MAST10013
UMEP Mathematics
for
High Achieving Students
Bridging Notes
for Vector Calculus
*
Department of Mathematics and Statistics
© University of Melbourne 2009
This compilation has been made in accordance with the
provisions of Part VB of the Copyright Act (1968) for the teaching purposes of the
University of Melbourne. No part of this publication may be reproduced or
transmitted by any form, except as permitted under this act.
Preface
The notes and problems in this booklet are intended for students who have completed MAST10013 UMEP
Mathematics and plan to enter Vector Calculus without having done Accelerated Mathematics 2 or Calculus 2
first. The notes cover Hyperbolic Functions and Techniques of Integration. At the end of each chapter there are
exercises given. Students are advised to do as many of these exercises as possible making sure that all topics
are practised.
Students are welcome to ask questions about this material. Please see your Vector Calculus lecturer if you want
any help.
Acknowledgement Some of the materials in these notes have been provided by Christine Mangelsdorf and
Paul Pearce.
1
2
Contents
1 Hyperbolic Functions
1.1 Introduction to Hyperbolic Functions . . . .
1.2 Differentiation . . . . . . . . . . . . . . . .
1.3 Inverse Hyperbolic Functions . . . . . . . .
1.4 Derivatives of Inverse Hyperbolic Functions
1.5 Exercises for Hyperbolic Functions . . . . .
1.6 Answers to Exercises . . . . . . . . . . . . .
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2 Techniques of Integration
2.1 Why Integration? . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Basic Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Double Angle Formulae . . . . . . . . . . . . . . . . . . . . . .
2.4 Derivative Substitution . . . . . . . . . . . . . . . . . . . . . . .
2.5 Change of Variable . . . . . . . . . . . . . . . . . . . . . . . . .
2.6 Trigonometric and Hyperbolic Substitutions – sin, cos, sinh and
2.7 Substitution with tan and tanh . . . . . . . . . . . . . . . . . .
2.8 Products of Trigonometric and Hyperbolic Functions . . . . . .
2.9 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . .
2.10 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . .
2.11 Complex Exponential (Revision from UMEP Mathematics) . .
2.12 Integration Exercises . . . . . . . . . . . . . . . . . . . . . . . .
2.13 Answers to Exercises . . . . . . . . . . . . . . . . . . . . . . . .
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5
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15
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cosh
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4
CONTENTS
Chapter 1
Hyperbolic Functions
1.1
Introduction to Hyperbolic Functions
Hyperbolic functions are the analogues of the trigonometric functions, sine and cosine. The basic functions
are hyperbolic sine and hyperbolic cosine abbreviated sinh and cosh respectively. From these we can derive
hyperbolic tangent (tanh) and so forth.
Just as cosine and sine (abbreviated sin and cos respectively) are used to parametrize the circle, hyperbolic
sine and hyperbolic cosine are used to parametrize the hyperbola. They also have application to techniques of
integration, modelling of hanging cables, electromagnetic theory, heat transfer and special relativity.
Abbreviations and Pronunciation
These hyperbolic functions are abbreviated and pronounced as follows:
Function
Abbreviation
Pronunciation
hyperbolic sine
sinh
shine which is the UK pronunciation (or, in the US, sinch)
hyperbolic cosine
cosh
cosh
hyperbolic tangent
tanh
than to rhyme with pan (or tanch)
hyperbolic cotangent
coth
coth to rhyme with moth
hyperbolic secant
sech
sheck (or setch)
hyperbolic cosecant
cosech
cosheck (or cosetch)
Definition
Before defining the hyperbolic functions we recall the identities of sin and cos that arise from the complex
exponential and Euler’s equation, eiθ = cos θ + i sin θ. The identities are:
cos θ =
eiθ + e−iθ
eiθ − e−iθ
and sin θ =
.
2
2i
The definition of hyperbolic sin and cos is very similar:
Definition 1.1 The hyperbolic functions are defined by
1. cosh : R → R such that
cosh x =
ex + e−x
2
sinh x =
ex − e−x
2
2. sinh : R → R such that
5
6
CHAPTER 1. HYPERBOLIC FUNCTIONS
3. tanh : R → R such that
tanh x =
sinh x
e2x − 1
1 − e−2x
= 2x
=
cosh x
e +1
1 + e−2x
coth x =
cosh x
e2x + 1
1 + e−2x
= 2x
=
sinh x
e −1
1 − e−2x
4. coth : R\{0} → R such that
5. sech : R → R such that
sech x =
1
2
= x
cosh x
e + e−x
6. cosech : R\{0} → R such that
1
2
= x
sinh x
e − e−x
cosech x =
Graphs
Below find the graphs of sinh x and cosh x. Asymptotically
1. sinh x →
ex
e−x
as x → ∞ and sinh x → −
as x → −∞
2
2
2. cosh x →
ex
e−x
as x → ∞ and cosh x →
as x → −∞
2
2
coshHxL
sinhx
3.5
3
3.0
2
ex
2.5
1
2.0
2
2
1
1
e
x
1
2
x
1.5
e-x
2
2
2
1.0
ex
2
0.5
3
-2
-1
It is clear from the graph of cosh x that the range of cosh is [1, ∞).
Identities
The fundamental identity is
cosh2 x − sinh2 x
1
(ex + e−x )2 − (ex − e−x )2
4
1
=
(e2x + 2 + e−2x ) − (e2x − 2 + e−2x )
4
= 1
=
From this it follows that
1 − tanh2 x = sech2 x and coth2 x − 1 = cosech2 x.
Parametrising the Hyperbola
1
2
x
1.1. INTRODUCTION TO HYPERBOLIC FUNCTIONS
7
The main identity, cosh2 t − sinh2 t = 1, gives a natural way to parametrise the hyperbola. If we set x = cosh t
and y = sinh t, t ∈ R we obtain the right branch of the hyperbola x2 − y 2 = 1. Setting x = − cosh t gives the
left branch.
y
3
2
2
2
Hx,yL: x - y = 1
y = SinhHtL
1
-3
-2
-1
1
2
3
x = CoshHtL
x
-1
-2
-3
More generally we have the following:
x2 y 2
−
= 1 and
a2 b2
Addition/Double Angle Formulas
(1) x = a cosh t and y = b sinh t for
(2) x = b sinh t and y = a cosh t for
y 2 x2
−
= 1.
a2 b2
The addition formulae are
• sinh(x + y)
=
sinh x cosh y + cosh x sinh y
• cosh(x + y)
=
cosh x cosh y + sinh x sinh y
• sinh(x − y)
=
sinh x cosh y − cosh x sinh y
• cosh(x − y)
=
cosh x cosh y − sinh x sinh y
The double angle formulae are
• sinh 2x
=
2 sinh x cosh x
• cosh 2x
=
cosh2 x + sinh2 x
=
2 cosh2 x − 1
=
2 sinh2 x − 1
Examples
Example 1.1 (Catenary) A flexible, heavy cable of uniform mass per unit length ρ has the shape
ρgx T
y=
cosh
ρg
T
where g is the acceleration due to gravity and T is the tension in the cable at the lowest point.
Example 1.2 Write (cosh x − sinh x)7 in terms of exponentials.
Solution
1 x
1
(cosh x − sinh x) =
(e + e−x ) − (ex − e−x )
2
2
7
7
= (e−x )7 = e−7x
8
CHAPTER 1. HYPERBOLIC FUNCTIONS
Example 1.3 If cosh x =
13
12
and x < 0, find sinh x and tanh x.
169
25
2
Solution Now cosh2 x − sinh2 x = 1 so sinh2 x = cosh2 x − 1 = ( 13
12 ) − 1 = 144 − 1 = 144 .
q
25
5
5
Hence sinh x = ± 144
= ± 12
. But x < 0 so sinh x < 0, so we must have sinh x = − 12
.
sinh x
(−5/12)
5
So finally, tanh x =
=
=− .
cosh x
(13/12)
13
Example 1.4 Write sinh(2 log x) as an algebraic function in x.
Solution
sinh(2 log x) =
2
−2
1 2 log x
1
1
[e
− e−2 log x ] = [elog x − elog x ] = (x2 − x−2 )
2
2
2
Example 1.5 Write cosh3 x in terms of cosh nx with n ∈ N.
Solution
cosh3 x
=
=
1
1
3
1
[ (ex + e−x )]3 = [e3x + 3ex + 3e−x + e−3x ] = (e3x + e−3x ) + (ex + e−x )
2
8
8
8
3
1
cosh 3x + cosh x
4
4
Example 1.6 Prove the first addition formula.
Solution
RHS
=
=
sinh x cosh y + cosh x sinh y
1 x
1
1
1
(e − e−x ) (ey + e−y ) + (ex + e−x ) (ey − e−y )
2
2
2
2
1 x+y
+ ex−y − e−x+y − e−x−y + ex+y − ex−y + e−x+y − e−x−y )
4 (e
=
x+y
1
4 (2e
=
− 2e−x−y )
=
1 x+y
(e
− e−(x+y) )
2
sinh(x + y)
=
LHS
=
Example 1.7 Prove the first double angle formula.
Solution
LHS
=
sinh 2x
=
sinh(x + x)
=
sinh x cosh x + cosh x sinh x
=
2 sinh x cosh x
=
RHS
1.2. DIFFERENTIATION
1.2
9
Differentiation
The derivatives of the hyperbolic functions are as follows:
d
(cosh x)
dx
d
(sinh x)
dx
d
(tanh x)
dx
d
(coth x)
dx
d
(sech x)
dx
d
(cosech x)
dx
=
sinh x
=
cosh x
=
sech2 x
= − cosech2 x
= − tanh x sech x
= − coth x cosech
Example 1.8 Prove the first and third derivative formulas.
Solution
d
(cosh x)
dx
d
tanh x
dx
=
=
=
=
=
d 1 x
(e + e−x )
dx 2
1 x
=
(e − e−x )
2
= sinh x
=
d sinh x dx cosh x
d
d
cosh x dx
(sinh x) − sinh x dx
(cosh x)
cosh2 x
cosh2 x − sinh2 x
cosh2 x
1
cosh2 x
sech2 x
Example 1.9 Find the derivative of y =
√
sinh 6x for x > 0.
Solution By the chain rule
dy
dx
=
=
=
1
1
d
√
(sinh 6x)
2 sinh 6x dx
6 cosh 6x
√
2 sinh 6x
3 cosh 6x
√
sinh 6x
(quotient rule)
10
CHAPTER 1. HYPERBOLIC FUNCTIONS
1.3
Inverse Hyperbolic Functions
Recall the following: Assume that U = Domain(f ) = Range(g) and V = Domain(g) = Range(f ).
Definition 1.2 If the functions f : U → V and g : V → U , satisfy two conditions
1. g(f (x)) = x for all x ∈ U and
2. f (g(y)) = y for all y ∈ V
then we say that g is an inverse of f and f is an inverse of g, that is f and g are inverse functions.
We write g = f −1 .
Theorem 1.3 (Inverse Function) If the domain of the function f is an interval, and f is either monotonic
increasing or decreasing on that domain (that is f is one to one on its domain) then the inverse function f −1
exists.
Notation for Inverse Trigonometric and Hyperbolic Functions
1
To avoid the confusion that arises because the inverse function sin−1 x may be confused with
the notation
sin x
adopted in University of Melbourne calculus courses is as follows:
Function
Inverse Function
Function
Inverse Function
cos
arccos
sin
arcsin
tan
arctan
cot
arccot
sec
arcsec
cosec
arcsec
cosh
arccosh
sinh
arcsinh
tanh
arctanh
coth
arccoth
sech
arcsech
cosech
arccosech
Inverse Hyperbolic Functions
The hyperbolic function sinh : R → R is monotonic increasing so the inverse function arcsinh exists.
However the hyperbolic function cosh : R → R is not monotonic. In order to define the inverse function of cosh
we therefore restrict its domain to [0, ∞) so that it is monotonic increasing and the inverse function arccosh is
well defined.
Definition 1.4
1. The inverse function of cosh is
arccosh : [1, ∞) → R
and
2. the inverse function of sinh is
arcsinh : R → R
The range of arccosh is [0, ∞). The graphs are as follows:
ArccoshHxL
ArcsinhHxL
1.5
1.5
1.0
0.5
1.0
-3
-2
-1
1
-0.5
2
3
x
0.5
-1.0
-1.5
0.5
1.0
1.5
2.0
2.5
3.0
x
1.3. INVERSE HYPERBOLIC FUNCTIONS
11
We can express inverse hyperbolic functions in terms of log as follows:
•
arcsinh x
•
arccosh x
•
arctanh x
=
log(x +
√
√
x2 + 1),
x∈R
x2
x≥1
log(x +
− 1),
1
1 + x
,
log
=
2
1−x
=
−1 < x < 1
Example 1.10 Derive a formula for arcsinh in terms of natural logarithms.
Solution
y = arcsinh x
⇐⇒
sinh y =
1 y
(e − e−y ) = x
2
Solve for y. Multiplying by 2ey we obtain a quadratic equation in ey
(ey )2 − 2x(ey ) − 1 = 0
with solution
ey =
p
p
1
2x ± 4x2 + 4 = x ± x2 + 1
2
Since ey > 0 we must choose the plus sign since
√
x2 + 1 > x. Taking the logarithm gives
y = arcsinh x = log x +
p
x2 + 1
Inverse Hyperbolic Examples
Example 1.11 Find the exact value of sinh(arccosh 3).
√
Now arccosh 3 = log(3 + 32 − 1). So
sinh(arccosh 3)
=
=
=
√ √ 1 log(3+√8)
− e− log(3+ 8)
8) =
e
2
√
√
√
1
1 1
1
3 − 8
√
√
√
3+ 8−
=
3+ 8−
2
2
3+ 8
3+ 8 3− 8
√
√ √
√
1
3 + 8 − (3 − 8) = 8 = 2 2
2
sinh log(3 +
Example 1.12 Express cosh(arctanh x) as an algebraic function of x for −1 < x < 1.
Solution
Let y = arctanh x so that x = tanh y. Now
cosh2 y =
1
1
1
=
=
2
2
1
−
x2
sech y
1 − tanh y
Hence
cosh(arctanh x) = cosh y = √
1
,
1 − x2
where we need the positive square root since cosh y ≥ 1 > 0.
−1 < x < 1
12
CHAPTER 1. HYPERBOLIC FUNCTIONS
1.4
Derivatives of Inverse Hyperbolic Functions
The derivatives are:
d
(arcsinh x)
dx
=
√
1
d
(arccosh x)
dx
=
√
d
(arctanh x)
dx
=
1
,
1 − x2
x2
+1
1
x2
−1
,
x∈R
,
x>1
−1 < x < 1
Example 1.13 Obtain the derivative of arcsinh x
Solution
Let y = arcsinh x so that x = sinh y. Then
dx
dy
dx
dy
=
cosh y
= ±p
1
sinh2 y + 1
1
= ±√
,
x2 + 1
x ∈ R.
But cosh y ≥ 1 > 0 so we require the positive square root.
Example 1.14 Find the derivative of arcsinh(x3 ).
Solution
By the chain rule
d arcsinh(x3 ) =
dx
1.5
1
d 3
3x2
p
(x ) = √
x6 + 1
(x3 )2 + 1 dx
Exercises for Hyperbolic Functions
Introductory Questions
1. Calculating Hyperbolic Functions Find the exact numerical value of each expression.
(a) sinh (log 3)
(b) cosh (− log 2)
(c) tanh (2 log 5)
2. Hyperbolic Expressions Write the following as algebraic expressions in x.
(a) sinh (log x)
(b) cosh (−3 log x)
3. Hyperbolic Functions
(a) If cosh x = 54 , what are the possible values of sinh x and tanh x?
(b) If sinh x = − 25 , compute cosh x, tanh x, coth x, sech x and cosech x.
(c) tanh (2 log x)
1.5. EXERCISES FOR HYPERBOLIC FUNCTIONS
13
4. Standard Hyperbolic Identities
Use the definitions of cosh x and sinh x to verify the following identities, where n is any integer.
(a) cosh x − sinh x = e−x
(b) cosh2 x + sinh2 x = cosh 2x
(c) (cosh x + sinh x)n = cosh nx + sinh nx
5. Hyperbolic Limits Evaluate the limit:
lim tanh x
x→∞
6. Sketching Hyperbolics
Sketch the graphs of
(a) tanh x,
(b) coth x,
(c) sech x,
(d) cosech x
clearly marking key features of the graphs such as intercepts, maxima and minima and asymptotic behaviour.
7. Manipulating Hyperbolic Functions
(a) Express the following functions in terms of hyperbolic sines and/or hyperbolic cosines of multiples
of x.
i. sinh5 x
ii. cosh6 x
(b) Express the following functions in terms of powers of sinh x and cosh x.
i. cosh 4x
ii. sinh 3x
8. Standard Hyperbolic Derivatives Using the derivatives of sinh x, cosh x and tanh x, show that
d
(coth x) = −cosech2 x, x 6= 0
dx
d
(b)
(sech x) = −sech x tanh x
dx
d
(c)
(cosech x) = −cosechx cothx, x 6= 0
dx
9. Hyperbolic Derivatives Find the derivatives of the following functions. Check domains.
(a)
√
(a) sinh (ex )
√
(b) cosh ( x )
(c)
(d) tanh(x2 − 1)
(e) tanh(sin 3x)
(f) x sinh(1/x)
cosh x
Revision from School: Inverse Trigonometric Functions
10. Exact Values Find the exact values of
(a) arccos(−1)
(b) arctan(−1)
√ !#
3
(c) cos arcsin −
2
"
11. Inverse Trigonometric Functions Write the following as algebraic expressions in x.
(a) cos(arcsin x)
(b) sin(arctan x)
(c) tan(arccos x)
14
CHAPTER 1. HYPERBOLIC FUNCTIONS
12. Sketching Trigonometric Functions On the same graph sketch the following functions:
x
x
x
y = arctan
, y = 3 + arctan
, y = −2 arctan
+2
2
2
2
13. Derivatives of Inverse Functions Find the derivative of the following functions.
(a) (arcsin x)2
(b) (1 + x2 ) arctan x
(c) arctan(ex )
x arcsin x
(d) 1 − √
1 − x2
14. More Derivatives of Inverse Functions Using implicit differentiation, show that
(a)
−1
d
(arccosec x) = √
,
dx
x x2 − 1
(b)
d
−1
,
(arccot x) =
dx
1 + x2
x>1
x 6= 0
15. Differentiation involving Inverse Functions
Find
dy
by implicit differentiation:
dx
(a) x3 + x arctan y = ey
(b) arcsin(xy) = 6 + x
Inverse Hyperbolic Functions
16. Functions and Inverses Express the following as algebraic functions of x:
(a) sinh(arccosh x)
(b) sinh2 (arctanh x)
(c) tanh(arccosh x)
17. Sketching Inverse Hyperbolics On the same graph sketch the following functions:
x
x
x
, y = cosech
, y = arccosech
y = sinh
3
3
3
18. Hyperbolic Identity If −1 < x < 1, show that:
x
arcsinh √
= arctanh x
1 − x2
19. Inverse Hyperbolic Functions
(a) Prove that, for x ≥ 1,
arccosh x = log(x +
p
x2 − 1)
(b) Find the derivative
d
(arccosh x)
dx
i. using the formula in part (a)
ii. using implicit differentiation
(c) Calculate the following limits:
i. lim (arccosh x − log x)
x→∞
ii. lim+ arccosh x
x→1
1.6. ANSWERS TO EXERCISES
15
20. Inverse Hyperbolic Derivatives Show that
(a)
d
1
,
(arctanh x) =
dx
1 − x2
(b)
d
−1
,
(arcsech x) = √
dx
x 1 − x2
−1 < x < 1
0<x<1
21. More Inverse Hyperbolic Derivatives Find the derivatives of the following functions:
(a) x3 arcsinh (ex )
√
(b) arccosh ( x)
(c) log(arccosh 4x)
(d)
1.6
1
arctanh x
Answers to Exercises
1.
(a)
4
3
2.
(a)
1
2
3.
(a) sinh x = ± 34 , tanh x = ± 53
√
√
(b) cosh x = 29/5,
tanh x = −2/ 29,
√
cosech x = −5/2
sech x = 5/ 29,
x−
1
x
(b)
5
4
(b)
1
2
x−3 + x3
(c)
312
313
(c)
x2 −x−2
x2 +x−2
√
coth x = − 29/2
4. Requires proof.
5. 1
6. Graphs required.
7.
(a) (i)
1
16
(sinh 5x − 5 sinh 3x + 10 sinh x) (ii)
(b) (i) 8 cosh4 x − 8 cosh2 x + 1 (ii) 4 sinh3 x +
1
32
(cosh 6x + 6 cosh 4x + 15 cosh 2x + 10)
3 sinh x
8. Proof required.
9.
(a) ex cosh(ex ), all x
(c)
(b)
sinh x
√
, for all x
2 cosh x
(d) 2x sech2 (x2 − 1), all x
(e) 3 cos 3x sech2 (sin 3x), for all x
10.
(a) π
11.
√
(a) 1√− x2 , −1 ≤ x ≤ 1
(b) x/ 1 + x2 , for all x
√
sinh( x)
√
for x > 0
2 x
(f) sinh(1/x) − (1/x) cosh(1/x), for x 6= 0
(b) − π4
(c)
(c)
√
1
2
1 − x2 /x, x 6= 0, −1 ≤ x ≤ 1
12. Graph required.
13.
√
(a) (2 arcsin x)/ 1 − x2
(b) 1 + 2x arctan x
(c) ex /(1 + e2x )
(d) −x/(1 − x2 ) − arcsin x/(1 − x2 )3/2
14. Requires proof.
15.
(a)
(1 + y 2 )(3x2 + arctan y)
(1 + y 2 )ey − x
p
1 − x2 y 2 − y
(b)
x
16
16.
CHAPTER 1. HYPERBOLIC FUNCTIONS
(a)
√
x2 − 1
(b) x2 /(1 − x2 )
√
(c)
x2 − 1
x
17. Graph required.
18. Proof required.
19.
(a) Proof required.
(b) √
1
x2 − 1
(c) (i) log 2
20. Requires proof.
21.
(a) 3x2 arcsinh (ex ) + √
(c)
x3 ex
1 + e2x
4
√
,x>
(arccosh 4x) 16x2 − 1
1
,x>1
√ √
2 x x−1
−1
(d)
, |x| < 1
(1 − x2 ) arctanh2 x
(b)
1
4
(ii) 0
Chapter 2
Techniques of Integration
You will find some exercises in this chapter. Answers are not provided, but can easily be checked by differentiating the result.
2.1
Why Integration?
One of the greatest challenges for ancient mathematicians was to find the area of objects bounded by curves.
The methods for finding such areas were rather basic until the 17th century when Isaac Newton and Gottfried
Leibniz independently discovered the relationship between derivatives and areas.
y
f(x)
b
a
x
Theorem 2.1 (The Fundamental Theorem of Calculus) The area under the graph y = f (x) between x =
a and x = b is
Z
b
f (x) dx = F (b) − F (a)
a
where f (x) =
d
F (x) and f (x) ≥ 0.
dx
In order to use this theorem we need to study the reverse of differentiation, antidifferentiation.
Definition 2.2 A function F is called an antiderivative of f on an interval I if f (x) =
d
F (x) for all x ∈ I.
dx
Alternatively F (x) is the indefinite integral of f (x).
Once we know an antiderivative of a function we can use it find areas. (Note: F is only defined up to a
constant.)
Integration and Differential Equations
Finding area is not the only application of integration. Integration is also the key to solving differential equations.
These had their origins in physics.
17
18
CHAPTER 2. TECHNIQUES OF INTEGRATION
Source: earthquake.usgs.gov
∆v
For example acceleration = g =
where v is velocity, x is distance travelled and t is time taken. Thus we
∆t
R
dv
have
= g so v = g dt = gt + c.
dt
Z
dx
∆x
= v giving velocity =
= v. Thus x = gt + c dt = gt2 + ct + d.
Then
∆t
dt
Given the height of the tree we can solve for c and d and then calculate the velocity and displacement of the
apple when it hits Newton’s head.
The above is just one sample of a differential equation. There are many other examples in physics, engineering,
chemistry, commerce and biology.
2.2
Basic Integrals
It is assumed that all students know the following integrals.
f (x)
xn
x−1
Integral
R
f (x) dx
1
xn+1 + c, for n 6= −1
n+1
log |x| + c
ekx
(**)Note
1 kx
e +c
k
Note that log x = ln x throughout these notes. Also note well the modulus signs in (**). This reflects the fact
d
d
that
log(−x) =
log(x).
dx
dx
Other Basic Integrals Including Some New Ones
R
R
f (x)
Integral f (x) dx
f (x)
Integral f (x) dx
1
1
sin kx
− cos kx + c
sinh kx
cosh kx + c
k
k
cos kx
1
sin kx + c
k
cosh kx
1
sinh kx + c
k
sec2 kx
1
tan kx + c
k
sech2 kx
1
tanh kx + c
k
Inverse Trigonometric and Hyperbolic Functions
2.3. DOUBLE ANGLE FORMULAE
f (x)
1
√
a2 − x2
19
R
Integral f (x) dx
x
+ c,
arcsin
a
f (x)
1
a2 + x2
R
Integral f (x) dx
x
1
+c
arctan
a
a
|x| < |a|
√
1
x2 − a 2
arccosh
x
a
+ c,
a2
1
− x2
x
1
+ c,
arctanh
a
a
|x| > |a|
√
2.3
1
x2 + a 2
arcsinh
x
a
|x| < |a|
+c
Double Angle Formulae
It is assumed that you know the following formulae:
cos 2x = cos2 x − sin2 x
cosh 2x = cosh2 x + sinh2 x
cos 2x = 2 cos2 x − 1
cosh 2x = 2 cosh2 x − 1
cos2 x =
1 1
+ cos 2x
2 2
cos 2x = 1 − 2 sin2 x
sin2 x =
1 1
− cos 2x
2 2
sin 2x = 2 sin x cos x
2.4
cosh2 x =
1 1
+ cosh 2x
2 2
cosh 2x = 1 + 2 sinh2 x
sinh2 x =
1
1
cosh 2x −
2
2
sinh 2x = 2 sinh x cosh x
Derivative Substitution
It is assumed that students know the following from school.
Z
f (g(x))g 0 (x) dx = F (g(x)) + c where F 0 (x) = f (x).
Z
Example 2.1 Find
cot(x) dx.
Solution:
Z
Z
cot(x) dx
=
=
cos x
dx
sin x
log | sin x| + C
20
CHAPTER 2. TECHNIQUES OF INTEGRATION
2.5
Change of Variable
We can extend the method of derivative substitution to other examples where the composition f (g(x)) appears
in the integrand. This is particularly useful where some awkward powers of linear functions appear.
The general method is as follows:
1. Look for some composition f (g(x)) within the integrand and let u = g(x) so that du = g 0 (x) dx. Then see
if this produces an integral expressed entirely in terms of u. (This may not work).
2. If Step 1 works, try to evaluate the integral in terms of u.
3. If Step 2 works, replace u by g(x) to express your final answer in terms of x.
Z
1
Example 2.2 Find
dx.
x2 + 8x + 25
Solution: We first complete the square:
x2 + 8x + 25 = x2 + 8x + 16 + 9 = (x + 4)2 + 9.
Let u = x + 4 so that
du
= 1 ⇒ du = dx. Then
dx
Z
Z
√
Z
1
dx
9 + (x + 4)2
Z
1
=
du
9 + (u)2
u
1
=
arctan
+C
3
3
x+4
1
arctan
+C
=
3
3
=
1
√ dx.
x+ x
Example 2.3 Find
Solution: Let t =
1
dx
2
x + 8x + 25
dx
= 2t ⇒ 2t dt = dx. Then
dt
Z
Z
1
1
√ dx =
(2t) dt
t2 + t
x+ x
Z
2
=
dt
t+1
= 2 log |t + 1| + C
√
= 2 log(1 + x) + C
x so that t2 = x and
(2x)2
dx.
(2x + 1)5
Z
Example 2.4 Find
Solution: Let u = 2x + 1 ⇒ u − 1 = 2x. Then du = 2dx ⇒ dx =
Z
(2x)2
dx
(2x + 1)5
1
du so
2
(u − 1)2 1
du
u5
2
Z 2
1
u
u
1
=
− 2 5 + 5 du
5
2
u
u
u
1
1
2
1
=
− 2 + 3 − 4 +C
2
2u
3u
4u
1
1
1
= −
+
−
+C
4(2x + 1)2
3(2x + 1)3
8(2x + 1)4
Z
=
2.6. TRIGONOMETRIC AND HYPERBOLIC SUBSTITUTIONS – SIN, COS, SINH AND COSH
Z
21
0
Example 2.5 Find
x2 (x + 1)10 dx.
−1
Solution: Let x + 1 = t. Then x = t − 1 and dx = dt. When x = −1, t = 0 and when x = 0, t = 1. Thus
Z
0
Z
x2 (x + 1)10 dx
1
(t − 1)2 t10 dt
=
−1
0
Z
=
1
t12 − 2t11 + t10 dt
0
t13
t12
t11
−2
+
13
12
11
1
1
1
−2 +
13
12 11
287
858
=
=
=
2.6
1
0
Trigonometric and Hyperbolic Substitutions – sin, cos, sinh and
cosh
2
2
Another area where change of variable may be useful is where there are√
rational functions
involving
√
√ ±a ± x .
2
2
2
2
2
2
Initially we will look at integrals that contain expressions of the form a − x , a + x or x − a . The
correct substitution comes from the identities
• cosh2 θ = 1 + sinh2 θ
• cos2 θ = 1 − sin2 θ
• sinh2 θ = cosh2 θ − 1.
Picking a substitution
•
p
p
p
√
a2 − x2 : Let x = a sin θ. Then a2 − x2 = a2 − a2 sin2 θ = a2 cos2 θ = a cos θ.
•
p
p
a2 + x2 : Let x = a sinh θ. Then a2 − x2 = a cosh θ.
•
p
p
x2 + a2 : Let x = a cosh θ. Then x2 − a2 = a sinh θ.
Z
Example 2.6 Use a substitution to verify that
Solution: Let x = 2 sinh t, then
Z
√
1
x2
+4
dx = arcsinh
x
+c
2
dx
= 2 cosh t ⇒ dx = 2 cosh t dt and
dt
√
Z
1
x2
+4
dx
=
1
p
4 sinh2 t + 4
Z
=
1 dt
=
t+c
=
arcsinh
x
+c
2
2 cosh t dt
22
CHAPTER 2. TECHNIQUES OF INTEGRATION
Z
Example 2.7 Find
1
(x2
3
+ 4) 2
dx.
Solution: Let x = 2 sinh t, then dx = 2 cosh t dt and
Z
1
√
dx
2
( x + 4)3
Z
=
=
=
=
=
=
Example 2.8 Find
Z p
1
p
2 cosh t dt
( 4 sinh2 t + 4)3
Z
1
2 dt
4
cosh
t
Z
1
sech2 t dt
4
1
tanh t + c
4
1 sinh t
+c
4 cosh t
x
1
q 2 +c
2
4
1 + x2
9 − x2 dx.
Solution: Let x = 3 sin t so that dx = 3 cos t dt. Then
Z p
Z p
9 − x2 dx =
9 − 9 sin2 t(3 cos t) dt
Z
=
9 cos2 t dt
Z
9
=
(1 + cos(2t)) dt
2
9
1
=
(t − sin(2t)) + c
2
2
9
=
(t − sin t cos t) + c
2
!
r
x x
x2
9
arcsin
1−
+c
=
−
2
3
3
9
x xp
9
=
arcsin
−
9 − x2 + c
2
3
2
Exercise: Use Example 3 to find the area of a circle of radius 3.
2.7
Substitution with tan and tanh
Z
This type of substitution is useful to find
1
and
2
a + x2
Z
a2
1
.
− x2
The correct substitution comes from the identities sec2 θ = 1 + tan2 θ and sech2 θ = 1 − tanh2 θ.
•
•
a2
1
: Let x = a tan θ. Then a2 + x2 = a2 sec2 θ.
+ x2
1
: Let x = a tanh θ. Then a2 − x2 = a2 sech2 θ. Note that this method is an alternative to the
a2 − x2
method of partial fractions (see later)..
2.8. PRODUCTS OF TRIGONOMETRIC AND HYPERBOLIC FUNCTIONS
Z
Example 2.9 Find
1
dx.
(4 − x2 )2
Solution: Let x = 2 tanh t. Then dx = 2 sech2 t dt and
Z
Z
1
2 sech2 t
dx
=
dt
2
2
(4 − x )
16 sech4 t
Z
1
=
cosh2 t dt
8
Z
1 1 1
=
( + cosh(2t)) dt
8 2 2
1 t
1
=
( + sinh(2t)) + c
8 2 4
1
=
(t + cosh t sinh t) + c
16 tanh t
1
t+
=
+c
16
sech2 t
1
tanh t
=
t+
+c
16
1 − tanh2 t
2.8
=
1
16
=
1
16
x
x
2
arctanh( ) +
2
2
1 − x4
!
x
2x
arctanh( ) +
2
4 − x2
+c
+c
Products of Trigonometric and Hyperbolic Functions
Z
1. Method for
sinm x cosn x dx
• If n is odd split off a factor of cos x and use cos2 x = 1 − sin2 x. Let u = sin x.
We obtain cosn x = cosn−1 x cos x = (1 − sin2 x)
n−1
2
cos x.
• If m is odd split off a factor of sin x and use sin2 x = 1 − cos2 x. Let u = cos x.
We obtain sinm x = sinm−1 x sin x = (1 − cos2 x)
m−1
2
sin x.
• If m and n are both even use double angle formulae.
Z
2. Method for
sinhm x coshn x dx
• If n is odd split off a factor of cosh x and use cosh2 x = 1 + sinh2 x. Let u = sinh x.
We will obtain coshn x = coshn−1 x cosh x = (1 + sinh2 x)
n−1
2
2
cosh x.
2
• If m is odd split off a factor of sinh x and use sinh x = cosh x − 1. Let u = cosh x.
We will obtain sinhm x = sinhm−1 x sinh x = (cosh2 x − 1)
• If m and n are both even use double angle formulae.
Z
3. Method for
tanm x secn x dx with n even.
• Split off a factor of sec2 x and use sec2 x = 1 + tan2 x.
Let u = tan x.
=⇒ secn x = secn−2 x sec2 x = (1 + tan2 x)
n−2
2
sec2 x.
m−1
2
sinh x.
23
24
CHAPTER 2. TECHNIQUES OF INTEGRATION
Z
4. Method for
tanhm x sechn x dx with n even.
• Split off a factor of sech2 x and use sech2 x = 1 − tanh2 x.
Let u = tanh x. Note:
d
tanh x = sech2 x.
dx
=⇒ sechn x = sechn−2 x sech2 x = (1 − tanh2 x)
n−2
2
sech2 x.
5. Note: A useful trick is the following
Z
Z
sec2 x + tan x sec x
dx
sec x dx =
sec x + tan x
= log | sec x + tan x| + c
Z
Example 2.10 Find
sinh2 x cosh2 x dx.
Solution:
Z
Z
Example 2.11 Find
Z
1
sinh2 (2x) dx
4
Z
1
(cosh(4x) − 1) dx
=
8
1 1
=
sinh(4x) − x + c
8 4
sinh2 x cosh2 x dx =
sinh3 x cosh2 x dx.
Solution:
Z
3
Z
2
sinh x cosh x dx =
Z
=
=
=
Z
Example 2.12 Find
(cosh2 x − 1) cosh2 x sinh x dx
(u2 − 1)u2 du
where u = cosh x
u5
u3
−
+c
5
3
cosh5 x cosh3 x
−
+c
5
3
tanh x sech4 x dx.
Solution:
Z
Z
4
tanh x sech x dx =
=
2.9
tanh2 x tanh4 x
−
+c
2
4
Partial Fractions
Z
Finding
tanh x(1 − tanh2 x) sech2 x dx
1
dx or
(x − 2)2
Z
1
dx is straightforward. But what about
x+1
Z
4x − 5
dx? The trick is
(x − 2)2 (x + 1)
to use
Z
4x − 5
dx =
(x − 2)2 (x + 3)
Z 1
1
1
+
−
2
(x − 2)
x−2 x+1
dx.
2.9. PARTIAL FRACTIONS
25
When do we use partial fractions?
When we have
Z
P (x)
dx
Q(x)
where
• P (x) and Q(x) are polynomials of degree p and q respectively,
• p < q and
• Q(x) can be factorised.
The Method
1. Factorise the denominator, Q(x).
2. Work out numerator for each term in the partial fraction expansion.
3. Write down partial fraction expansion.
4. Find unknown coefficients.
5. Integrate each partial fraction
Choosing The Numerators in the Partial Fraction Expansion
Factor in Denominator
Partial Fraction Expansion
A
x−a
x−a
(x − a)n , n = 2, 3, . . .
A1
A2
An
+
+ ... +
x − a (x − a)2
(x − a)n
Irreducible quadratics
x2 + bx + c
Bx + C
x2 + bx + c
(x2 + bx + c)n
Bn x + C n
B1 x + C1
+ ... + 2
,
x2 + bx + c
(x + bx + c)n
n = 2, 3, . . .
Z
Finding
P (x)
dx when degree Q(x) = q ≤ degree P (x) = p
Q(x)
When degree of P (x) is greater than degree of Q(x) divide P (x) by Q(x) and then use standard integrals and
partial fractions as appropriate.
Example 2.13 Find the integral of
4
.
(x + 1)2 (x + 3)
Solution: Put
4
A
B
(A + C)x2 + (4A + B + 2C)x + (3A + 3B + C)
C
=
+
=
.
+
(x + 1)2 (x + 3)
x + 1 (x + 1)2
x+3
(x + 1)2 (x + 3)
Equating the coefficients of x2 and x and the constant terms we obtain
and find that A = −1, B = 2 and C = 1.
A+C
=
0
4A + B + 2C
=
0
3A + 3B + C
=
4
26
CHAPTER 2. TECHNIQUES OF INTEGRATION
So
Z
1
4
2
2
−1
+
dx
=
+
dx = − log |x + 1| −
+ log(x + 3) + c.
(x + 1)2 (x + 3)
x + 1 (x + 1)2
x+3
x+1
Z
5x
Example 2.14 Find
dx.
2
(x + 4)(x − 1)
Z
Solution: Put
Ax + B
C
(Ax + B)(x − 1) + C(x2 + 4)
5x
=
+
=
.
(x2 + 4)(x − 1)
x2 + 4
x−1
(x2 + 4)(x − 1)
Equating the coefficients of x2 and x and the constant terms we obtain
A+C
=
0
−A + B
=
5
−B + 4C
=
0
to give A = −1, B = 4, C = 1 so
Z
5x
dx
(x2 + 4)(x − 1)
−x
4
1
+
+
dx
x2 + 4 x2 + 4 x − 1
x
1
4
− log |x2 + 4| + arctan
+ log |x − 1| + c
2
2
2
1
x
− log(x2 + 4) + 2 arctan
+ log |x − 1| + c
2
2
Z
=
=
=
2x3 + 9x2 + 12x + 3
dx.
x2 + 4x + 3
Z
Example 2.15 Find
Solution: As degree of numerator is greater than degree of denominator we first divide.
2x
2
x + 4x + 3
+
1
3
+
9x2
+
2x3
+
8x2
+
6x
2
+
6x +
3
x2
+
4x +
3
) 2x
x
12x +
3
2x
Thus
2x3 + 9x2 + 12x + 3
2x
= 2x + 1 + 2
x2 + 4x + 3
x + 4x + 3
and we find
Z
Solving
x2
2x3 + 9x2 + 12x + 3
dx
x2 + 4x + 3
Z =
2x + 1 +
2x
2
x + 4x + 3
dx
2x
A
B
=
+
yields A = −1 and B = 3 so
+ 4x + 3
x+1 x+3
Z
2x3 + 9x2 + 12x + 3
dx
x2 + 4x + 3
=
Z 2x + 1 −
=
x2 + x − log |x + 1| + 3 log |x + 3| + C
1
3
+
x+1 x+3
dx
2.10. INTEGRATION BY PARTS
27
Example 2.16 Write down the partial fraction expansion for
5x
. Do not solve for the
(x2 + 2x + 4)2 (x − 1)3
constants.
Solution:
5x
Ax + B
E
G
Cx + D
F
= 2
+
+
+
+
(x2 + 2x + 4)2 (x − 1)3
x + 2x + 4 (x2 + 2x + 4)2
x − 1 (x − 1)2
(x − 1)3
Notice that the degree of the denominator on the left is 7 which equals the number of constants in the partial
fraction expansion.
2.10
Integration by Parts
The primary goal of this method of integration is to find a method to integrate functions of the form f (x)g(x).
The key to this is to use the product rule:
d
du
dv
(uv) =
v+u
dx
dx
dx
or
dv
d
du
=
(uv) −
v.
dx
dx
dx
Integrating both sides of the secont form gives the formula for integration by parts:
Z
Z
dv
du
u dx = uv −
v dx.
dx
dx
u
Strategy There are no set rules for doing integration by parts. Success is mainly a matter of experience that
comes from doing lots of problems. However there are some useful ways of starting.
• choose u so that
du
is simpler than u and
dx
• choose
dv
so that it can be integrated.
dx
Some good choices for u and v:
Z
Example 2.17 Find
f (x)g(x)
u
dv
dx
du
dx
v
x cos x
x
cos x
1
sin x
xn loge x
loge x
xn
1
x
xn+1
(n + 1)
arcsin x
arcsin x
1
√
1
1 − x2
x
arctan x
arctan x
1
1
(x2 + 1)
x
x3 log x dx.
Solution:
Z
3
x log x dx =
=
Z 4
x4
x 1
log x −
dx
4
4 x
x4
x4
log x −
+c
4
16
28
CHAPTER 2. TECHNIQUES OF INTEGRATION
Z
xe2x dx.
Example 2.18 Find
Solution:
Z
xe
2x
Z 2x
xe2x
e
−
dx
2
2
xe2x
e2x
−
+C
2
4
=
=
Z
π
x cos nx dx where n ∈ Z.
Example 2.19 Find
0
Solution: Assume n 6= 0. Then
Z π
x cos nx dx
0
π
Z
π
1
sin nx dx
0 n
0
π
1
1
=
π sin nπ − − 2 cos nx
n
n
0
1
= 0 + 2 (cos nπ − 1)
 n

0,
n even
=

 2
− n2 , n odd
=
1
x sin nx
n
−
If n = 0 then the answer will be 0.
Z
Example 2.20 Find
x2 sinh x dx.
Solution:
Z
x2 sinh x dx = x2 cosh x −
Z
2x cosh x dx
Z
2
= x cosh x − 2x sinh x + 2 sinh x dx
= x2 cosh x − 2x sinh x + 2 cosh x + C
Z
Example 2.21 Find
log x dx.
Solution:
Z
Z
x
dx
x
= x log x − x + C
log x dx = x log x −
Z
Example 2.22 What is
e2x sin x dx?
Solution:
Z
e2x sin x dx
Z
1
2x
= −e cos x − − e2x cos x dx
2
Z
= −e2x cos x + 2e2x sin x − 4 e2x sin x dx
I
=
= −e2x cos x + 2e2x sin x − 4I
Solving for I gives
Z
I=
e2x sin x dx =
1
−e2x cos x + 2e2x sin x + C.
5
2.11. COMPLEX EXPONENTIAL (REVISION FROM UMEP MATHEMATICS)
2.11
29
Complex Exponential (Revision from UMEP Mathematics)
There is another way than integration by parts to do Example 2.22 – using the complex exponential.
Solution: Using cos x = Re(eix ) we obtain
Z
e2x sin x dx =
Z
Im
Z
=
Im
e2x eix dx
e(2+i)x dx
1 (2+i)x
e
+C
2+i
2 − i 2x
= Im
e (cos x + i sin x) + C
5
1 2x
= Im e (2 cos x + sin x − i cos x + 2i sin x) + C
5
1 2x
=
e (− cos x + 2 sin x) + C
5
=
2.12
Im
Integration Exercises
1. Basic Integration Evaluate the following integrals:
Z
Z
sin kx
√
dx (k 6= 0)
(a) (2x − 5)3 + (2x + 5)3 dx
(b)
2 + cos kx
Z
Z
1
1
√ dx
(d)
dx
(c)
x2 + 4x + 13
5x − x
Z
Z
3x
(e)
cos2 7x dx
(f)
dx
(x − 2)(x + 4)
Z
Z
1
ex
(g)
dx
(h)
dx
x log x
ex + 1
Z
(i)
Z
(k)
Z
4x + 17
dx
2
x + 10x + 25
3
(j)
−1
Z
sin6 x cos3 x dx
(l)
1
dx
2x + 3
√
1
dx
48 − 8x − x2
2. Trigonometric and Hyperbolic Substitutions Using an appropriate trigonometric or hyperbolic substitution,
find the indefinite integrals of the following functions:
p
p
1
(b) 4 − x2
(c)
(a) 1 + 4x2
3
(x2 − 1) 2
3. Hyperbolic Powers Find the indefinite integrals of the following powers of hyperbolic functions:
(a) sinh6 x cosh x
(b) cosh2 3x
(d) sinh3 4x
(e) cosh4 x
(c) sinh2 x cosh3 x
4. Hyperbolic Tangent Integrals Evaluate the following integrals, where k > 0 is a constant:
Z
Z
Z
sech2 kx
(a) tanh x sech2 x dx
(b)
dx
(c) tanh2 3x dx
2 + tanh kx
30
CHAPTER 2. TECHNIQUES OF INTEGRATION
5. Partial Fractions Find the indefinite integrals of the following:
(a)
1
(x + 2)(x2 + 1)
(b)
37 − 11x
(x + 1)(x − 2)(x − 3)
(c)
4x + 3
(x2 + 1)(x2 + 2)
(d)
x + 13
x3 + 2x2 − 5x − 6
x3 + 3x − 2
x2 − x
6. By Parts Evaluate the following integrals:
Z
Z
(a) x cos 3x dx
(b) arcsin x dx
(e)
Z
1
(c)
x arctan x dx
0
Z
Z
2
(d)
x cosh x dx
(e)
arcsinh x dx
7. By Parts Again Using integration by parts, evaluate the following integrals:
Z
Z e
7
(a) x log x dx, x > 0
(b)
log(x4 ) dx
1
Z
(c)
Z
ex cos 3x dx
(d)
e−2x sin 11x dx
8. Complex Exponential (Revision) Find the indefinite integrals of the following functions using the complex
exponential:
(b) e−2x sin 11x
(a) ex cos 3x
(c) e5t cos 7t
9. Mixed Integrals Find the indefinite integrals of the following functions:
(a) cosh3 x sinh3 x
(b)
x2 − 2
− 6x − 2
√
(g) x x + 3
(d)
1.
x log x,
(e) √
x3
2.13
√
x>0
(c)
1
9 − 4x2
x2 + 3x + 4
x2 + x
(f) cosh5 x sinh4 x
√
5
(h) cosh x sinh x cosh 2x + 55
(i) e2x sin 3x
Answers to Exercises
(a)
(c)
(e)
1
1
4
8 (2x − 5) + 8 (2x +
√
2
5 log |5 x − 1| + C
1
1
2 x + 28 sin 14x + C
√
(b) − k2 2 + cos kx + C
(d) 13 arctan x+2
+C
3
5)4 + C
(f) log |x − 2| + 2 log |x + 4| + C
(h) log(ex + 1) + C
(g) log | log x| + C
(i) 4 log |x + 5| +
(k)
1
7
7
sin x −
1
9
3
x+5
9
+C
(j)
sin x + C
√
2. (a) 14 arcsinh 2x + 2x 1 + 4x2 + C
−x
(c) √
+C
x2 − 1
3. (a)
(c)
(e)
4. (a)
7
1
7 sinh x + C
3
5
1
1
3 sinh x + 5 sinh x + C
3
1
1
8 x + 4 sinh 2x + 32 sinh 4x
1
2
tanh2 x + C
(c) x −
1
3
tanh 3x + C
1
2
log 9
(l) arcsin x+4
+C
8
√
(b) x2 4 − x2 + 2 arcsin
x
2
(d)
1
12 sinh 6x + C
3
1
1
12 cosh 4x − 4 cosh 4x
(b)
1
k
+C
(b) 21 x +
+C
log |2 + tanh kx| + C
+C
2.13. ANSWERS TO EXERCISES
5.
31
1
(a) 51 log |x + 2| − 10
log(x2 + 1) + 25 arctan x + C
(b) 4 log |x + 1| − 5 log |x − 2| + log |x − 3| + C
(c) 2 log(x2 + 1) + 3 arctan x − 2 log(x2 + 2) −
√3
2
arctan
x
√
2
(d) −2 log |x + 1| + log |x + 3| + log |x − 2| + C
(e) 12 x2 + x + 2 log |x| + 2 log |x − 1| + C
6. (a) 13 x sin 3x +
(c) arctan 1 −
1
9
1
2
(e) x arcsinh x −
7. (a) 18 x8 log x −
(c)
8. (a)
(c)
9.
x
e
10 (cos 3x
cos 3x + C
√
(b) x arcsin x +
√
+C
1 − x2 + C
(d) x2 sinh x − 2x cosh x + 2 sinh x + C
x2 + 1 + C
1 8
64 x
+C
+ 3 sin 3x) + C
ex
10 (cos 3x + 3 sin 3x) + C
e5t
74 (5 cos 7t + 7 sin 7t) + C
(b) 4
(d)
e−2x
125 (−11 cos 11x
− 2 sin 11x) + C
(b)
e−2x
125 (−11 cos 11x
− 2 sin 11x) + C
1
1
1
1
1
1
sinh4 x + sinh6 x + C = cosh6 x − cosh4 x + D =
cosh3 2x −
cosh 2x + E
4 6
6
4
48
16
2
4
(b) x3/2
log x −
+C
(c) x + 4 log |x| − 2 log |x + 1| + C
3
9
1
2x
1
(e) arcsin
+C
(d) log |x3 − 6x − 2| + C
3
2
3
5
3
1
2
1
2
(f) sinh9 x + sinh7 x + sinh5 x + C
(g) (x + 3) 2 − 2(x + 3) 2 + C
9
7
5
5
e2x
5
(h) 24
(cosh 2x + 55)6/5 + C
(i)
(−3 cos 3x + 2 sin 3x) + C
13
(a)