© Boardworks Ltd 2006 1 of 66 © Boardworks Ltd 2006 1 of 66 A2-Level Maths: Core 4 for Edexcel C4.6 Integration 2 This icon indicates the slide contains.

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A2-Level Maths: Core 4for Edexcel

C4.6 Integration 2

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Using trigonometric identities in integration

Using partial fractions in integration

First-order differential equations

Differential equations with separable variables

Using differential equations to model real-life situations

The trapezium rule

Examination-style questions

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In these cases, it may be possible to rewrite the expression using an appropriate trigonometric identity.For example:

Many expressions involving trigonometric functions cannot be integrated directly using standard integrals.

sin2 2sin cosx x x

Find . sin cosx x dx

So, we can write:

Using the double angle formula for sin 2x:

12sin cos = sin2x x dx x dx

14= cos2 +x c


Integrating cos2 x and sin2

x

There are two ways of writing this involving sin2 x and cos2 x:

We can rewrite these with sin2 x and cos2 x as the subject:

To integrate functions involving even powers of cos x and sin x we can use the double angle formulae for cos 2x.

2cos2 2cos 1x x

2cos2 1 2sinx x

2 12cos (1+ cos2 )x x 1

2 12sin (1 cos2 )x x 2


Integrating cos2 x and sin2

x

Find . 2cos x dx2 1

2cos = (1+ cos2 )x dx x dx Using 1

1 12 2= ( + sin2 )+x x c

Find . 2sin 2x dxUsing and replacing x with 2x gives: 2

2 12sin 2 = (1 cos4 )x dx x dx 1 1

42= ( sin4 )+x x c


Integrating even powers of cos x and sin x

We can extend the use of these identities to integrate any even power of cos x or sin x. For example:

Find . 4 12cos x dx

This can be written in terms of cos2 x as: 12

4 2 21 12 2cos = (cos )x dx x dx

212= ( (1+ cos ))x dx

214= (1+ 2cos + cos )x x dx1 14 2= (1+ 2cos + (1+ cos2 ))x x dx

31 14 2 2= ( + 2cos + cos2 )x x dx

31 14 42= ( + 2sin + sin2 )+x x x c


Integrating odd powers of cos x and sin x

Odd powers of cos x and sin x can be integrated using the identity cos2 x + sin2 x = 1.

Find . 3sin x dx

2 12cos (1+ cos2 )x x 1

2 12sin (1 cos2 )x x 2

3 2sin = sin sinx dx x x dx Using 2 2= (1 cos )sinx x dx

2= (sin cos sin )x x x dx



The first part, sin x, integrates to give –cos x.

3 2(cos ) = 3cos sind

x x xdx

This is now in a form that we can integrate.

The second part, cos2 x sin x, can be recognized as the product of two functions.

Remember the chain rule for differentiation:

1= = n ndyy n

dy

The derivative of cos x is –sin x and so:

where is f (x) and is f ’(x).



2 313cos sin = cos +x x dx x c

So, returning to the original problem:

Therefore,

3 2sin = (sin cos sin )x dx x x x dx 31

3= cos + cos +x x c

213= cos (cos 3)+x x c







The trapezium rule


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Integrating rational functions

We have seen that rational functions of the form can be integrated using:

In particular, if f(x) is a linear function then:

Suppose we want to integrate a rational function with more than one linear factor in the denominator.

For example:

ff'( )( )

xx

ff

f

'( )= ln ( )

( )

xdx x

x

1 1= ln + +

+dx ax b c

ax b a

2 1

( 2)( 1)

xdx

x x



We can integrate this by first splitting into partial fractions.

2 1

( 2)( 1)

x

x x

Let2 1

+( 2)( 1) 2 1

x A B

x x x x

Multiplying through by (x – 2)(x –1):

4 1= A

= 3A

2 1= B

= 1B

Substituting x = 2 into : 1

Substituting x = 1 into :1

2 1 ( 1)+ ( 2)x A x B x 1



So, 2 1 3 1

( 2)( 1) 2 1

x

x x x x

We can now integrate:

2 1 3 1=

( 2)( 1) 2 1

xdx dx dx

x x x x

= 3ln 2 ln 1 +x x c

3( 2)= ln +

1

xc

x

3= ln ( 2) ln 1 +x x c



Find .

The denominator involves the difference between two squares and so we can write:

2

6

4 9dx

x

2

6 6

4 9 (2 + 3)(2 3)dx dx

x x x

Let6

+(2 + 3)(2 3) 2 + 3 2 3

A B

x x x x

Multiplying through by (2x + 3)(2x – 3):

6 (2 3)+ (2 + 3)A x B x 1



Substituting x = into :132

326 = (2( )+ 3)B

6 = 6B

=1B


326 = (2( ) 3)A

6 = 6A

= 1A

So, 2

6 1 1

4 9 2 3 2 + 3x x x




2

6 1 1=

4 9 2 3 2 + 3dx dx dx

x x x

1 12 2= ln 2 3 ln 2 + 3 +x x c

12

2 3= ln +

2 + 3

xc

x



Find . 2

2

8 + 21 +13

(2 +1)( + 2)

x xdx

x x

Let2

2 2

8 + 21 +13+ +

(2 +1)( + 2) 2 +1 + 2 ( + 2)

x x A B C

x x x x x

Multiplying through by (2x + 1)(x + 2)2:

2 28 + 21 +13 ( + 2) + (2 +1)( + 2)+ (2 +1)x x A x B x x C x 1


921422 +13 = A

9 942 = A

= 2A



Comparing the coefficients of x2 :

8 = + 2A B

8 = 2 + 2B

= 3B

Substituting x = –2 into :1

32 42 +13 = 3C

3 = 3C

= 1C

So,2

2 2

8 + 21 +13 2 3 1+

(2 +1)( + 2) 2 +1 + 2 ( + 2)

x x

x x x x x



We can now integrate:2

2 2

8 + 21 +13 2 3 1= +

(2 +1)( + 2) 2 +1 + 2 ( + 2)

x xdx dx dx dx

x x x x x

1= ln 2 +1 + 3ln + 2 + +

+ 2x x c

x

3 1= ln 2 +1 + ln + 2 + +

+ 2x x c

x

The integral of (x + 2)–2

is –(x +2)–1.

3 1= ln (2 +1)( + 2) + +

+ 2x x c

x



To integrate an improper fraction we need to rewrite it in proper form before integrating. For example:

Find . 4

2 + 3

xdx

x

4 2(2 + 3) 6=

2 + 3 2 + 3

x x

x x

2(2 + 3) 6=

2 + 3 2 + 3

x

x x

The improper fraction in this example can be written in proper form by rewriting the numerator.

6= 2

2 + 3x



More difficult examples may require us to set up an inequality or use polynomial long division. For example:

4 6= 2

2 + 3 2 + 3

xdx dx dx

x x

62= 2 ln 2 + 3 +x x c


= 2 3ln 2 + 3 +x x c

Find . 3

2

4 +10 + 4

2 +

x xdx

x x

Let34 +10 + 4

+ + +(2 +1) 2 +1

x x C DAx B

x x x x



Multiplying through by x(2x + 1):34 +10 + 4 ( + ) (2 1)+ (2 +1)+x x Ax B x x C x Dx 1

Substituting x = into :11248 25 + 4 = D

32 2= D

= 3D

Substituting x = 0 into :1

4 = C

= 4C

A and B can now be found by comparing coefficients.



Comparing coefficients of x3:

4 = 2A

Comparing coefficients of x2:

0 = + 2A B

2 = 2B

= 2A

= 1B

34 +10 + 4 4 32 1+ +

(2 +1) 2 +1

x xx

x x x x So,



We can now integrate:34 +10 + 4 4 3

= (2 1) + +(2 +1) 2 +1

x xdx x dx dx dx

x x x x

2 32= + 4ln + ln 2 +1 +x x x x c

322 4= + ln + ln (2 +1) +x x x x c

322 4= + ln (2 +1) +x x x x c







The trapezium rule


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Differential equations

In this course we will only be looking at first-order differential equations.

A differential equation in two variables x and y is an equation that contains derivatives of y with respect to x. For example:

2= 4 +1,dy

xdx

3= ,dy

xydx

2 + = 5 .dy

x y xydx

The order of a differential equation is given by the highest order of derivative that occurs in it.

First-order differential equations contain terms in ,dydx

second-order differential equations contain terms in ,2

2d ydx

third-order differential equations contain terms in , etc.3

3d ydx


Differential equations

For example, suppose we have the differential equation:

The solution to a differential equation in x and y will take the form y = f(x).

The simplest differential equations are those of the form:

f= ( )dy

xdx

Differential equations of this form can be solved by integrating both sides with respect to x to give:

f= ( )y x dx

= 4 +1dy

xdx


Differential equations of the form = f(x)

Integrating both sides with respect to x gives:

dydx

= (4 +1)y x dx2= 2 + +y x x c

Since the constant c can take any value, this represents a whole family of solutions as shown here:

This is called the general solution

to the equation .dy

xdx

= 4 +1


Finding a particular solution

Suppose that as well as being given the differential equation:

Substituting x = 1 and y = 4:

we are also told that when x = 1, y = 4.

We can use this additional information to find the value of the arbitrary constant c in the general solution:

y x x c2= 2 + +

4 = 2 + 1 + c

c = 1

This gives us the particular solution:

y x x2= 2 + +1

= 4 +1dy

xdx


Solving first-order differential equations

Find the particular solution to the differential equation

given that y = 6 when x = 0.

Divide both sides by (x2 + 1):

Integrate both sides with respect to x:

dyx x

dx2( +1) = 4

2

4=

+1

dy x

dx x

xy

x 2

2= 2

+1

y x c2= 2ln( +1)+

We can use brackets because x2 + 1 > 0.

Writing the quotientin the form .f

f'( )( )

xx


Solving first-order differential equations

Substitute x = 0 and y = 6:

The particular solution is therefore:

c6 = 2ln(1)+

c = 6

y x2= 2ln( +1)+ 6


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Separable variables






The trapezium rule



Separable variables

Differential equations that can be arranged in the form

can be solved by the method of separating the variables.

This method works by collecting all the terms in y, including the ‘dy’, on one side of the equation, and all the terms in x, including the ‘dx’, on the other side, and then integrating.

f g( ) = ( )dy

y xdx

f g( ) = ( )y dy x dx Although the dy and the dx have been separated it is important to remember that is not a fraction.dy

dx

For example, avoid writing:

f g( ) = ( )y dy x dx


Separable variables

Here is an example:

Find the general solution to .+ 2

=dy x

dx y

= ( + 2)y dy x dx 2 2

= + 2 +2 2

y xx c

2 2= + 4 +y x x A

We only need a ‘c’ on one side of the equation.

You can miss out the step

and use the fact that

to separate the dy from the dx directly.

... = ... dy

dx dydx

= ( + 2)dy

y dx x dxdx

2= + 4 +y x x A

Separate the variables and integrate:

Rearrange to give: = + 2dy

y xdx


Separable variables

Separating the variables and integrating with respect to x gives:3=y xe dy e dx 31

3= +y xe e c

Using the laws of indices this can be written as:

3= x ydye e

dx

313= ln( + )xy e c

Take the natural logarithms of both sides:

Find the particular solution to the differential equation

given that y = ln when x = 0.

3= x ydye

dx

73


Separable variables


Given that y = ln when x = 0:73

7 13 3ln = ln( + )c

= 2c

313= ln( + 2)xy e


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Modelling real-life situations






The trapezium rule




For example, suppose we hypothesize that the rate at which a particular type of plant grows is proportional to the difference between its current height, h, and its final height, H.

The word “rate” in this context refers to the change in height with respect to time. We can therefore write:

Since these situations involve derivatives they are modelled using differential equations.

Many real-life situations involve the rate of change of one variable with respect to another.

Remember, the rate of change of one variable, say s, with respect to another variable, t, is .ds

dt

( )dh

H hdt



The general solution to this differential equation can be found by separating the variables and integrating.

We can write this relationship as an equation by introducing a positive constant k :

= ( )dh

k H hdt

1=dh k dt

H h ln( ) =H h kt c

ln( ) =H h kt c

= kt cH h e

= kt ch H e e

= kth H Ae where A = ec

Remember the minus sign, because we have –h. (H is a constant).



If we are given further information then we can determine the value of the constants in the general solution to give a particular solution.

This is the general solution to the differential equation:

= ( )dh

k H hdt

= 20 kth Ae

For example, suppose we are told that the height of a plant is 5 cm after 7 days and that its final height is 20 cm.

We can immediately use this value for H to write:

Also, assuming that when t = 0, h = 0:

0 = 20 A

= 20A



And finally using the fact that when t = 7, h = 5:75 = 20 20 ke

Take the natural logarithms of both sides:

This gives the particular solution:

720 =15ke

7 34=ke

347 = ln( )k

34ln( )

=7

k

34ln

7= 20 20t

h e



Find the height of the plant after 21 days.

Using t = 21 in the particular solution gives

Comment on the suitability of this model as the plant reaches its final height.

343ln

= 20 20h e33

4= 20 20( )Using the fact that

343ln 33

4= ( )e9

16=11 cm

Using this model the plant will reach its final height when:34ln

7 = 0t

eSince ex never equals 0 this model predicts that the plant will get closer and closer to its final height without ever reaching it.

This will never happen.


Exponential growth

Remember, exponential growth occurs when a quantity increases at a rate that is proportional to its size.

For example, suppose that the rate at which an investment grows is proportional to the size of the investment, P, after t years.

This gives us the differential equation:

The most common situations that are modelled by differential equations are those involving exponential growth and decay.

We can write this as:dP

Pdt

=dP

kPdt

where k is a positive constant.


Exponential growth

Integrating both sides with respect to t gives:

If the initial investment is £1000 and after 5 years the balance is £1246.18, find the particular solution to this differential equation.

=dP

kPdt

1=

dPk

P dt

1=dP k dt

P ln = +P kt c

We don’t need to write |P| because P > 0.

+= kt cP e

= kt cP e e


Exponential growth

Also when t = 5, P = 1246.18:

Now, using the fact that when t = 0, P = 1000:

This is the general solution to .=dP

kPdt

= where = kt cP Ae A e

01000 = Ae

=1000A

51246.18 =1000 ke5 =1.24618ke

5 = ln1.24618k

= 0.044 (to 3 s.f.)k


Exponential growth


Find the value of the investment after 10 years.

0.044=1000 tP e

When t = 10: 0.44=1000P e

= £1552.71P

How long will it take for the initial investment to double?

Substitute P = 2000 into the particular solution:0.0442000 =1000 te

0.044 = ln2tln2

=0.044

t

15.75 years


Exponential decay

Remember, exponential decay occurs when a quantity decreases at a rate that is proportional to its size.

For example, suppose the rate at which the concentration of a certain drug in the bloodstream decreases is proportional to the amount of the drug, m, in the bloodstream at time t.

Since the rate is decreasing we write:

This gives us the differential equation:

dmm

dt

=dm

kmdt

where k is a positive constant.


Exponential decay

Separating the variables and integrating gives:

1=dm k dt

m

Suppose a patient is injected with 5 ml of the drug.

ln = +m kt c+= kt cm e

= kt cm e e

= where = kt cm Ae A e

This is the general solution to the differential equation .

=dm

kmdt


Exponential decay

There is 4 ml of the drug remaining in the patient’s bloodstream after 1 hour. How long after the initial dose is administered will there be only 1 ml remaining?

The initial dose (when t = 0) is 5 ml and so we can write directly:

= 5 ktm e

Also, given that m = 4 when t = 1 we have:

4 = 5 ke

45=ke

45= ln( )k

This gives us the particular solution:45

ln( )= 5 tm e

We could also write this as

45= 5( )tm


Exponential decay

When m = 1 we have:

So it will be about 7 hours and 12 minutes before the amount of drug in the bloodstream reduces to 1 ml.

45

ln( )1= 5 te45

ln( ) 15=te

4 15 5ln( ) = ln( )t

15

45

ln( )=

ln( )t

7.2t







The trapezium rule


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The trapezium rule


The trapezium rule

Previously in the course we used the trapezium rule as a method for approximating the area under a curve.

x

y

a b

Suppose, for example, that we wish to find the area under a curve, y = f(x), between x = a and x = b.

y0 y1 y2 y3 y4

We can divide the area into four trapeziums of equal width, h.The parallel sides of the four trapeziums are given by the five ordinates y0, y1, y2, y3 and y4.

hhhh

In general, if there are n trapeziums there will be n + 1 ordinates.


The trapezium rule

The approximate area using the trapezium rule is:

f 1 1 1 10 1 1 2 2 3 3 42 2 2 2( ) ( + )+ ( + )+ ( + )+ ( + )

b

ax dx h y y h y y h y y h y y

10 1 1 2 2 3 3 42= ( + + + + + + + )h y y y y y y y y

10 1 2 3 42= ( + 2 + 2 + 2 + )h y y y y y

The ordinates have to be spaced out evenly so that the width of each trapezium is the same.

For n trapeziums of equal width h : =b a

hn

In general, the trapezium rule with n trapeziums is:

f 10 1 2 12( ) ( + 2 + 2 +...+ 2 + )

b

n nax dx h y y y y y


The trapezium rule

Use the trapezium rule with four trapeziums to estimate the value of

to 3 significant figures.By calculating the actual value of I, find the percentage error given using the trapezium rule with four trapeziums.

2 2

0= xI e dx

Using a table to record the value of each ordinate to 3 s.f.:

The width h of each trapezium =2 0

4

1

210x

2= xy e

y4y3y2y1y0

1e

12

32

2e 3e 4e

1=

2


The trapezium rule

We can now work out the area using

f 0 1 2 112( ) ( + 2 + 2 +...+ 2 + )

b

n nax dx h y y y y y

2 2

0

xe dx 1 2 3 411+ 2( )+ 2( )+ 2( )+

4e e e e

= 0.531 (to 3 s.f.)

We can find the actual value of I using integration.

with h = and the ordinates given by the table. 12

2 22 2

00

12=x xe dx e

4 01 12 2= +e e

= 0.491 (to 3 significant figures)


Calculating the percentage error

The percentage error is given by:

This gives us the percentage error:

estimated value actual value×100%

actual value

0.531 0.491×100%

0.491

= 8.15%

This percentage is fairly large.

Remember that a greater degree of accuracy can be achieved by using more trapeziums.







The trapezium rule


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Examination-style question 1

a) Find the value of the constants A and B such that:

b) The region R is bound by the curve y = f(x), the coordinate axes and the line x = 2.

f7 +11

( ) =(4 )(3 +1)

xx

x x

f ( ) = +4 3 +1

A Bx

x x

a) Let7 +11

+(4 )(3 +1) 4 3 +1

x A B

x x x x

Find the area of the region R, writing your solution in the form ln a, where a is given to 3 significant figures.



Multiplying through by (4 – x)(3x + 1):

39 =13A

= 3A

7 13 3+11= (4 + )B

26 133 3= B

Substituting x = 4 into : 1

7 +11 (3 +1)+ (4 )x A x B x 1

Substituting x = – into :113

= 2B



b) The area of the region R is given by:

2 2

0 0

7 +11 3 2= +

(4 )(3 +1) 4 3 +1

xdx dx

x x x x

2

023= 3ln(4 )+ ln(3 +1)x x

2 23 3= 3ln(2)+ ln(7)+ 3ln(4) ln1

233 3= ln(2 )+ ln(7 )+ ln(4 )

23 3

3

7 4= ln

2

= ln(29.3)



Water is leaking from the bottom of a tank of height 1 m. The rate at which the water leaks from the tank is given by the differential equation.

where h is the depth of the water in the tank at time t and A is a positive constant.

=dh

A hdt

a) Solve this differential equation given that the tank is initially filled to full capacity.

b) After three hours the depth of the water is 0.25 m. Find the depth of the water in the tank after four hours.

c) How long does it take for the tank to empty from full capacity?



Separating the variables and integrating gives:

=dh

A hdt

1=dh Adt

h

2 = +h At c

=2

c Ath

2

=2

c Ath

a)



Given that when t = 0, h = 1:

The height of the water in the tank is therefore given by:

2

1=2

c

=12

c

= 2c

22

=2

Ath

2

= 12

At



b) When t = 3, h = 0.25 so:2

30.25 = 1

2

A

30.5 =1

2

A

3= 0.5

2

A

13=A

The particular solution is:21

3= 12

th

2

= 16

t



After four hours, when t = 4: 2

4= 1

6h

21

=3

1=

9

The height of the water after 4 hours is 0.11 m (to 2 d.p.).



The tank will take 6 hours to empty from full capacity.

1 = 06

t

=16

t

= 6t

c) The tank will be empty when h = 0, that is when:

2

0 = 16

t

© Boardworks Ltd 2006 1 of 66 © Boardworks Ltd 2006 1 of 66 A2-Level Maths: Core 4 for Edexcel C4.6 Integration 2 This icon indicates the slide contains.

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