using trigonometric identities in integration using partial fractions in integration first-order...

$: Using trigonometric identities in integration Using partial fractions in integration First-order differential equations Differential equations with separable$
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 questionsCont

ents

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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

Cont

ents


Separable variables






The trapezium rule

Examination-style questions

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

dxFor 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

The particular solution is therefore:

Given that y = ln when x = 0:73

7 13 3ln = ln( + )c

= 2c

313= ln( + 2)xy e

Cont

ents


Modelling real-life situations






The trapezium rule

Examination-style questions


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

Modelling real-life situationsAnd 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

Modelling real-life situationsFind 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 growthThe particular solution is therefore:

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 decayRemember, 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 decaySeparating 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 decayThere 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 decayWhen 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

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