Download 24 Calculus using cosx and sinx (OCR)

“Teach A Level Maths”
Vol. 2: A2 Core Modules
24: Differentiation and
Integration with Trig
Functions
© Christine Crisp
Module C4
OCR
This presentation follows on from
“Differentiating some Trig
Functions”
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In C3 you learnt to differentiate using the following
methods:
•
the product rule
•
the quotient rule
You also learnt to integrate some compound functions
using :
•
inspection
This presentation gives you some practice using the
above methods with trig functions.
Reminder:
A function such as y  3 sin x is a product, BUT we
don’t need the product rule.
When we differentiate, a constant factor just “tags
along” multiplying the answer to the 2nd factor.
e.g.
y  3 sin x

dy
 3 cos x
dx
However, the product rule will work even though you
shouldn’t use it
u  3 and
du
0
dx
N.B.
v  sin x
Reminder:
A function such as y  3 sin x is a product, BUT we
don’t need the product rule.
When we differentiate, a constant factor just “tags
along” multiplying the answer to the 2nd factor.
e.g.
y  3 sin x

dy
 3 cos x
dx
However, the product rule will work even though you
shouldn’t use it
so,
u  3 and v  sin x
du
dv
0
 cos x
dx
dx
dy
dy
du
dv

 3 cos x
v
u
dx
dx
dx
dx
as before.
Product Rule
Exercise
Use the product rule to differentiate the following.
1.
y  x 4 sin x
2.
y  e x cos x
Product Rule
Solutions:
1.
y  x 4 sin x
Let
dy

 4 x 3 sin x  x 4 cos x
dx
factors:
 x 3 (4 sin x  x cos x )
dy
du
dv
v
u
dx
dx
dx
Remove common
u  x 4 and v  sin x
du
dv
3
 4x
 cos x
dx
dx
Product Rule
2.
y  e x cos x
Let
dy
du
dv
v
u
dx
dx
dx
ue
du
x
e
dx
x
and
v  cos x
dv
  sin x
dx
dy

 e x cos x  e x (  sin x )
dx
dy

 e x cos x  e x sin x
dx
dy

 e x (cos x  sin x )
dx
Product Rule or Chain Rule?
For products we use the product rule and for
functions of a function we use the chain rule.
Decide how you would differentiate each of the
following ( but don’t do them ):
(c)
y  sin 2 x
y  x sin x
y  2 sin x
This is a simple function
(d)
y  sin 3 x
Chain rule
(a)
(b)
Chain rule
Product rule
 y  (sin x ) 
3
Product Rule or Chain Rule?
Exercise
Decide with a partner how you would differentiate the
following ( then do them if you need the practice ):
Write C for the Chain rule and P for the Product Rule
1.
y  x sin x
P
3.
2.
y  sin 2 x
y  sin x 2
C or P
C
Product Rule or Chain Rule?
Solutions
1. y  x sin x P
u x
v  sin x
du
dv
1
 cos x
dx
dx
dy
du
dv
dy
v
 u
 sin x  x cos x

dx
dx
dx
dx
2.
y  sin x 2
C
u  x 2  y  sin u
du
dy
 2x
 cos u  cos x 2
dx
du
dy
dy dy du



 2 x cos x 2
dx du dx
dx
Product Rule or Chain Rule?
3.
y  sin 2 x
u  sin x  y  u 2
Either C
du
dy
 cos x
 2u  2 sin x
dx
du
dy
dy dy du



 2 sin x cos x
dx du dx
dx
Or P
u  sin x
v  sin x
du
dv
 cos x
 cos x
dx
dx
dy
dy
du
dv
v
 u
 sin x cos x  sin x cos x

dx
dx
dx
dx
dy

 2 sin x cos x
dx
Quotient Rule
The quotient rule is
u
y
v

du
dv
v
 u
dy
 dx 2 dx
dx
v
Exercise
Use the quotient rule to differentiate the following.
1.
x
y
sin x
2.
y
cos x
x
4
Quotient Rule
Solution:
1.
x
y
sin x
u x
du
1
dx
u
y
v

du
dv
v
 u
dy
 dx 2 dx
dx
v
v  sin x
dv
 cos x
dx
sin x  x cos x
 y
2
sin x
and
Quotient Rule
2.
y
cos x
x
Solution:
4
u
y
v

du
dv
v
 u
dy
 dx 2 dx
dx
v
u
y   u  cos x and v  x 4
v
dv
du
3

4x
  sin x
dx
dx
dy  x 4 sin x  4 x 3 cos x


4 2
dx
(x )
 ( x sin x  4 cos x )
dy  x 3 ( x sin x  4 cos x )



dx
x5
x8 5
x
Quotient Rule
We can now differentiate the trig function
by writing
sin x
y  tan x 
cos x
u
u  sin x
y

v
du
 cos x
dx
y  tan x
v  cos x
dv
  sin x
dx
dy (cos x )(cos x )  (sin x )(  sin x )


dx
(cos x ) 2

cos2 x  sin2 x
cos2 x
Quotient Rule
So,
y  tan x 
dy cos2 x  sin2 x

dx
cos2 x
This answer can be simplified:
dy
1

cos x  sin x  1 
dx cos 2 x
1
Also,
is defined as secx
cos x
2
So,
2
y  tan x 
dy
 sec2 x
dx
Quotient Rule
Exercise
Use the quotient rule ( or, for (a) and (b), the
chain rule ) to find the derivatives with respect to x
of
(a)
y  cosec x (b) y  sec x
(c) y  cot x
Before you check the solutions, look in your formula
books to see the forms used for the answers. Try
to get your answers into these forms.
Quotient Rule
(a) Solution:
1
y  cosec x 
sin x
dy sin x(0)  1(cos x )

dx
(sin x ) 2
cos x

sin x sin x
cos x
1


sin x sin x
  cot x cosec x (   cosec x cot x )
Quotient Rule
(b) Solution:
1
y  sec x 
cos x
dy cos x(0)  1(  sin x )

2
dx
(cos x )
sin x

cos x cos x
sin x
1


cos x cos x
 tan x sec x
(  sec x tan x )
Quotient Rule
(c) Solution:
cos x
y  cot x 
sin x
dy sin x(  sin x )  cos x cos x

2
dx
(sin x )


 (sin 2 x  cos 2 x )
sin 2 x
1
sin 2 x
  cosec2 x
Integration
Since integration is the reverse of differentiation,
for the trig ratios we have
y
cos x
sin x

y dx
sin x  C
 cos x  C
Integration
e.g. 1


cos x dx
0
  sin x 0

  sin 
0

Radians!

 sin 0 
The definite integral can give an area, so this result
may seem surprising. However, the graph shows us
why it is correct.
The areas above
This part gives a positive integral
and below the axis
y  cos x
are equal, . . .
but the integral
for the area
below is negative.
This part gives a negative integral
Integration
e.g. 1


cos x dx
0
  sin x 0

  sin 
0

Radians!

 sin 0 
The definite integral can give an area, so this result
may seem surprising. However, the graph shows us
why it is correct.
How would you find
This part gives a positive integral
the area?
y  cos x
This part gives a negative integral
Ans: Find the
integral from 0 to

and double it.
2
Integration
2.


0
sin x dx    cos x  0
   cos 
    cos 0 
N.B. cos 0  0
 1     1 
2
y  sin x
The area is
above the axis,
so the integral
gives the
entire area.
Integration
Before we try to integrate compound functions, we
need to be able to recognise them, and know the
rule for differentiating them.
If
dy dy du
y  f ( g( x )) ,


dx du dx
where u  g( x ),
the inner function.
We saw that in words this says:
 differentiate the inner function
 multiply by the derivative of the outer function
e.g. For
y  sin 3 x
we get
dy
 3  cos 3 x
dx
 3 cos 3 x
Integration
Since indefinite integration is the reverse of
differentiation, we get

cos 3 x dx 

3 cos 3 x dx 
So,
The rule is:
sin 3 x  C
sin 3 x
C
3
If we divide C by 3, we get
 integrate the outer another
functionconstant, say C1, but
we usually just write C.
Integration
Since indefinite integration is the reverse of
differentiation, we get

cos 3 x dx 

3 cos 3 x dx  sin 3 x  C
So,
sin 3 x
C
3
The rule is:
 integrate the outer function
 divide by the derivative of the inner function
Integration
Since indefinite integration is the reverse of
differentiation, we get

cos 3 x dx 

3 cos 3 x dx  sin 3 x  C
So,
sin 3 x
C
3
The rule is:
 integrate the outer function
 divide by the derivative of the inner function
Integration
Since indefinite integration is the reverse of
differentiation, we get

cos 3 x dx 

3 cos 3 x dx  sin 3 x  C
So,
sin 3 x
C
3
The rule is:
 integrate the outer function
 divide by the derivative of the inner function
Tip: We can check the answer by differentiating it.
We should get the function we wanted to integrate.
Integration
However, we can’t integrate all compound functions
in this way.
Let’s try the rule on another example:
e.g.

cos x dx 
2
sin x 2  C
2x
THIS IS
WRONG !
 integrate the outer function
 divide by the derivative of the inner function
Integration
However, we can’t integrate all compound functions
in this way.
Let’s try the rule on another example:
e.g.

cos x dx 
2
sin x 2  C
2x
THIS IS
WRONG !
The rule has given us a quotient, which, if we
differentiate it, gives:
du
dv
v
 u
2
2
dx
dx  2 x( 2 x cos x )  2 (sin x )
(2 x ) 2
v2
. . . nothing like the function we wanted to integrate.
Integration
What is the important difference between
 cos 3 x dx
and
 cos x dx
2
?
When we differentiate the inner function of the 1st
example, we get 3, a constant.
Dividing by the 3 does NOT give a quotient of the
u
form
( since v is a function of x ).
v
The 2nd example gives 2x,which is a function of x.
Integration
What is the important difference between
 cos 3 x dx
and
 cos x dx
2
?
When we differentiate the inner function of the 1st
example, we get 3, a constant.
Dividing by the 3 does NOT give a quotient of the
u
form
( since v is a function of x ).
v
The 2nd example gives 2x,which is a function of x.
So, the important difference is that the 1st example has
an inner function that is linear; it differentiates to a
constant. The 2nd is non-linear. We can’t integrate it.
Integration
Exercises
Find

1.
sin 4 x dx

2.

2
cos 3 x dx
0
Solutions:
1.
2.


 cos 4 x
sin 4 x dx 
C
4

2
0

 sin 3 x 
cos 3 x dx  

 3 0
2
 sin 3( 2 ) sin 0 



3 
 3
1

3
Integration
We found earlier that
y  tan x
so,


dy
 sec2 x
dx
sec2 x dx  tan x  C
You need to remember this exception to the
general rule that we can only integrate directly if
the inner function is linear.
We also have, for example,

tan 2 x
sec 2 x dx 
C
2
2
Integration
There are 3 more important trig integrals.
e.g.
Find
2
sin
 x dx
We have a function of a function . . .
 ( sin x )
2
dx
but the inner function . . .
Integration
There are 3 more important trig integrals.
e.g.
Find
2
sin
 x dx
We have a function of a function . . .
 ( sin x )
2
dx
but the inner function . . .
is not linear.
However, we can use a trig formula to convert
the function into one that we can integrate.
Integration
e.g.
Find
 sin
2
x dx
Which double angle formula can we use to change
the function so that it can be integrated?
ANS: cos 2 A  1  2 sin 2 A
      ( 2b)
Rearranging the formula:

So,
cos 2 A  1  2 sin 2 A
sin 2 A  12 (1  cos 2 A)
2
sin
 x dx 
1
2
 (1  cos 2 x ) dx
sin 2 x 
1

 x 
2 
2
C
Integration
The previous example is an important application of
a double angle formula.
The next 2 are also important. Try them yourself.
Exercise
1.
Find
2.
Find
 cos
2
x dx
 sin x cos x dx
Integration
Exercise
1.
Find
Solution:

So,
2
cos
 x dx
cos 2 A  2 cos 2 A  1
1
(1  cos 2 A)
2
 cos 2 A
2
cos
 x dx 
1
2
 (1  cos 2 x ) dx
sin 2 x 

1
 x
C
2
2 
Integration
2.
Find
 sin x cos x dx
sin 2 A  2 sin A cos A
 12 sin 2 A  sin A cos A
Solution:
So,
 sin x cos x dx 
1
2
 sin 2 x dx
1  cos 2 x 
 
C
2
2 
cos 2 x

C
4
Integration
SUMMARY
The rearrangements of the double angle formulae
for cos 2 A are
sin 2 A  12 (1  cos 2 A)
cos 2 A  12 (1  cos 2 A)
They are important in integration so you should
either memorise them or be able to obtain them
very quickly.
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.
The Product Rule
SUMMARY
To differentiate a product:
 Check if it is possible to multiply out. If so, do it
and differentiate each term.
 Otherwise use the product rule:
If y  uv , where u and v are both functions of x
dy
du
dv
v
 u
dx
dx
dx
Product Rule or Chain Rule?
For products we use the product rule and for
functions of a function we use the chain rule.
For example,
(c)
y  sin 2 x
y  x sin x
y  2 sin x
This is a simple function
(d)
y  sin x
Chain rule
(a)
(b)
3
Chain rule
Product rule
 y  (sin x ) 
3
Quotient Rule
The quotient rule is
u
y
v

du
dv
v
 u
dy
 dx 2 dx
dx
v
Quotient Rule
With the quotient rule we can differentiate the
trig function
y  tan x
sin x
y  tan x 
cos x
by writing
u
y

v
v  cos x
u  sin x
dv
du
  sin x
 cos x
dx
dx
dy (cos x )(cos x )  (sin x )( sin x )


dx
(cos x )(cos x )

cos2 x  sin2 x
cos2 x
Quotient Rule
So,
y  tan x 
dy cos2 x  sin2 x

dx
cos2 x
This answer can be simplified:
dy
1

cos x  sin x  1 
dx cos 2 x
1
Also,
is defined as secx
cos x
2
So,
2
y  tan x 
dy
 sec2 x
dx
Integration
Since integration is the reverse of differentiation,
for the trig ratios we have
y
cos x
sin x

y dx
sin x  C
 cos x  C
e.g. 1

Integration

cos x dx 
0
 sin x

0
  sin   
0
Radians!
 sin 0
The definite integral can give an area, so this result
may seem surprising. However, the graph shows us
why it is correct.
This part gives a positive integral
y  cos x
To find the area,
find the integral

from 0 to
2
and double it.
This part gives a negative integral
Integration
Before we try to integrate compound functions, we
need to be able to recognise them, and know the
rule for differentiating them.
If
dy dy du
y  f ( g( x )) ,


dx du dx
where u  g( x ),
the inner function.
We saw that in words this says:
 differentiate the inner function
 multiply by the derivative of the outer function
e.g. For
y  sin 3 x
we get
dy
 3  cos 3 x
dx
 3 cos 3 x
Integration
Since indefinite integration is the reverse of
differentiation, we get

cos 3 x dx 

3 cos 3 x dx  sin 3 x  C
So,
sin 3 x
C
3
The rule is:
 integrate the outer function
 divide by the derivative of the inner function
Tip: We can check the answer by differentiating it.
We should get the function we wanted to integrate.
Integration
What is the important difference between

cos 3 x dx
and

2
cos x dx
?
When we differentiate the inner function of the 1st
example, we get 3, a constant.
Dividing by the 3 does NOT give a quotient of the
u
form
( since v is a function of x ).
v
The 2nd example gives 2x,which is a function of x.
So, the important difference is that the 1st example has
an inner function that is linear; it differentiates to a
constant. The 2nd is non-linear. We can’t integrate it.
Integration
We found earlier that
y  tan x
so,


dy
 sec2 x
dx
sec2 x dx  tan x  C
You need to remember this exception to the
general rule that we can only integrate directly if
the inner function is linear.
We also have, for example,

tan 2 x
sec 2 x dx 
C
2
2
Integration
There are 3 more important trig integrals.
e.g.
Find
2
sin
 x dx
We have a function of a function . . .
 ( sin x )
2
dx
but the inner function is not linear
However, we can use a trig formula to covert
the function into one that we can integrate.
Integration
e.g.
Find
 sin
2
x dx
Which double angle formula can we use to change
the function so that it can be integrated?
ANS: cos 2 A  1  2 sin 2 A
      ( 2b)
Rearranging the formula:

So,
cos 2 A  1  2 sin 2 A
sin 2 A  12 (1  cos 2 A)
2
sin
 x dx 
1
2
 (1  cos 2 x ) dx
sin 2 x 
1

 x 
2 
2
C
Integration
SUMMARY
The rearrangements of the double angle formulae
for cos 2 A are
sin 2 A  12 (1  cos 2 A)
cos 2 A  12 (1  cos 2 A)
They are important in integration so you should
either memorise them or be able to obtain them
very quickly.