bridging notes for vector calculus · hyperbolic sine sinh shine which is the uk pronunciation (or,...
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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 UMEPMathematics and plan to enter Vector Calculus without having done Accelerated Mathematics 2 or Calculus 2first. The notes cover Hyperbolic Functions and Techniques of Integration. At the end of each chapter there areexercises given. Students are advised to do as many of these exercises as possible making sure that all topicsare practised.
Students are welcome to ask questions about this material. Please see your Vector Calculus lecturer if you wantany help.
Acknowledgement Some of the materials in these notes have been provided by Christine Mangelsdorf andPaul Pearce.
1
2
Contents
1 Hyperbolic Functions 51.1 Introduction to Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.3 Inverse Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.4 Derivatives of Inverse Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.5 Exercises for Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.6 Answers to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2 Techniques of Integration 172.1 Why Integration? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.2 Basic Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.3 Double Angle Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.4 Derivative Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.5 Change of Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.6 Trigonometric and Hyperbolic Substitutions – sin, cos, sinh and cosh . . . . . . . . . . . . . . . . 212.7 Substitution with tan and tanh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.8 Products of Trigonometric and Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . 232.9 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.10 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272.11 Complex Exponential (Revision from UMEP Mathematics) . . . . . . . . . . . . . . . . . . . . . 292.12 Integration Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.13 Answers to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
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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 functionsare hyperbolic sine and hyperbolic cosine abbreviated sinh and cosh respectively. From these we can derivehyperbolic tangent (tanh) and so forth.
Just as cosine and sine (abbreviated sin and cos respectively) are used to parametrize the circle, hyperbolicsine and hyperbolic cosine are used to parametrize the hyperbola. They also have application to techniques ofintegration, 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 complexexponential and Euler’s equation, eiθ = cos θ + i sin θ. The identities are:
cos θ =eiθ + e−iθ
2and sin θ =
eiθ − e−iθ
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
coshx =ex + e−x
2
2. sinh : R→ R such that
sinhx =ex − e−x
2
5
6 CHAPTER 1. HYPERBOLIC FUNCTIONS
3. tanh : R→ R such that
tanhx =sinhx
coshx=e2x − 1
e2x + 1=
1− e−2x
1 + e−2x
4. coth : R\{0} → R such that
cothx =coshx
sinhx=e2x + 1
e2x − 1=
1 + e−2x
1− e−2x
5. sech : R→ R such that
sechx =1
coshx=
2
ex + e−x
6. cosech : R\{0} → R such that
cosechx =1
sinhx=
2
ex − e−x
Graphs
Below find the graphs of sinhx and coshx. Asymptotically
1. sinhx→ ex
2as x→∞ and sinhx→ −e
−x
2as x→ −∞
2. coshx→ ex
2as x→∞ and coshx→ e−x
2as x→ −∞
ex
2
�ex
2
�2 �1 1 2x
�3
�2
�1
1
2
3
sinh�x�
ex
2
e-x
2
-2 -1 1 2x
0.5
1.0
1.5
2.0
2.5
3.0
3.5
coshHxL
It is clear from the graph of coshx that the range of cosh is [1,∞).
Identities
The fundamental identity is
cosh2 x− sinh2 x =1
4
((ex + e−x)2 − (ex − e−x)2
)=
1
4
((e2x + 2 + e−2x)− (e2x − 2 + e−2x)
)= 1
From this it follows that
1− tanh2 x = sech2 x and coth2 x− 1 = cosech2 x.
Parametrising the Hyperbola
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 tand y = sinh t, t ∈ R we obtain the right branch of the hyperbola x2 − y2 = 1. Setting x = − cosh t gives theleft branch.
x2- y2
= 1Hx,yL:
x = CoshHtL
y = SinhHtL
-3 -2 -1 1 2 3x
-3
-2
-1
1
2
3
y
More generally we have the following:
(1) x = a cosh t and y = b sinh t forx2
a2− y
2
b2= 1 and (2) x = b sinh t and y = a cosh t for
y2
a2− x
2
b2= 1.
Addition/Double Angle Formulas
The addition formulae are
• sinh(x+ y) = sinhx cosh y + coshx sinh y
• cosh(x+ y) = coshx cosh y + sinhx sinh y
• sinh(x− y) = sinhx cosh y − coshx sinh y
• cosh(x− y) = coshx cosh y − sinhx sinh y
The double angle formulae are
• sinh 2x = 2 sinhx coshx
• 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
y =T
ρgcosh
(ρgxT
)where g is the acceleration due to gravity and T is the tension in the cable at the lowest point.
Example 1.2 Write (coshx− sinhx)7 in terms of exponentials.
Solution
(coshx− sinhx)7 =
[1
2(ex + e−x)− 1
2(ex − e−x)
]7= (e−x)7 = e−7x
8 CHAPTER 1. HYPERBOLIC FUNCTIONS
Example 1.3 If coshx = 1312 and x < 0, find sinhx and tanhx.
Solution Now cosh2 x− sinh2 x = 1 so sinh2 x = cosh2 x− 1 = (1312 )2 − 1 = 169
144 − 1 = 25144 .
Hence sinhx = ±√
25144 = ± 5
12 . But x < 0 so sinhx < 0, so we must have sinhx = − 512 .
So finally, tanhx =sinhx
coshx=
(−5/12)
(13/12)= − 5
13.
Example 1.4 Write sinh(2 log x) as an algebraic function in x.
Solution
sinh(2 log x) =1
2[e2 log x − e−2 log x] =
1
2[elog x
2
− elog x−2
] =1
2(x2 − x−2)
Example 1.5 Write cosh3 x in terms of coshnx with n ∈ N.
Solution
cosh3 x = [1
2(ex + e−x)]3 =
1
8[e3x + 3ex + 3e−x + e−3x] =
1
8(e3x + e−3x) +
3
8(ex + e−x)
=1
4cosh 3x+
3
4coshx
Example 1.6 Prove the first addition formula.
Solution
RHS = sinhx cosh y + coshx sinh y
=1
2(ex − e−x)
1
2(ey + e−y) +
1
2(ex + e−x)
1
2(ey − e−y)
= 14 (ex+y + ex−y − e−x+y − e−x−y + ex+y − ex−y + e−x+y − e−x−y)
= 14 (2ex+y − 2e−x−y)
=1
2(ex+y − e−(x+y))
= sinh(x+ y)
= LHS
Example 1.7 Prove the first double angle formula.
Solution
LHS = sinh 2x
= sinh(x+ x)
= sinhx coshx+ coshx sinhx
= 2 sinhx coshx
= RHS
1.2. DIFFERENTIATION 9
1.2 Differentiation
The derivatives of the hyperbolic functions are as follows:
d
dx(coshx) = sinhx
d
dx(sinhx) = coshx
d
dx(tanhx) = sech2 x
d
dx(cothx) = − cosech2 x
d
dx(sechx) = − tanhx sechx
d
dx(cosechx) = − cothx cosech
Example 1.8 Prove the first and third derivative formulas.
Solution
d
dx(coshx) =
d
dx
[12
(ex + e−x)]
=1
2(ex − e−x)
= sinhx
d
dxtanhx =
d
dx
( sinhx
coshx
)=
coshx ddx (sinhx)− sinhx d
dx (coshx)
cosh2 x(quotient rule)
=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
2
1√sinh 6x
d
dx(sinh 6x)
=6 cosh 6x
2√
sinh 6x
=3 cosh 6x√
sinh 6x
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 monotonicincreasing 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
To avoid the confusion that arises because the inverse function sin−1 x may be confused with1
sinxthe notation
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 coshwe therefore restrict its domain to [0,∞) so that it is monotonic increasing and the inverse function arccosh iswell defined.
Definition 1.4 1. The inverse function of cosh is
arccosh : [1,∞)→ R
and
2. the inverse function of sinh isarcsinh : R→ R
The range of arccosh is [0,∞). The graphs are as follows:
-3 -2 -1 1 2 3x
-1.5
-1.0
-0.5
0.5
1.0
1.5
ArcsinhHxL
0.5 1.0 1.5 2.0 2.5 3.0x
0.5
1.0
1.5
ArccoshHxL
1.3. INVERSE HYPERBOLIC FUNCTIONS 11
We can express inverse hyperbolic functions in terms of log as follows:
• arcsinhx = log(x+√x2 + 1), x ∈ R
• arccoshx = log(x+√x2 − 1), x ≥ 1
• arctanhx =1
2log(1 + x
1− x), −1 < x < 1
Example 1.10 Derive a formula for arcsinh in terms of natural logarithms.
Solution
y = arcsinhx ⇐⇒ sinh y =1
2(ey − e−y) = x
Solve for y. Multiplying by 2ey we obtain a quadratic equation in ey
(ey)2 − 2x(ey)− 1 = 0
with solution
ey =1
2
(2x±
√4x2 + 4
)= x±
√x2 + 1
Since ey > 0 we must choose the plus sign since√x2 + 1 > x. Taking the logarithm gives
y = arcsinhx = log(x+
√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) = sinh(
log(3 +√
8))
=1
2
[elog(3+
√8) − e− log(3+
√8)]
=1
2
[3 +√
8− 1
3 +√
8
]=
1
2
[3 +√
8− 1
3 +√
8
3−√
8
3−√
8
]=
1
2
[3 +√
8− (3−√
8)]
=√
8 = 2√
2
Example 1.12 Express cosh(arctanhx) as an algebraic function of x for −1 < x < 1.
Solution
Let y = arctanhx so that x = tanh y. Now
cosh2 y =1
sech2 y=
1
1− tanh2 y=
1
1− x2
Hence
cosh(arctanhx) = cosh y =1√
1− x2, −1 < x < 1
where we need the positive square root since cosh y ≥ 1 > 0.
12 CHAPTER 1. HYPERBOLIC FUNCTIONS
1.4 Derivatives of Inverse Hyperbolic Functions
The derivatives are:
d
dx(arcsinhx) =
1√x2 + 1
, x ∈ R
d
dx(arccoshx) =
1√x2 − 1
, x > 1
d
dx(arctanhx) =
1
1− x2, −1 < x < 1
Example 1.13 Obtain the derivative of arcsinhx
Solution
Let y = arcsinhx so that x = sinh y. Then
dx
dy= cosh y
dx
dy= ± 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
dx
[arcsinh(x3)
]=
1√(x3)2 + 1
d
dx(x3) =
3x2√x6 + 1
1.5 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) (c) tanh (2 log x)
3. Hyperbolic Functions
(a) If coshx = 54 , what are the possible values of sinhx and tanhx?
(b) If sinhx = − 25 , compute coshx, tanhx, cothx, sechx and cosechx.
1.5. EXERCISES FOR HYPERBOLIC FUNCTIONS 13
4. Standard Hyperbolic Identities
Use the definitions of coshx and sinhx to verify the following identities, where n is any integer.
(a) coshx− sinhx = e−x
(b) cosh2 x+ sinh2 x = cosh 2x
(c) (coshx+ sinhx)n = coshnx+ sinhnx
5. Hyperbolic Limits Evaluate the limit:
limx→∞
tanhx
6. Sketching Hyperbolics
Sketch the graphs of
(a) tanhx, (b) cothx, (c) sechx, (d) cosechx
clearly marking key features of the graphs such as intercepts, maxima and minima and asymptotic be-haviour.
7. Manipulating Hyperbolic Functions
(a) Express the following functions in terms of hyperbolic sines and/or hyperbolic cosines of multiplesof x.
i. sinh5 x ii. cosh6 x
(b) Express the following functions in terms of powers of sinhx and coshx.
i. cosh 4x ii. sinh 3x
8. Standard Hyperbolic Derivatives Using the derivatives of sinhx, coshx and tanhx, show that
(a)d
dx(cothx) = −cosech2x, x 6= 0
(b)d
dx(sechx) = −sechx tanhx
(c)d
dx(cosechx) = −cosechx cothx, x 6= 0
9. Hyperbolic Derivatives Find the derivatives of the following functions. Check domains.
(a) sinh (ex) (b) cosh (√x ) (c)
√coshx
(d) tanh(x2 − 1) (e) tanh(sin 3x) (f) x sinh(1/x)
Revision from School: Inverse Trigonometric Functions
10. Exact Values Find the exact values of
(a) arccos(−1) (b) arctan(−1)(c) cos
[arcsin
(−√
3
2
)]
11. Inverse Trigonometric Functions Write the following as algebraic expressions in x.
(a) cos(arcsinx) (b) sin(arctanx) (c) tan(arccosx)
14 CHAPTER 1. HYPERBOLIC FUNCTIONS
12. Sketching Trigonometric Functions On the same graph sketch the following functions:
y = arctan(x
2
), y = 3 + arctan
(x2
), y = −2 arctan
(x2
+ 2)
13. Derivatives of Inverse Functions Find the derivative of the following functions.
(a) (arcsinx)2 (b) (1 + x2) arctanx
(c) arctan(ex) (d) 1− x arcsinx√1− x2
14. More Derivatives of Inverse Functions Using implicit differentiation, show that
(a)d
dx(arccosecx) =
−1
x√x2 − 1
, x > 1
(b)d
dx(arccotx) =
−1
1 + x2, x 6= 0
15. Differentiation involving Inverse Functions
Finddy
dxby implicit differentiation:
(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(arccoshx) (b) sinh2(arctanhx) (c) tanh(arccoshx)
17. Sketching Inverse Hyperbolics On the same graph sketch the following functions:
y = sinh(x
3
), y = cosech
(x3
), y = arccosech
(x3
)
18. Hyperbolic Identity If −1 < x < 1, show that:
arcsinh
(x√
1− x2
)= arctanhx
19. Inverse Hyperbolic Functions
(a) Prove that, for x ≥ 1,
arccoshx = log(x+√x2 − 1)
(b) Find the derivatived
dx(arccoshx)
i. using the formula in part (a) ii. using implicit differentiation
(c) Calculate the following limits:
i. limx→∞
(arccoshx− log x) ii. limx→1+
arccoshx
1.6. ANSWERS TO EXERCISES 15
20. Inverse Hyperbolic Derivatives Show that
(a)d
dx(arctanhx) =
1
1− x2, −1 < x < 1
(b)d
dx(arcsechx) =
−1
x√
1− x2, 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
arctanhx
1.6 Answers to Exercises
1. (a) 43 (b) 5
4 (c) 312313
2. (a) 12
(x− 1
x
)(b) 1
2
(x−3 + x3
)(c) x2−x−2
x2+x−2
3. (a) sinhx = ± 34 , tanhx = ± 3
5
(b) coshx =√
29/5, tanhx = −2/√
29, cothx = −√
29/2
sechx = 5/√
29, cosechx = −5/2
4. Requires proof.
5. 1
6. Graphs required.
7. (a) (i) 116 (sinh 5x− 5 sinh 3x+ 10 sinhx) (ii) 1
32 (cosh 6x+ 6 cosh 4x+ 15 cosh 2x+ 10)
(b) (i) 8 cosh4 x − 8 cosh2 x + 1 (ii) 4 sinh3 x + 3 sinhx
8. Proof required.
9. (a) ex cosh(ex), all x (b)sinh(
√x)
2√x
for x > 0
(c)sinhx
2√
coshx, for all x (d) 2x sech2(x2 − 1), all x
(e) 3 cos 3x sech2(sin 3x), for all x (f) sinh(1/x)− (1/x) cosh(1/x), for x 6= 0
10. (a) π (b) −π4 (c) 12
11. (a)√
1− x2, −1 ≤ x ≤ 1(b) x/
√1 + x2, for all x
(c)√
1− x2/x, x 6= 0, −1 ≤ x ≤ 1
12. Graph required.
13. (a) (2 arcsinx)/√
1− x2 (b) 1 + 2x arctanx
(c) ex/(1 + e2x) (d) −x/(1− x2)− arcsinx/(1− x2)3/2
14. Requires proof.
15.(a)
(1 + y2)(3x2 + arctan y)
(1 + y2)ey − x (b)
√1− x2y2 − y
x
16 CHAPTER 1. HYPERBOLIC FUNCTIONS
16. (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 (ii) 0
20. Requires proof.
21.(a) 3x2 arcsinh (ex) +
x3ex√1 + e2x
(b)1
2√x√x− 1
, x > 1
(c)4
(arccosh 4x)√
16x2 − 1, x > 1
4 (d)−1
(1− x2) arctanh2 x, |x| < 1
Chapter 2
Techniques of Integration
You will find some exercises in this chapter. Answers are not provided, but can easily be checked by differenti-ating 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 GottfriedLeibniz independently discovered the relationship between derivatives and areas.
y
xa b
f(x)
Theorem 2.1 (The Fundamental Theorem of Calculus) The area under the graph y = f(x) between x =a and x = b is
∫ b
a
f(x) dx = F (b)− F (a)
where f(x) =d
dxF (x) and f(x) ≥ 0.
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
dxF (x) for all x ∈ I.
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 aconstant.)
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
For example acceleration = g =∆v
∆twhere v is velocity, x is distance travelled and t is time taken. Thus we
havedv
dt= g so v =
∫g dt = gt+ c.
Then∆x
∆t= v giving velocity =
dx
dt= v. Thus x =
∫gt+ c dt = gt2 + ct+ d.
Given the height of the tree we can solve for c and d and then calculate the velocity and displacement of theapple 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) Integral∫f(x) dx
xn1
n+ 1xn+1 + c, for n 6= −1
x−1 log |x|+ c (**)Note
ekx1
kekx + c
Note that log x = lnx throughout these notes. Also note well the modulus signs in (**). This reflects the fact
thatd
dxlog(−x) =
d
dxlog(x).
Other Basic Integrals Including Some New Ones
f(x) Integral∫f(x) dx f(x) Integral
∫f(x) dx
sin kx −1
kcos kx+ c sinh kx
1
kcosh kx+ c
cos kx1
ksin kx+ c cosh kx
1
ksinh kx+ c
sec2 kx1
ktan kx+ c sech2 kx
1
ktanh kx+ c
Inverse Trigonometric and Hyperbolic Functions
2.3. DOUBLE ANGLE FORMULAE 19
f(x) Integral∫f(x) dx f(x) Integral
∫f(x) dx
1√a2 − x2
arcsin(xa
)+ c,
1
a2 + x21
aarctan
(xa
)+ c
|x| < |a|
1√x2 − a2
arccosh(xa
)+ c,
1
a2 − x21
aarctanh
(xa
)+ c,
|x| > |a| |x| < |a|
1√x2 + a2
arcsinh(xa
)+ c
2.3 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
2+
1
2cos 2x cosh2 x =
1
2+
1
2cosh 2x
cos 2x = 1− 2 sin2 x cosh 2x = 1 + 2 sinh2 x
sin2 x =1
2− 1
2cos 2x sinh2 x =
1
2cosh 2x− 1
2
sin 2x = 2 sinx cosx sinh 2x = 2 sinhx coshx
2.4 Derivative Substitution
It is assumed that students know the following from school.∫f(g(x))g′(x) dx = F (g(x)) + c where F ′(x) = f(x).
Example 2.1 Find
∫cot(x) dx.
Solution: ∫cot(x) dx =
∫cosx
sinxdx
= log | sinx|+ 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)) appearsin 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′(x) dx. Then seeif 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.
Example 2.2 Find
∫1
x2 + 8x+ 25dx.
Solution: We first complete the square:
x2 + 8x+ 25 = x2 + 8x+ 16 + 9 = (x+ 4)2 + 9.
Let u = x+ 4 so thatdu
dx= 1⇒ du = dx. Then
∫1
x2 + 8x+ 25dx =
∫1
9 + (x+ 4)2dx
=
∫1
9 + (u)2du
=1
3arctan
(u3
)+ C
=1
3arctan
(x+ 4
3
)+ C
Example 2.3 Find
∫1
x+√xdx.
Solution: Let t =√x so that t2 = x and
dx
dt= 2t⇒ 2t dt = dx. Then∫
1
x+√xdx =
∫1
t2 + t(2t) dt
=
∫2
t+ 1dt
= 2 log |t+ 1|+ C
= 2 log(1 +√x) + C
Example 2.4 Find
∫(2x)2
(2x+ 1)5dx.
Solution: Let u = 2x+ 1⇒ u− 1 = 2x. Then du = 2dx⇒ dx =1
2du so
∫(2x)2
(2x+ 1)5dx =
∫(u− 1)2
u51
2du
=1
2
∫u2
u5− 2
u
u5+
1
u5du
=1
2
(− 1
2u2+
2
3u3− 1
4u4
)+ C
= − 1
4(2x+ 1)2+
1
3(2x+ 1)3− 1
8(2x+ 1)4+ C
2.6. TRIGONOMETRIC AND HYPERBOLIC SUBSTITUTIONS – SIN, COS, SINH AND COSH 21
Example 2.5 Find
∫ 0
−1x2(x+ 1)10 dx.
Solution: Let x+1 = t. Then x = t−1 and dx = dt. When x = −1, t = 0 and when x = 0, t = 1. Thus
∫ 0
−1x2(x+ 1)10 dx =
∫ 1
0
(t− 1)2t10 dt
=
∫ 1
0
t12 − 2t11 + t10dt
=
[t13
13− 2
t12
12+t11
11
]10
=1
13− 2
1
12+
1
11
=287
858
2.6 Trigonometric and Hyperbolic Substitutions – sin, cos, sinh andcosh
Another area where change of variable may be useful is where there are rational functions involving ±a2 ± x2.Initially we will look at integrals that contain expressions of the form
√a2 − x2,
√a2 + x2 or
√x2 − a2. The
correct substitution comes from the identities
• cos2 θ = 1− sin2 θ • cosh2 θ = 1 + sinh2 θ • sinh2 θ = cosh2 θ − 1.
Picking a substitution
•√a2 − x2: Let x = a sin θ. Then
√a2 − x2 =
√a2 − a2 sin2 θ =
√a2 cos2 θ = a cos θ.
•√a2 + x2: Let x = a sinh θ. Then
√a2 − x2 = a cosh θ.
•√x2 + a2: Let x = a cosh θ. Then
√x2 − a2 = a sinh θ.
Example 2.6 Use a substitution to verify that
∫1√
x2 + 4dx = arcsinh
x
2+ c
Solution: Let x = 2 sinh t, thendx
dt= 2 cosh t⇒ dx = 2 cosh t dt and
∫1√
x2 + 4dx =
∫1√
4 sinh2 t+ 42 cosh t dt
=
∫1 dt
= t+ c
= arcsinhx
2+ c
22 CHAPTER 2. TECHNIQUES OF INTEGRATION
Example 2.7 Find
∫1
(x2 + 4)32
dx.
Solution: Let x = 2 sinh t, then dx = 2 cosh t dt and∫1
(√x2 + 4)3
dx =
∫1
(√
4 sinh2 t+ 4)32 cosh t dt
=
∫1
4 cosh2 tdt
=
∫1
4sech2 t dt
=1
4tanh t+ c
=1
4
sinh t
cosh t+ c
=1
4
x2√
1 +(x2
)2 + c
Example 2.8 Find
∫ √9− x2 dx.
Solution: Let x = 3 sin t so that dx = 3 cos t dt. Then∫ √9− x2 dx =
∫ √9− 9 sin2 t(3 cos t) dt
=
∫9 cos2 t dt
=
∫9
2(1 + cos(2t)) dt
=9
2(t− 1
2sin(2t)) + c
=9
2(t− sin t cos t) + c
=9
2
(arcsin
(x3
)− x
3
√1− x2
9
)+ c
=9
2arcsin
(x3
)− x
2
√9− x2 + c
Exercise: Use Example 3 to find the area of a circle of radius 3.
2.7 Substitution with tan and tanh
This type of substitution is useful to find
∫1
a2 + x2and
∫1
a2 − x2.
The correct substitution comes from the identities sec2 θ = 1 + tan2 θ and sech2 θ = 1− tanh2 θ.
• 1
a2 + x2: Let x = a tan θ. Then a2 + x2 = a2 sec2 θ.
• 1
a2 − x2: Let x = a tanh θ. Then a2 − x2 = a2 sech2 θ. Note that this method is an alternative to the
method of partial fractions (see later)..
2.8. PRODUCTS OF TRIGONOMETRIC AND HYPERBOLIC FUNCTIONS 23
Example 2.9 Find
∫1
(4− x2)2dx.
Solution: Let x = 2 tanh t. Then dx = 2 sech2 t dt and∫1
(4− x2)2dx =
∫2 sech2 t
16 sech4 tdt
=
∫1
8cosh2 t dt
=
∫1
8(1
2+
1
2cosh(2t)) dt
=1
8(t
2+
1
4sinh(2t)) + c
=1
16(t+ cosh t sinh t) + c
=1
16
(t+
tanh t
sech2 t
)+ c
=1
16
(t+
tanh t
1− tanh2 t
)+ c
=1
16
(arctanh(
x
2) +
x2
1− x2
4
)+ c
=1
16
(arctanh(
x
2) +
2x
4− x2
)+ c
2.8 Products of Trigonometric and Hyperbolic Functions
1. Method for
∫sinm x cosn x dx
• If n is odd split off a factor of cosx and use cos2 x = 1− sin2 x. Let u = sinx.
We obtain cosn x = cosn−1 x cosx = (1− sin2 x)n−12 cosx.
• If m is odd split off a factor of sinx and use sin2 x = 1− cos2 x. Let u = cosx.
We obtain sinm x = sinm−1 x sinx = (1− cos2 x)m−1
2 sinx.
• If m and n are both even use double angle formulae.
2. Method for
∫sinhm x coshn x dx
• If n is odd split off a factor of coshx and use cosh2 x = 1 + sinh2 x. Let u = sinhx.
We will obtain coshn x = coshn−1 x coshx = (1 + sinh2 x)n−12 coshx.
• If m is odd split off a factor of sinhx and use sinh2 x = cosh2 x− 1. Let u = coshx.
We will obtain sinhm x = sinhm−1 x sinhx = (cosh2 x− 1)m−1
2 sinhx.
• If m and n are both even use double angle formulae.
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 = tanx.
=⇒ secn x = secn−2 x sec2 x = (1 + tan2 x)n−22 sec2 x.
24 CHAPTER 2. TECHNIQUES OF INTEGRATION
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 = tanhx. Note:d
dxtanhx = sech2 x.
=⇒ sechn x = sechn−2 x sech2 x = (1− tanh2 x)n−22 sech2 x.
5. Note: A useful trick is the following∫secx dx =
∫sec2 x+ tanx secx
secx+ tanxdx
= log | secx+ tanx|+ c
Example 2.10 Find
∫sinh2 x cosh2 x dx.
Solution: ∫sinh2 x cosh2 x dx =
∫1
4sinh2(2x) dx
=
∫1
8(cosh(4x)− 1) dx
=1
8
(1
4sinh(4x)− x
)+ c
Example 2.11 Find
∫sinh3 x cosh2 x dx.
Solution: ∫sinh3 x cosh2 x dx =
∫(cosh2 x− 1) cosh2 x sinhx dx
=
∫(u2 − 1)u2 du where u = coshx
=u5
5− u3
3+ c
=cosh5 x
5− cosh3 x
3+ c
Example 2.12 Find
∫tanhx sech4 x dx.
Solution: ∫tanhx sech4 x dx =
∫tanhx(1− tanh2 x) sech2 x dx
=tanh2 x
2− tanh4 x
4+ c
2.9 Partial Fractions
Finding
∫1
(x− 2)2dx or
∫1
x+ 1dx is straightforward. But what about
∫4x− 5
(x− 2)2(x+ 1)dx? The trick is
to use
∫4x− 5
(x− 2)2(x+ 3)dx =
∫ (1
(x− 2)2+
1
x− 2− 1
x+ 1
)dx.
2.9. PARTIAL FRACTIONS 25
When do we use partial fractions?
When we have ∫P (x)
Q(x)dx
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
x− a A
x− a
(x− a)n, n = 2, 3, . . .A1
x− a+
A2
(x− a)2+ . . .+
An(x− a)n
Irreducible quadratics
x2 + bx+ cBx+ C
x2 + bx+ c
(x2 + bx+ c)nB1x+ C1
x2 + bx+ c+ . . .+
Bnx+ Cn(x2 + bx+ c)n
,
n = 2, 3, . . .
Finding
∫P (x)
Q(x)dx when degree Q(x) = q ≤ degree P (x) = p
When degree of P (x) is greater than degree of Q(x) divide P (x) by Q(x) and then use standard integrals andpartial fractions as appropriate.
Example 2.13 Find the integral of4
(x+ 1)2(x+ 3).
Solution: Put
4
(x+ 1)2(x+ 3)=
A
x+ 1+
B
(x+ 1)2+
C
x+ 3=
(A+ C)x2 + (4A+B + 2C)x+ (3A+ 3B + C)
(x+ 1)2(x+ 3).
Equating the coefficients of x2 and x and the constant terms we obtain
A+ C = 0
4A+B + 2C = 0
3A+ 3B + C = 4
and find that A = −1, B = 2 and C = 1.
26 CHAPTER 2. TECHNIQUES OF INTEGRATION
So ∫4
(x+ 1)2(x+ 3)dx =
∫−1
x+ 1+
2
(x+ 1)2+
1
x+ 3dx = − log |x+ 1| − 2
x+ 1+ log(x+ 3) + c.
Example 2.14 Find
∫5x
(x2 + 4)(x− 1)dx.
Solution: Put5x
(x2 + 4)(x− 1)=Ax+B
x2 + 4+
C
x− 1=
(Ax+B)(x− 1) + C(x2 + 4)
(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
∫5x
(x2 + 4)(x− 1)dx =
∫−x
x2 + 4+
4
x2 + 4+
1
x− 1dx
= −1
2log |x2 + 4|+ 4
2arctan
(x2
)+ log |x− 1|+ c
= −1
2log(x2 + 4) + 2 arctan
(x2
)+ log |x− 1|+ c
Example 2.15 Find
∫2x3 + 9x2 + 12x+ 3
x2 + 4x+ 3dx.
Solution: As degree of numerator is greater than degree of denominator we first divide.
2x + 1
x2 + 4x+ 3 ) 2x3 + 9x2 + 12x + 3
2x3 + 8x2 + 6x
x2 + 6x + 3
x2 + 4x + 3
2xThus
2x3 + 9x2 + 12x+ 3
x2 + 4x+ 3= 2x+ 1 +
2x
x2 + 4x+ 3
and we find
∫2x3 + 9x2 + 12x+ 3
x2 + 4x+ 3dx =
∫ (2x+ 1 +
2x
x2 + 4x+ 3
)dx
Solving2x
x2 + 4x+ 3=
A
x+ 1+
B
x+ 3yields A = −1 and B = 3 so
∫2x3 + 9x2 + 12x+ 3
x2 + 4x+ 3dx =
∫ (2x+ 1− 1
x+ 1+
3
x+ 3
)dx
= x2 + x− log |x+ 1|+ 3 log |x+ 3|+ C
2.10. INTEGRATION BY PARTS 27
Example 2.16 Write down the partial fraction expansion for5x
(x2 + 2x+ 4)2(x− 1)3. Do not solve for the
constants.
Solution:
5x
(x2 + 2x+ 4)2(x− 1)3=
Ax+B
x2 + 2x+ 4+
Cx+D
(x2 + 2x+ 4)2+
E
x− 1+
F
(x− 1)2+
G
(x− 1)3
Notice that the degree of the denominator on the left is 7 which equals the number of constants in the partialfraction 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
dx(uv) =
du
dxv + u
dv
dxor
udv
dx=
d
dx(uv)− du
dxv.
Integrating both sides of the secont form gives the formula for integration by parts:∫udv
dxdx = uv −
∫du
dxv dx.
Strategy There are no set rules for doing integration by parts. Success is mainly a matter of experience thatcomes from doing lots of problems. However there are some useful ways of starting.
• choose u so thatdu
dxis simpler than u and • choose
dv
dxso that it can be integrated.
Some good choices for u and v:
f(x)g(x) udv
dx
du
dxv
x cosx x cosx 1 sinx
xn loge x loge x xn1
x
xn+1
(n+ 1)
arcsinx arcsinx 11√
1− x2x
arctanx arctanx 11
(x2 + 1)x
Example 2.17 Find
∫x3 log x dx.
Solution: ∫x3 log x dx =
x4
4log x−
∫x4
4
1
xdx
=x4
4log x− x4
16+ c
28 CHAPTER 2. TECHNIQUES OF INTEGRATION
Example 2.18 Find
∫xe2x dx.
Solution: ∫xe2x =
xe2x
2−∫e2x
2dx
=xe2x
2− e2x
4+ C
Example 2.19 Find
∫ π
0
x cosnx dx where n ∈ Z.
Solution: Assume n 6= 0. Then∫ π
0
x cosnx dx
=
[1
nx sinnx
]π0
−∫ π
0
1
nsinnx dx
=1
nπ sinnπ −
[− 1
n2cosnx
]π0
= 0 +1
n2(cosnπ − 1)
=
0, n even
− 2n2 , n odd
If n = 0 then the answer will be 0.
Example 2.20 Find
∫x2 sinhx dx.
Solution: ∫x2 sinhx dx = x2 coshx−
∫2x coshx dx
= x2 coshx− 2x sinhx+
∫2 sinhx dx
= x2 coshx− 2x sinhx+ 2 coshx+ C
Example 2.21 Find
∫log x dx.
Solution: ∫log x dx = x log x−
∫x
xdx
= x log x− x+ C
Example 2.22 What is
∫e2x sinx dx?
Solution:
I =
∫e2x sinx dx
= −e2x cosx−∫−1
2e2x cosx dx
= −e2x cosx+ 2e2x sinx− 4
∫e2x sinx dx
= −e2x cosx+ 2e2x sinx− 4I
Solving for I gives
I =
∫e2x sinx dx =
1
5
(−e2x cosx+ 2e2x sinx
)+ C.
2.11. COMPLEX EXPONENTIAL (REVISION FROM UMEP MATHEMATICS) 29
2.11 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 cosx = Re(eix) we obtain
∫e2x sinx dx = Im
∫e2xeix dx
= Im
∫e(2+i)x dx
= Im
(1
2 + ie(2+i)x
)+ C
= Im
(2− i
5e2x(cosx+ i sinx)
)+ C
= Im1
5e2x (2 cosx+ sinx− i cosx+ 2i sinx) + C
=1
5e2x(− cosx+ 2 sinx) + C
2.12 Integration Exercises
1. Basic Integration Evaluate the following integrals:
(a)
∫(2x− 5)3 + (2x+ 5)3 dx (b)
∫sin kx√
2 + cos kxdx (k 6= 0)
(c)
∫1
5x−√xdx (d)
∫1
x2 + 4x+ 13dx
(e)
∫cos2 7x dx (f)
∫3x
(x− 2)(x+ 4)dx
(g)
∫1
x log xdx (h)
∫ex
ex + 1dx
(i)
∫4x+ 17
x2 + 10x+ 25dx (j)
∫ 3
−1
1
2x+ 3dx
(k)
∫sin6 x cos3 x dx (l)
∫1√
48− 8x− x2dx
2. Trigonometric and Hyperbolic Substitutions Using an appropriate trigonometric or hyperbolic substitution,find the indefinite integrals of the following functions:
(a)√
1 + 4x2 (b)√
4− x2 (c)1
(x2 − 1)32
3. Hyperbolic Powers Find the indefinite integrals of the following powers of hyperbolic functions:
(a) sinh6 x coshx (b) cosh2 3x (c) sinh2 x cosh3 x
(d) sinh3 4x (e) cosh4 x
4. Hyperbolic Tangent Integrals Evaluate the following integrals, where k > 0 is a constant:
(a)
∫tanhx sech2 x dx (b)
∫sech2 kx
2 + tanh kxdx (c)
∫tanh2 3x dx
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
(e)x3 + 3x− 2
x2 − x6. By Parts Evaluate the following integrals:
(a)
∫x cos 3x dx (b)
∫arcsinx dx (c)
∫ 1
0
x arctanx dx
(d)
∫x2 coshx dx (e)
∫arcsinhx dx
7. By Parts Again Using integration by parts, evaluate the following integrals:
(a)
∫x7 log x dx, x > 0 (b)
∫ e
1
log(x4) dx
(c)
∫ex cos 3x dx (d)
∫e−2x sin 11x dx
8. Complex Exponential (Revision) Find the indefinite integrals of the following functions using the complexexponential:
(a) ex cos 3x (b) e−2x sin 11x (c) e5t cos 7t
9. Mixed Integrals Find the indefinite integrals of the following functions:
(a) cosh3 x sinh3 x (b)√x log x, x > 0 (c)
x2 + 3x+ 4
x2 + x
(d)x2 − 2
x3 − 6x− 2(e)
1√9− 4x2
(f) cosh5 x sinh4 x
(g) x√x+ 3 (h) coshx sinhx
5√
cosh 2x+ 55 (i) e2x sin 3x
2.13 Answers to Exercises
1. (a) 18 (2x− 5)4 + 1
8 (2x+ 5)4 + C (b) − 2k
√2 + cos kx+ C
(c) 25 log |5
√x− 1|+ C (d) 1
3 arctan(x+23
)+ C
(e) 12x+ 1
28 sin 14x+ C (f) log |x− 2|+ 2 log |x+ 4|+ C
(g) log | log x|+ C (h) log(ex + 1) + C
(i) 4 log |x+ 5|+ 3x+5 + C (j) 1
2 log 9
(k) 17 sin7 x− 1
9 sin9 x+ C (l) arcsin(x+48
)+ C
2. (a) 14
(arcsinh 2x+ 2x
√1 + 4x2
)+ C (b) x
2
√4− x2 + 2 arcsin
(x2
)+ C
(c)−x√x2 − 1
+ C
3. (a) 17 sinh7 x+ C (b) 1
2x+ 112 sinh 6x+ C
(c) 13 sinh3 x+ 1
5 sinh5 x+ C (d) 112 cosh3 4x− 1
4 cosh 4x+ C
(e) 38x+ 1
4 sinh 2x+ 132 sinh 4x+ C
4. (a) 12 tanh2 x+ C (b) 1
k log |2 + tanh kx|+ C
(c) x− 13 tanh 3x+ C
2.13. ANSWERS TO EXERCISES 31
5. (a) 15 log |x+ 2| − 1
10 log(x2 + 1) + 25 arctanx+ C
(b) 4 log |x+ 1| − 5 log |x− 2|+ log |x− 3|+ C
(c) 2 log(x2 + 1) + 3 arctanx− 2 log(x2 + 2)− 3√2
arctan(x√2
)+ C
(d) −2 log |x+ 1|+ log |x+ 3|+ log |x− 2|+ C(e) 1
2x2 + x+ 2 log |x|+ 2 log |x− 1|+ C
6. (a) 13x sin 3x+ 1
9 cos 3x+ C (b) x arcsinx+√
1− x2 + C
(c) arctan 1− 12 (d) x2 sinhx− 2x coshx+ 2 sinhx+ C
(e) x arcsinhx−√x2 + 1 + C
7. (a) 18x
8 log x− 164x
8 + C (b) 4
(c) ex
10 (cos 3x+ 3 sin 3x) + C (d) e−2x
125 (−11 cos 11x− 2 sin 11x) + C
8. (a) ex
10 (cos 3x+ 3 sin 3x) + C (b) e−2x
125 (−11 cos 11x− 2 sin 11x) + C
(c) e5t
74 (5 cos 7t+ 7 sin 7t) + C
9. (a)1
4sinh4 x+
1
6sinh6 x+ C =
1
6cosh6 x− 1
4cosh4 x+D =
1
48cosh3 2x− 1
16cosh 2x+ E
(b) x3/2(
2
3log x− 4
9
)+ C (c) x+ 4 log |x| − 2 log |x+ 1|+ C
(d)1
3log |x3 − 6x− 2|+ C (e)
1
2arcsin
2x
3+ C
(f)1
9sinh9 x+
2
7sinh7 x+
1
5sinh5 x+ C (g)
2
5(x+ 3)
52 − 2(x+ 3)
32 + C
(h) 524 (cosh 2x+ 55)6/5 + C (i)
e2x
13(−3 cos 3x+ 2 sin 3x) + C