Using trigonometric identities in integration
Using trigonometric identities in integration
Using partial fractions in integration
Contents
First-order differential equations
Differential equations with separable variables
Using differential equations to model real-life situations
The trapezium rule
Examination-style questions
1 of 66
© Boardworks Ltd 2006
Using trigonometric identities in
integration
Many expressions involving trigonometric functions cannot be integrated directly using
standard integrals.
In these cases, it may be possible to rewrite the expression using an appropriate
trigonometric identity.
For example:
Find
. s x dx
 sin x co
Using the double angle formula for sin 2x:
s in 2 x  2 s in x c o s x
So, we can write:
 sin x co s x dx =
1
2
=
 sin 2 x dx
1
4
co s 2 x + c
Integrating cos2 x and sin2 x
To integrate functions involving even powers of cos x and sin x we can use the double angle
formulae for cos 2x.
There are two ways of writing this involving sin2 x and cos2 x:
cos 2 x  2 cos x  1
2
cos 2 x  1  2 sin x
2
We can rewrite these with sin2 x and cos2 x as the subject:
co s x 
1
2
(1 + co s 2 x )
1
sin x 
1
2
(1  co s 2 x )
2
2
2
Integrating cos2 x and sin2 x
Find
Using
Find
Using
 co s. x dx
2
1
 co s x dx =
1
2
 (1 + co s 2 x ) dx
=
1
2
(x +
 sin 2 x dx =
1
2
 (1  co s 4 x ) dx
=
1
2
(x 
2
1
2
sin 2 x ) + c
2
sin
 . 2 x dx
and
2 replacing x with 2x gives:
2
1
4
sin 4 x ) + c
Integrating even powers of cosx and
sinx
We can extend the use of these identities to integrate any even power of cos x or sin x. For
example:
Find
 co s .
4 1
2
x dx
This can be written in terms of cos2 x as:
 co s
4 1
2
x dx =
1
2
 (co s
2 1
2
2
x ) dx
=
1
(
 2 (1 + co s x )) dx
=
1
4
 (1 + 2 co s x + co s x ) dx
=
1
4
 (1 + 2 co s x +
=
1
4
3
(
 2 + 2 co s x +
=
1
4
( 32 x + 2 sin x +
2
2
1
2
(1 + co s 2 x )) dx
1
2
1
4
co s 2 x ) dx
sin 2 x ) + c
Integrating odd powers of cosx and sin
x
Odd powers of cos x and sin x can be integrated using the identity cos2 x + sin2 x = 1.
co s x 
1
2
(1 + co s 2 x )
1
sin x 
1
2
(1  co s 2 x )
2
2
2
Find
 sin . x dx
3
 sin x dx =
3
Using
2
 sin x sin x dx
2
=
 (1  co s x ) sin x dx
=
 (sin x  co s x sin x ) dx
2
2
Integrating odd powers of cosx and sin
x
This is now in a form that we can integrate.
The first part, sin x, integrates to give –cos x.
The second part, cos2 x sin x, can be recognized as the product of two functions.
Remember the chain rule for differentiation:
n
y=

dy
n 1
=n
dy
where
is f (x) and
is f ’(x).
The derivative of cos x is –sin x and so:
d
dx
(co s x ) =  3 co s x sin x
3
2
Integrating odd powers of cosx and sin
x
Therefore,
 co s x sin x dx = 
2
1
3
3
co s x + c
So, returning to the original problem:
 sin x dx =
3
 (sin x  co s x sin x ) dx
2
=  co s x +
=
1
3
1
3
3
co s x + c
co s x (co s x  3 ) + c
2
Separable variables
Using trigonometric identities in integration
Using partial fractions in integration
Contents
First-order differential equations
Differential equations with separable variables
Using differential equations to model real-life situations
The trapezium rule
Examination-style questions
9 of 66
© Boardworks Ltd 2006
Separable variables
Differential equations that can be arranged in the form
f ( y)
dy
= g( x )
dx
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 ( y ) dy =  g ( x ) dx
Although the dy and the dx have been separated it is important to remember that
a fraction.
dy
dx
For example, avoid writing:
f ( y ) dy = g ( x ) dx
is not
Separable variables
Here is an example:
Find the general solution to
dy
.
=
dx
Rearrange to give:
y
dy
x+2
y
= x+2
dx
You can miss out the step
Separate the variables and integrate:

We only need a ‘c’ on
one side of the equation.
y dy =
y
2
=
2
2
 ( x + 2 ) dx
x
2
2
+ 2x + c
2
y = x + 4x + A
y=
dy
dx =  ( x + 2 ) dx
and use
the
fact that
dx
y
2
x + 4x + A
dy
...
dx = ... dy
to separate
dx the dy from the
dx directly.
Separable variables
Find the particular solution to the differential equation
dy
given that y = ln when x = 0.
=e
3 x y
dx
7
3
Using the laws of indices this can be written as:
dy
3x
=e e
y
dx
Separating the variables and integrating with respect to x gives:
 e dy =
y
y
e =
 e dx
3x
1
3
e
3x
+c
Take the natural logarithms of both sides:
y = ln ( 31 e
3x
+ c)
Separable variables
Given that y = ln when 7x = 0:
3
ln
7
3
= ln( 31 + c )
c=2
The particular solution is therefore:
y = ln ( 31 e
3x
+ 2)
Modelling real-life situations
Using trigonometric identities in integration
Using partial fractions in integration
Contents
First-order differential equations
Differential equations with separable variables
Using differential equations to model real-life situations
The trapezium rule
Examination-style questions
14 of 66
© Boardworks Ltd 2006
Modelling real-life situations
Remember, the rate of change of one variable, say s, with respect to another variable, t, is
.
ds
dt
Many real-life situations involve the rate of change of one variable with respect to
another.
Since these situations involve derivatives they are modelled using differential equations.
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:
dh
dt
 (H  h)
Modelling real-life situations
We can write this relationship as an equation by introducing a positive constant k :
dh
= k(H  h)
dt
The general solution to this differential equation can be found by separating the variables
and integrating.
Remember the minus
sign, because we have –
h. (H is a constant).
H
1
h
dh =
 k dt
 ln( H  h ) = kt  c
ln( H  h ) =  kt  c
H h=e
 kt  c
h=H e
 kt
h = H  Ae
e
c
 kt
where A = ec
Modelling real-life situations
This is the general solution to the differential equation:
dh
= k(H  h)
dt
If we are given further information then we can determine the value of the constants in the
general solution to give a particular solution.
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:
h = 20  Ae
Also, assuming that when t = 0, h = 0:
0 = 20  A
A = 20
 kt
Modelling real-life situations
And finally using the fact that when t = 7, h = 5:
5 = 20  20 e
20 e
e
7 k
7 k
7 k
= 15
=
3
4
Take the natural logarithms of both sides:
 7 k = ln( 34 )
k =
ln ( 34 )
7
This gives the particular solution:
t ln 34
h = 20  20e
7
Modelling real-life situations
Find the height of the plant after 21 days.
Using t = 21 in the particular solution gives
h = 20  20 e
= 20 
4
3 ln 3
3
2 0( 34 )
Using the fact that
e
4
3 ln 3
= ( 34 )
3
= 1 1 196 cm
Comment on the suitability of this model as the plant reaches its final height.
Using this model the plant will reach its final height when:
t ln 34
e
7
This will never happen.
=0
Since ex never equals 0 this model predicts that the plant will get closer and closer to its
final height without ever reaching it.
Exponential growth
The most common situations that are modelled by differential equations are those involving
exponential growth and decay.
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.
We can write this as:
dP
 P
dt
This gives us the differential equation:
dP
dt
where k is a positive constant.
= kP
Exponential growth
If the initial investment is £1000 and after 5 years the balance is £1246.18, find the
particular solution to this differential equation.
dP
= kP
dt
1 dP
=k
P dt
Integrating both sides with respect to t gives:
1
 P dP =  k dt
ln P = kt + c
P =e
kt + c
kt
P=e e
c
We don’t need to write
|P| because P > 0.
Exponential growth
P = Ae
kt
w h e re A = e
c
This is the general solution to
dP
.
= kP
dt
Now, using the fact that when t = 0, P = 1000:
1000 = Ae
0
A = 1000
Also when t = 5, P = 1246.18:
1246.18 = 1000 e
e
5k
5k
= 1.24618
5 k = ln 1 .2 4 6 1 8
k = 0.044 (to 3 s.f.)
Exponential growth
The particular solution is therefore:
0.044 t
P = 1000 e
Find the value of the investment after 10 years.
When t = 10:
P = 1000 e
0.44
P = £ 1 5 5 2 .7 1
How long will it take for the initial investment to double?
Substitute P = 2000 into the particular solution:
2000 = 1000 e
0.044 t
0 .0 4 4 t = ln 2
ln 2
t=
0 .0 4 4
 1 5 .7 5 ye a rs
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:

dm
 m
dt
This gives us the differential equation:
dm
dt
where k is a positive constant.
=  km
Exponential decay
Separating the variables and integrating gives:
1
 m dm =   k dt
ln m =  kt + c
m=e
m=e
 kt + c
 kt
m = Ae
e
c
 kt
This is the general solution to the differential equation
w here A = e
.
c
dm
dt
Suppose a patient is injected with 5 ml of the drug.
=  km
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:
m = 5e
 kt
Also, given that m = 4 when t = 1 we have:
4 = 5e
e
k
=
k
4
5
 k = ln( 54 )
This gives us the particular solution:
m = 5e
t ln ( 4 )
5
We could also write this
as
m = 5( 54 )
t
Exponential decay
When m = 1 we have:
1 = 5e
e
t ln ( 54 )
=
t ln ( 54 )
1
5
t ln( 54 ) = ln( 51 )
t=
ln ( 51 )
ln ( 54 )
t  7 .2
So it will be about 7 hours and 12 minutes before the amount of drug in the bloodstream
reduces to 1 ml.