MATH1231 II Integration Slides Summer 2014 (1)

8/10/2019 MATH1231 II Integration Slides Summer 2014 (1)

1/62

MATH 1241 S2 2011: Calculus

For use in Dr Chris Tisdells lectures

Section 2: Techniques of integration.

Created and compiled by Chris Tisdell

S1: Motivation

S2: What you should already knowS3: Integrals of trig functions

S4: Reduction formulae

S5: Trig & hyperbolic substitutions

S6: Partial fractions

S7: Rationalising substitutionsS8: Weierstrass substitutions

S9: Appendix (more on what you should already know)

Images fromThomas calculusby Thomas, Wier, Hass & Giordano, 2008, Pearson Education, Inc.

1


2/62

1. Motivation.

Why study integration?

The theory of integration is a cornerstone of calculus. Integration finds

a useful place in many disciplines, such as: engineering; physics; biology;

economics and beyond.

Where are we going?

We will develop a number of important integration techniques that will

be useful in the study of applied problems.

Throughout our discussions we will see HOW integration naturally arises

in the analysis of applied problems and in modelling. This is critical to

motivate the ideas and also to build the intuition.

2


3/62

2. What you should already know.

You studied integration at school and in MATH 1131. You should becomfortable with:

using a table of integrals

integrating by inspection

integration by parts

integration by simple substitution.

Just in case youve forgotten these techniques, I have included someexamples to refresh your memory and to test your skills in the Appendix.

You will find this coursemuch easierif you can integrate with ease. Yourhard work will payoff later in the session!!

3


4/62

4


5/62

5


6/62

3. Integrals of trig functions.

You should be familiar with:

The above integrals are derived by using the double angle formulas:

cos2x = cos2 x

sin2 x

= 2 cos2 x 1= 1 2sin2 x

and for future reference

sin2x= 2 sin x cos x.

6


7/62

Integrals of products ofsin and/or cos.

We first discuss how such integrals naturally arise.

Applications matter! SocalledFourier seriesplay an important role

in the study ofheat flow.

A finite Fourier series of a functionf is given by the sum

f(x) := N

n=1

an sin nx (1)

= a1 sin x + a2 sin2x + aNsin N xwhere thean are some numbers that are to be determined. That is, we

aim to write f(x) as a sum ofsinfunctions.

7


8/62

8


9/62

We now see how integration of trig functions is central to the method.

We claim that themth coefficientamis given by

am =1

f(x)sin mx dx. (2)

If we multiply both sides of (1) by sin mx and then integrate over

[, ]we obtain

f(x)sin mx dx=

Nn=1

an sin nx sin mx dx

=N

n=1

an sin nx sin mx dx

.

Thus we have obtained an integral that involves a product ofsinfunc-

tions.

How do we evaulate such an integral?

9


10/62

Integrals of products ofsin and/or cos

These are done using the socalled product to sum formulae:

sin A cos B = [sin(AB) + sin(A + B)]/2sin A sin B = [cos(AB) cos(A + B)]/2

cos A cos B = [cos(AB) + cos(A + B)]/2Ex:sin4x cos2x dx=

Independent learning ex: Now return to, and prove, (2).

10


11/62

Integrals of the form

cosm x sinn x dx

Ifm orn are oddthen our aim is to transform the integral into one of

the types: sink x cos x dx =

1

k+ 1sink+1 x + C

cosk x sin x dx = 1k+ 1

cosk+1 x + C.

At the heart of the approach is to factor out a sin xor cos xfrom theodd power term and then use the identity

cos2 x + sin2 x= 1

to transform the integral into one which can be directly evalutated.

Q: How can you verify the above integrals?

11


12/62

Ex. Evaluate

I :=

cos4 x sin3 x dx.

12


13/62

Ex: Evaluate

I :=

cos5 x sin5 x dx

13


14/62

Ifbothm and n are even, then the integral is slightly harder. We willsee shortly a recursive method for dealing with such an integral, but forsmaller values ofmandnone can use the identities

cos2

x=

1 + cos 2x

2 sin2

x=

1

cos2x

2 .

Ex. EvaluateI :=

cos4 x sin2 x dx. We have

I =

(cos2 x)2 sin2 x dx

=

1 + c o s 2x2

2

1 cos2x2

dx

= 1

8

1 + c o s 2x cos2 2x cos3 2x dx

= 1

8 1 + c o s 2x cos

2 2x

(1

sin2 2x)cos2x dx

= 1

8

1 cos2 2x + sin2 2x cos2x dx

= 1

8

x

2 1

8sin4x +

1

6sin3 2x

+ C.

14


15/62

Applications matter! An important problem in applied math, engi-neering and physics is to calculate the total mass of an given object.Consider a thin plate occupying the unit quarterdisk in the first quad-rant.

If the density (x, y) of the plate at any point (x, y) is given by(x, y) =x2y4 then the total mass of the plate is given by

M= 1

0

1x2

0x2y4 dy

dx.

15


16/62

To make this integral easier, we switch to polar coordinates: x =

r cos ;y =r sin ;dy dx=r dr d with a substitution giving

M = /2

0 1

0 (r

2

cos

2

)(r

4

sin

4

) rdrd

= 1

0r7 dr

/20

(cos2 )(sin4 )d.

16


17/62

4. Reduction formulae

Repeated integration by parts can be a long process! We can shortenthe process by applying socalledreduction formulae.

Ex: The reduction formula tann x dx=

1

n 1 tann1 x

tann2 x dx ()

can be used in an iterative fashion to calculate

tan5 x dx.

IfIn:=

tann x dxthen (*) may be compactly written as

In:= 1

n 1 tann1 xIn2, n2.

17


18/62

Ex. LetIn :=/2

0 sinn x dx. Use the reduction formula

In :=n 1

n In2, n2

to calculate I7.

18


19/62

Ex: Construct a reduction formula for

In :=

xnex dx.

19


20/62

Ex: Find the reduction formula for

In:= /2

0sinn x dx.

The reduction formula for the integral ofcosn x is similar.

20


21/62

Ex: Construct a reduction formulae from

In:= /4

0secn x dx.

A similar method is used to obtain the reduction formula for the integral

In:=

/4

0 tann x dx.

21


22/62

Ex. [Q1, Class Test 1, 2002] Let

In := /4

0tann sec d.

Show that

In :=1

n

2 (n 1)In2

, forn2.

Note thatd

dsec = sec tan .

22


23/62

A Two parameter recurrence:

Consider

Im,n =

/2

0cosm x sinn x dx

= /2

0[cosm1 x][sinn x cos x]dx

and use integration by parts. Choose u = cosm1 x and v =

sinn x cos x. Thus,

u =(m 1) sin x cosm2 x, v = 1n + 1

sinn+1 x.

Integration by parts then leads to

Im,n=

m

1

m + nIm2,n, m2.

23


24/62

In a similar way, one could also obtain the recurrence

Im,n= n 1m + n

Im,n2. n2.

In applying the above formulae we must reach one ofI1,1 = 12, I1,0=I0,1 = 1orI0,0 =

2.

You are not expected to memorise this formula.

24


25/62

5. Trig & Hyperbolic Substitutions.

You will have already seen integration by simple substitution.

With integrals involving square roots of quadratics, the idea is to make

a suitable trigonometric or hyperbolic substitution that greatly simplifies

the integral.

25


26/62

Integrals involving square roots of quadratics can be worked out using

the following trigonometric or hyperbolic substitutions.

a2 x2 tryx=a sin

a2 + x2 try x=a tan orx=a sinh

x2 a2 try x = a sec orx=a cosh

26


27/62

27


28/62

Ex: Evaluate

I= 4

0x2

16 x2 dx.

28


29/62

Ex: Evaluate

I :=

dx

(a2 + x2)3/2.

29


30/62

Ex: Evaluate

I :=

dx

x2x2 1

via the substitutionx= sec

(This last integral can also be done using a cosh substitution but it

is not easy.)

30


31/62

Applications matter! An important problem in applied math, engi-

neering and physics is to calculate the total mass of an given object.

Consider a thin wire ofconstantdensity (mass per unit length) thatlies in the XYplane along the curve

y =f(x) :=x2/2, axb.It can be shown that the total mass Mof the wire is

M =

ba

1 + [f(x)]2 dx

= b

a

1 + x2 dx.

See how an integral of1 + x2 naturally arises?

31


32/62

6. Partial Fractions.

A rational function is a function of the form f(x)

g(x) where f and g are

polynomials.

Ex: Which of the following are rational functions?

(a) 2x + 1

3x2 4 (b)

cos x

x2 + 1

(c) x2

x 1 (d) ex

x + 1.

For the purpose of integration, we ensure that the degree of the numer-ator is less than the degree of the denominator. If not then we may need

to divide.

32


33/62

Building the intuition.

An integral like

dx

x(1 x)cannot be easily computed directly, however, we know that the integrals

dx

x,

dx

1

x

can both be directly evaluated.

Thus, can we write 1x(1x) as some linear combination of

1x and

11x?

33


34/62

34


35/62

35


36/62


37/62

Factors are Linear:

Ex. Evaluate

I :=

2x 1x2 + 5x + 6

dx=

2x 1(x + 3)(x + 2)

dx.

37


38/62

Ex. Evaluate 2x2 + 2x + 6

(x 1)(x + 1)(x 2)dx.

38


39/62

Repeated Linear Factors:

Ex: Evaluate

I := x + 3

(x + 2)3dx.

39


40/62

In general, each factor in the denominator of the form (xa)k givesrise to an expression of the form

A1

xa+

A2

(xa)2+ ..... +

Ak

(xa)k.

Ex: Evaluate

I :=

dx

x(x + 1)3.

40


41/62

Ex: [Q2, Class Test 1, 2003] Calculate 8x + 9

(x 2)(x + 3)2dx.

41


42/62

Irreducible Quadratic Factors:

Suppose the denominator of our rational function is a quadratic which

does not factor (over the real numbers). For example,

f(x) = 2x + 1

x2 + 4x + 5.

To find the integral of such a function, we manipulate the integrand so

that

the numerator equals the derivative of the denominator.

This leads to a logarithm and complete the square on the denominator

of the remaining term leading to an inverse tangent function.

42


43/62

Ex: Evaluate

I :=

2x + 1

x2 + 4x + 5dx.

43


44/62

If the quadratic does have real but irrational roots, one can use this same

method to avoid nasty partial fractions. The second factor will then lead

to an inverse tanh.

e.g. 2x + 1

x2 + 4x 6dx

= 2x + 4

x2

+ 4x 6+

3

10 (x + 2)2

dx

= ln |x2 + 4x 6| + 310

tanh1

x + 210

+ C.

44


45/62

Ex: [Q1, Class Test 1, 2002] Calculate x

x2 + 2x + 10dx.

You are given that du

u2 + a2=

1

atan1u

a+ C.

45


46/62

Combinations of the Rules:

These rules can be combined to cover a variety of situations.

Ex: Calculate

I :=

x + 6

x(x2 + 2x + 3)dx.

46


47/62

In general, an irreducible quadratic factor

x2 + ax + b

in the denominator will give rise to a term of the form

Ax + B

x2 + ax + b

and more generally again, a factor

(x2 + ax + b)k

in the denominator will give rise to terms of the form

A1x + B1x2 + ax + b

+ A2x + B2

(x2 + ax + b)2+ ... +

Akx + Bk(x2 + ax + b)k

.

47


48/62

Ex. Evaluate

I :=

x

(x + 1)2(x2 + 1)dx.

48


49/62

Applications matter! Where does an integral involving partial frac-

tions arise? Let P = P(t) denote the population size of a species

at time t. It can be shown thatP and dP/dt satisfy the differential

equation

dP

dt =P

1 P

M

(3)

whereM >0is a constant.

The challenge is to determine the unknown function P from (3) and

hencepredictwhat will happen to the population over time.

If we rearrange (3) and integrate both sides w.r.t t then we obtain

1

P

1 PMdP

dt

dt = 1dtwhich becomes

M

dP

P(MP) =t + K.

49


50/62


51/62

51


52/62

7. Rationalising substitutions

The above methods using partial fractions allow us to integrate (in prin-

ciple) any rational function, so we often make a change of variable in anintegral that will lead us to some rational function.

Ex: Evaluate

I := dx

1 +x

1/4.

52


53/62


54/62

8. Weierstrass substitution.

Weierstrasssubstitution is very useful for integrals involving a rational

expression insin xand/orcos x.

Weierstrasssubstitution is t= tan x2.

Recall, that sin x= 2t

1 + t2, cos x=

1 t21 + t2

, and that dx= 2 dt

1 + t2.

You should know how to derive these formulae. For example, if t =

tan(x/2)then

2t

1 + t2 =

2 tan(x/2)

1 + t a n2(x/2)

= 2 [sin(x/2)]/[cos(x/2)]

1 + [sin2(x/2)]/[cos2(x/2)]

= 2 sin(x/2) cos(x/2) = sin x.

Who was Weierstrass??

54


55/62

Ex: Calculate

I :=

dx

1 + c o s x + sin x.

55

/2 d


56/62

Ex: Show thatI=/2

0dx

2+cos x = 3

3.

We lett= tan(x/2), so that

cos x= 1 t21 + t2

dx= 2dt

1 + t2

in a similar way to the previous example. Substitution then leads to

I = 1

0

2

3 + t2dt

=

2

3tan1 t

3

10

= 3

3.

56


57/62

Applications matter! Where does an integral like d

1 + c o s + sin

arise? Consider a thin plate in the XYplane that occupies the region(in polars)

1r2, 0/2.Suppose at each point (x, y)= (0, 0)the plate has density function

(x, y) =

1x2 + y2 + x + y.

To evaluate the total mass Mof the plate we consider

M = 1

0

4y2

1y2 (x, y)dx

dy.

Now, making substitutions: x = r cos ; y = r sin ; dx dy =r dr d we obtain

M = 2

1dr

/20

d

1 + c o s + sin .

57


58/62

9. APPENDIX

MAPLE

The following examples show how to use MAPLE to perform partial

fraction decompositions.

> convert(x^2/(x+2), parfrac, x);

x 2 + 4x + 2

> convert(x/(x-b)^2, parfrac, x);

b

(xb)2 + 1

xb

58


59/62

Integration by Parts.

The analysis requires you to:

choose functions f andg;

then calculate f andg from your above choices;

and then apply the above formula.

Ex: Evaluate I=

xex dx.

We choose f = x and g = ex

. Thus, f = 1 and g = ex

. Usingour IBP formula

I=

xex dx = xex

1ex dx

= xex ex + C.59


60/62

Integration by inspection.

When confronted by an integral, you should always see if the integralcan be guessed and then finetuned by multiplying or dividing by a

constant.

Alternatively, multiply and divide the integral by a particular constant toput it into a useful form.

Ex: Evaluatex(x2 + 1)10 dx.

See thatxis almostthe derivative of(x2 + 1).

We make the guess(x2 +1)11 and then check our guess by evaluating

d

dx[(x2 + 1)11] = 11 2x(x2 + 1)10 = 22x(x2 + 1)10.

See that we have an unwanted 22 in the above. Thus, we divide ourinitial guess by 22to obtain our answer

x(x2 + 1)10 dx= 1

22(x2 + 1)11 + C.

60

O O f f f


61/62

Observe that this method ONLY works for integrals of the form f(g(x))g(x) dx

when the derivativeg(x)is present in the integrand up to a constant. sin t cos5 t dt =

d

dt

cos6 t6

dt

= cos6 t

6 .

The above methods should remind you of the chain rule!

x4(x3 + 1)10 dx

cannot be guessed and fine tuned since the derivative ofx3 + 1is notin the integrand.

61

O h f l i l


62/62

Other useful integrals.

1

a

2x

2dx = sin1x

a

1a2 + x2

dx = sinh1xa

1

a2 + x2dx =

1

atan1x

a

1

a2x

2dx =

1

a

tanh1x

a

These four should be learnt carefully.

62

Documents

MATH1231 II Integration Slides Summer 2014 (1)