− cos x) dx = −x cos x + sin x
= x 2 sin x − 2(−x cos x + sin x)
= x 2 sin x + 2x cos x − 2 sin x
Integration (3/4)
MS121: IT Mathematics
John Carroll
31 / 37
Integration (3/4)
MS121: IT Mathematics
John Carroll
32 / 37
Two Definite Integral Examples
Two Definite Integral Examples
Outline
1
Definite Integrals: Integration by Substitution Example
Z
Integration by Substitution: The Technique
4
Calculate
x
dx
+1
3x 2
1
2
Substitution Example
Integration by Substitution: Worked Examples
Substitution
3
Integration by Parts: The Technique
u = 3x 2 + 1
4
⇒
Integration by Parts: Worked Examples
Z
5
Integration (3/4)
MS121: IT Mathematics
Two Definite Integral Examples
John Carroll
33 / 37
Z
x
1
dx =
3x 2 + 1
6
Z
Therefore
Z 4
1
Z
⇒
1
du = x dx
6
1
du
u
John Carroll
34 / 37
John Carroll
36 / 37
Integration-by-parts Example
Definite Integrals: Integration by Parts Example
1
du
u
Z
Change the limits of integration
x = 1 ⇒ u =3x 2 + 1 = 4
du = 6x dx
MS121: IT Mathematics
Two Definite Integral Examples
Definite Integrals: Integration by Substitution Example
Substitution u = 3x 2 + 1 leads to
Integration (3/4)
Substitution Example
⇒
1
x
dx =
2
3x + 1
6
This means that
Two Definite Integral Examples
du
= 6x
dx
Calculate
3
r 3 ln r dr
1
x =4
u =3x 2 + 1 = 49
⇒
Integration by Parts
x
dx
2
3x + 1
4
Z
=
1
1
(x dx) =
2
3x + 1
Z
49
4
1
u
1
du
6
⇒
du =
dv = r 3 dr
⇒
1
v = r4
4
1 49 1
du
6 4 u
1
1
[ln u]49
(ln 49 − ln 4)
4 =
6
6
Z
=
=
Integration (3/4)
MS121: IT Mathematics
John Carroll
35 / 37
1
dr
r
u = ln r
Integration (3/4)
MS121: IT Mathematics
Two Definite Integral Examples
Integration-by-parts Example
Definite Integrals: Integration by Parts Example
Z
1
3
1
r 3 ln r dr with u = ln r and dv = r 3 dr ⇒ v = r 4
4
Using the integration by parts formula
3
Z
Z 3
Z
3
u dv =
r ln r dr = uv − v du
1
=
=
=
Integration (3/4)
r =1
1 4
r ln r
4
3
Z
3
1 3
r dr
1 4
1
81
1 1 4 3
ln 3 − 0 −
r
4
4 4
1
81
1
81
ln 3 − (81 − 1) =
ln 3 − 5
4
16
4
−
MS121: IT Mathematics
John Carroll
37 / 37