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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 questionsCont
ents
© Boardworks Ltd 20061 of 66
Using trigonometric identities in integration
Using trigonometric identities in integration
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
Integrating odd powers of cos x and sin x
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).
Integrating odd powers of cos x and sin 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
© Boardworks Ltd 20069 of 66
Separable variables
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
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
© Boardworks Ltd 200614 of 66
Modelling real-life situations
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
Modelling real-life situations
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
Modelling real-life situations
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).
Modelling real-life situations
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