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Chapter 3
Pure Integration
Techniques
Design is not making beauty, beauty emerges from selec-
tion, affinities, integration, love.
—Louis Kahn
We consider ODEs of the form
u
′ = f (t) ⇐⇒ du = f (t)dt (3.1)
(differential form)
Such an ODE is called a pure integration problem, because the solution u = u(t)
can directly be obtained as the result of an integration, according to the
Fundamental theorem of Calculus
(ODE) u
′ = f (t) ⇐⇒ u(t) =
f (t) dt
(IVP)
u
′ = f (t),
u(0) = u 0
⇐⇒ u(t) =
∫ (^) t
0
f (τ ) dτ + u 0
3.1 Simple integrals
t
n dt =
1 n+ t
n+
dt
t
= ln |t| + C
cos(nt)dt =
1 n
sin(nt) + C (n 6 = 0)
sin(nt)dt = −
1 n
cos(nt) + C (n 6 = 0)
e
at dt =
1 a
e
at
24 Chapter 3. Pure Integration Techniques
Algebraic relations/formulas can sometimes be used to reduce an integral to a simple
one.
Example 3.
sin t cos t dt
(1.4c) −−−−→ − 2
sin 2t dt
integrate −−−−−−−→ cos(2t) + C
For integrals which cannot readily be determined explicitly or transformed into a
simple one using relations/formulas, a number of techniques can be used to evaluate
them.
3.2 Substitution
Substitution transforms an integral into one which can be evaluated simply.
Example 3.
sin t cos t dt −2 sin
2 t + C
sin t = τ
cos t dt = dτ
y
x τ = sin t
τ dτ
integrate −−−−−−−→ − 2 τ 2
The result looks different from that of Example 3.1. However the two functions
cos(2t) and −2 sin
2 t only differ by a constant in view of (1.5b).
Example 3. ∫ ln t
t
dt
1 2 (ln t)
2
t = e τ
dt = e τ dτ
y
x τ = ln t
τ dτ
integrate −−−−−−−→
1 2
τ 2
Note that both t as a function of τ and τ as function of t are used in Example 3.3.
3.3 Integration by parts
The integration by parts formula
∫
f (t)g
′ (t)dt = f (t)g(t) −
f
′ (t)g(t)dt (3.2)
is simply a restatement of the differentiation product rule (see Table page 5) com-
bined with the fundamental Theorem of Calculus page 23 (move the second integral
in (3.2) to the left, then differentiate both sides). This formula is useful whenever
26 Chapter 3. Pure Integration Techniques
A factor like this in Q(s) ... yields a contribution like this in the PFD ...
(s − α)
a
s − α
(s − α)
2 a
s − α
b
(s − α)^2
(s − α) 2
a(s − α) + b
(s − α) 2
((s − α)
2
2 )
2 a(s − α) + b
(s − α) 2
c(s − α) + d
((s − α) 2
- Determine unknown coefficients. The set-up phase leads
P (s) Q(s) =PFD, i.e.,
P (s) = Q(s) PFD. Simplify and either substitute appropriate values for s
(Approach 1) or expand and compare powers of s (Approach 2). This
yields a system of linear equations for the unknown coefficients.
Example 3.6 Determine
F (s) ds with F (s) =
s 2
- Factor denominator: s
2
- Set-up:
(s + 1)(s + 2)
a
s + 1
b
s + 2
- Multiply by common denominator and simplify: 1 = a(s + 2) + b(s + 1).
Approach 1: substitute values
s = − 1 ⇒ 1 = a
s = − 2 ⇒ 1 = −b
Approach 2: compare powers of s:
s 0 : 1 = 2a + b
s 1 : 0 = a + b
a = 1
b = − 1
Thus F (s) =
s + 1
s + 2
and
F (s) ds = ln |s+1|−ln |s+2|+C = ln
s + 1
s + 2
+C.
Example 3.7 Determine
F (s) ds with F (s) =
s + 2
(s 2
- Factor denominator: (s
2
2
2
(s + 1)
2
- Set-up: F (s) =
a
s + 1
b
(s + 1) 2
c(s + 1) + d
(s + 1) 2
- Multiply by common denominator and simplify:
s + 2 = a(s + 1)
(s + 1)
2
(s + 1)
2
c(s + 1) + d
(s + 1)
2 .
Approach 1: substitute values
s = − 1 ⇒ 1 = b
s = 0 ⇒ 2 = 2a + 2b + c + d
s = − 2 ⇒ 0 = − 2 a + 2b − c + d
s = 1 ⇒ 3 = 10a + 5b + 8c + 4d
3.5. Trigonometric integrals 27
Pick values until you get as many equations as unknowns. The choice s = − 1
is obvious (most terms vanish), while other choices are somewhat arbitrary.
Approach 2: expand in powers of s, or, better here, (s + 1) = z (s = z − 1),
z + 1 = az(z
2
2
2 = (a + c)z
3
2
and compare powers of z:
z
0 : 1 = b
z
1 : 1 = a
z
2 : 0 = b + d
z
3 : 0 = a + c
a = 1
b = 1
c = − 1
d = − 1
We verify that (3.5) satisfies the system (3.4). Thus
F (s) =
s + 1
(s + 1) 2
s + 1
(s + 1) 2
(s + 1) 2
There is a strong incentive in leaving the last two terms of (3.6) separate. Indeed,
F (s) ds =
ds
s + 1
ds
(s + 1)^2
s + 1
(s + 1)^2 + 1
ds −
ds
(s + 1)^2 + 1
+ C
ln |s + 1| −
s + 1
ln((s + 1)
2
Approach 1 works best when all factors in the denominator are of the form (s−α).
Approach 2 is safer (the set-up is incorrect if no solution is found) but may lead
to systems which are more difficult to solve. A combination of both strategies may
be used. A matrix elimination technique for solving systems is also described in
Chapter ...
3.5 Trigonometric integrals
Trigonometric integrals are integrals involving (standard) trigonometric functions.
Some trigonometric integrals can be directly obtained from known formulas.
Example 3. ∫ dt
cos 2 t
sec
2 t dt = tan t + C.
Example 3.
∫ dt
cos t
sec t dt = ln | sec t + tan t| + C
(differentiate both sides to verify the result!)
3.5. Trigonometric integrals 29
Example 3.
∫ tan t dt
sin t + cos t ysimplify in terms of sin and cos ∫ sin t dt
cos t(sin t + cos t) yτ = tan
t
(τ = sin t, τ = cos t do not work)
∫ 4 τ dτ
(1 − τ 2 )(1 + 2τ − τ 2 ) yPFD^
a 1+τ +^
b 1 −τ +^ √ c 2+1−τ
√ d 2 −1+τ
∫
1 − τ
1 + τ
2 + 1 − τ
2 − 1 + τ
dτ
yintegrate
ln
1 + τ
1 − τ
ln
2 − 1 + τ √ 2 + 1 − τ
+ C
ysubstitute
ln
1 + tan
t 2
1 − tan
t 2
ln
2 − 1 + tan
t 2 √ 2 + 1 − tan
t 2
+ C
Typical mistakes and useful tips
- (^) [mistakes] Typical integration mistakes
wrong correct explanation
dx
x
= ln(x) + C
dx
x
= ln |x| + C errors with ln ∫ dx
x 2
= ln(x
2
dx
x 2
= arctan(x) + C
ln(x)dx =
x
+ C
ln(x)dx = x ln(x) − x + C do not differentiate
dx
x 2
x
dx
x 2
x
30 Chapter 3. Pure Integration Techniques
e
at cos(bt)dt = e
at
a
a 2
cos(bt) +
b
a 2
sin(bt)
e
at sin(bt)dt = e
at
−b
a 2
cos(bt) +
a
a 2
sin(bt)
will prove useful in this course and are worth memorizing. To verify the
identities simply differentiate both sides. In case you forget them they can be
obtained in a number of ways:
- [undetermined coefficients] Simply remember that both integrals are of
the form e
at (α cos(bt) + β sin(bt)) + C with α and β to be determined by
differentiating and comparing e
at cos(bt) and e
at sin(bt) terms.
- [double integration by parts] Integrate
e
at cos(bt)dt by parts (integrate
e
at , differentiate cos(bt)) and get a term
e
at sin(bt)dt. Another inte-
gration by parts (integrate e
at , differentiate sin(bt)) yields an equation
for
e
at cos(bt)dt. See Example 3.5.
- [using Euler’s formula] Express sin(bt) and cos(bt) in terms of e
±ibt .
Integrate the resulting exponentials and simplify the complex expression
obtained using Euler’s formula again (the result is a real function!).
32 Chapter 3. Pure Integration Techniques
0 1 2 3 4 5
0
1
2
3
Figure 3.1. The functions (3.9) (with C = 0, top) and (3.10) (with C = 0,
bottom) differ by a constant (
2 3
As an alternative to the substitution v = tan
θ 2 used above, consider the steps
2 sin θ
cos 2 θ(1 + sin θ)
dθ
y
∫ 2 sin θ(1 − sin θ)dθ
cos 2 θ(1 + sin θ)(1 − sin θ)
sin θdθ
cos 4 θ
tan
2 θ sec
2 θdθ
yu^ = cos^ θ^ in first int.,^ v^ = tan^ θ^ in second int.
du
u 4
v
2 dv =
u
− 3 −
v
3
ysubstitute
2 3
(1 + tan 2 θ) 3 / 2 − (tan 2 θ) 3 / 2
+ C =
2 3
(1 + t) 3 / 2 − t 3 / 2
+ C
Different integration strategies may lead to different but equally legitimate
forms of a result, some better than others for the purpose at hand. This
is especially true when dealing with trigonometric expressions. This chapter
gives you a few tools and ideas to get to at least one of these forms.
Exercises
In Problems 3.1-3.116 determine the given definite or indefinite integral.
Exercises 33
(t
2 / 5
t − 7 sin t)dt
te
t dt
ln(5t)
t
dt
t 4
t
dt
e
5 t dt
2 t − 9
t 2 − 6 t + 9
dt
t
3 ln(t) dt
arctan(
t) dt
t − e
dt
5 y
2
(y^2 + 1)(y + 1)
dy
sin(t)
3 + 2 cos(t) dt
t
2 e
t 3 dt
(u 2
2
√ u
du
t sin(2t) dt
(t + 1)e
t dt
e 1 /t
t 2
dt
(t
3
t − 6 sin(t)) dt
t
5
t
dt
∫ π 4
0
tan(t) sec
2 (t) dt
4
1
ln(4t)
t
dt
− 2
|t| dt
e
t ln(1 + e
t ) dt
t
4 ln(t) dt
sin(t) cos(t)e
sin(t) dt
arctan(t) dt
u √ 1 − u 2
du
4 w 2 − w + 2
(w 2
dw
t ln(t) dt
e
t
(e t
2
dt
t
e 2 dt
y + 1 √ y − 1
dy
4 t + 21
t 2
dt
(t
4 / 3 − 2
t + 3 cos(2t)) dt
t
4
t 2
dt
cos(2t) cos(sin(2t)) dt
t
t − 1 dt
2 e t
e t
dt
t
6 ln(2t) dt
1
e
3 t
e 3 t
dt
3 t + 1
(t 2
dt
t
1 + t 4
dt
cos(2t)
sin(2t) + 1
dt
Exercises 35
∫ (^) π/ 3
0
sec
2 β tan β dβ
π/ 3
0
sin
3 t dt
sec α tan α
(1 + sec α) 2
dα
sec t tan t dt
sec
5 θ tan θ dθ
e
√ t dt
t
dt
cos
2 t dt
sin
4 t dt
ds
s(1 + (ln s) 2 )
1
0
e t
dt
sin 5t sin 2t dt
sin
3 t cos
2 t dt
4 − t 2 dt
2 t − 2
(t 2 − 2 t + 3) 5
dt
3 y 2
y 3
dy
1
0
t
3
√ 1 − t 2
dt
cos t sin
3 t dt
t^2 − 6 t + 10
dt
r
r 2
π/ 3
0
sin t
cos^3 t
dt
0
z(z +
3
z) dz
2
0
2 t
(t − 3) 2
dt
t
dt
sin 3t cos t dt
cos 4t cos 2t dt
t
3 e
t 2 dt
tan
3 t sec
3 t dt
tan
2 t dt
t 2
t 2 − 9
dt
t
3
√ t 2
dt
1
2 t − t 2 dt
3.117. Using the substitution (3.7) show that
dt
cos t
= ln
1 + tan
t 2
1 − tan
t 2
+ C.
Verify that this expression is identical to the result obtained in Example 3.
(see Problem 1.55).
