620-190 bridging notes for vector calculus - university of melbourne

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

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

3

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


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


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.


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)


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. (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


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


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.


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.


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


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


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

620-190 bridging notes for vector calculus - university of melbourne

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