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Higher
Mathematics
UNIT 3 OUTCOME 2
Further Calculus
Contents
Further Calculus
1
2
3
4
5
6
Differentiating sinx and cosx
Integrating sinx and cosx
The Chain Rule
Special Cases of the Chain Rule
A Special Integral
Integrating sin(ax + b) and cos(ax + b)
HSN23200
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Higher Mathematics
Unit 3 – Further Calculus
OUTCOME 2
Further Calculus
1
Differentiating sinx and cosx
In order to differentiate expressions involving trigonometric functions, we
use the following rules:
d ( sin x ) = cos x ,
dx
d ( cos x ) = − sin x .
dx
These rules only work when x is an angle measured in radians. A form of
these rules is given in the exam.
EXAMPLES
1. Differentiate y = 3sin x with respect to x.
dy
= 3cos x .
dx
2. A function f is defined by f ( x ) = sin x − 2 cos x for x ∈ ℝ .
( )
Find f ′ π3 .
f ′ ( x ) = cos x − ( −2sin x )
= cos x + 2sin x
Remember
The exact value triangle:
( )
f ′ π3 = cos π3 + 2 sin π3
π
6
2
= 12 + 2 × 23
= 12 + 3.
3
π
3
1
3. Find the equation of the tangent to the curve y = sin x when x = π6 .
( )
(
)
When x = π6 , y = sin π6 = 12 . So the point is π6 , 12 .
We also need the gradient at the point where x = π6 :
dy
= cos x .
dx
( )
When x = π6 , mtangent = cos π6 = 23 .
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Higher Mathematics
Unit 3 – Further Calculus
(
)
Now we have the point π6 , 12 and the gradient mtangent = 23 , so:
y − b = m( x − a)
(
y − 12 = 23 x − π6
)
2 y − 1 = 3 x − 36π
3 x − 2 y − 36π + 1 = 0.
2
Integrating sinx and cosx
We know the derivatives of sin x and cos x , so it follows that the integrals are:
∫ cos x dx = sin x + c ,
∫ sin x dx = − cos x + c .
Again, these results only hold if x is measured in radians.
EXAMPLES
1. Find ∫ ( 5sin x + 2cos x ) dx .
∫ (5sin x + 2cos x ) dx = −5cos x + 2sin x + c .
2. Find
π
4
0
∫ ( 4 cos x + 2 sin x ) dx .
π
4
0
π
∫ ( 4 cos x + 2 sin x ) dx = [ 4 sin x − 2 cos x ]04
( ( ) ( ) ) − ( 4sin0 − 2 cos0)
= ( ( 4 × 1 ) − ( 2 × 1 ) ) − ( −2 )
2
2
= 4 sin π4 − 2 cos π4
= 4 − 2 +2
2
(2
2
2)
= 2 × 2 +2
= 2 + 2.
3. Find the value of
4
∫0
4
∫0 12 sin x dx .
1 sin x dx =  − 1 cos x  4
2
 2
0
= − 12 cos ( 4 ) + 12 cos ( 0 )
= 12 ( 0.654 + 1)
= 0.827 (to 3 d.p.).
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Note
It is good practice to
rationalise the
denominator.
Page 150
Remember
We must use radians
when integrating or
differentiating trig.
functions.
HSN23200
Higher Mathematics
3
Unit 3 – Further Calculus
The Chain Rule
We will now look at how to differentiate composite functions, such as
f ( g ( x ) ) . If the functions f and g are defined on suitable domains, then
d  f g ( x )) = f ′( g ( x )) × g ′( x ) .

dx  (
Stated simply: differentiate the outer function, the bracket stays the same,
then multiply by the derivative of the bracket.
This is called the chain rule. You will need to remember it for the exam.
EXAMPLE
(
)
If y = cos 5 x + π6 , find
(
y = cos 5 x + π6
dy
.
dx
)
Note
The “× 5 ” comes from
dy
= − sin 5x + π6 × 5
dx
(
)
(
(
)
d 5x + π .
6
dx
)
= −5sin 5 x + π6 .
4
Special Cases of the Chain Rule
We will now look at how the chain rule can be applied to particular types of
expression.
Powers of a Function
n
For expressions of the form [ f ( x )] , where n is a constant, we can use a
simpler version of the chain rule:
d  f ( x ) ) n  = n [ f ( x )]n −1 × f ′ ( x ) .
dx (

Stated simply: the power (n ) multiplies to the front, the bracket stays the
same, the power lowers by one (giving n − 1) and everything is multiplied by
the derivative of the bracket ( f ′ ( x )) .
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Higher Mathematics
Unit 3 – Further Calculus
EXAMPLES
1. A function f is defined on a suitable domain by f ( x ) = 2 x 2 + 3 x .
Find f ′ ( x ) .
1
f ( x ) = 2 x 2 + 3x = ( 2 x 2 + 3x ) 2
− 12
f ′ ( x ) = 12 ( 2 x 2 + 3 x )
× ( 4 x + 3)
(
= 12 ( 4 x + 3 ) 2 x 2 + 3 x
=
4x + 3
2 2x 2 + 3x
− 12
)
.
2. Differentiate y = 2 sin 4 x with respect to x.
y = 2sin 4 x = 2 ( sin x )4
dy
= 2 × 4 ( sin x )3 × cos x
dx
= 8sin 3 x cos x .
Powers of a Linear Function
n
The rule for differentiating an expression of the form ( ax + b ) , where a, b
and n are constants, is as follows:
d ( ax + b )n  = an ( ax + b ) n −1 .

dx 
EXAMPLES
3. Differentiate y = ( 5 x + 2 )3 with respect to x.
y = ( 5 x + 2 )3
dy
= 3 ( 5 x + 2 )2 × 5
dx
= 15 ( 5 x + 2 )2 .
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Higher Mathematics
4. If y =
y=
Unit 3 – Further Calculus
dy
1
.
3 , find
dx
(2x + 6)
1
−3
3 = ( 2x + 6)
( 2x + 6)
dy
−4
= −3 ( 2 x + 6 ) × 2
dx
−4
= −6 ( 2 x + 6 )
=−
6
.
( 2 x + 6 )4
5. A function f is defined by f ( x ) = 3 ( 3 x − 2 )4 for x ∈ ℝ . Find f ′ ( x ) .
4
f ( x ) = 3 ( 3 x − 2 )4 = ( 3 x − 2 ) 3
1
f ′ ( x ) = 34 ( 3 x − 2 ) 3 × 4
3
= 16
3 (3x − 2 ).
Trigonometric Functions
The following rules can be used to differentiate trigonometric functions.
d sin ax + b ) = a cos ( ax + b )

dx  (
d cos ax + b ) = −a sin ( ax + b )

dx  (
These are given in the exam.
EXAMPLE
6. Differentiate y = sin ( 9 x + π ) with respect to x.
dy
= 9cos ( 9 x + π ) .
dx
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Higher Mathematics
5
Unit 3 – Further Calculus
A Special Integral
n
The method for integrating an expression of the form ( ax + b ) is:
n
∫ ( ax + b )
( ax + b )n+1
dx =
a ( n + 1)
+c
where a ≠ 0 and n ≠ −1.
Stated simply: raise the power (n ) by one, divide by the new power and
also divide by the derivative of the bracket ( a ( n + 1)) , add c.
EXAMPLES
1. Find ∫ ( x + 4 ) dx .
7
8
∫ ( x + 4)
7
dx =
( x + 4)
+c
8 ×1
8
( x + 4)
=
+ c.
8
2. Find ∫ ( 2 x + 3 )2 dx .
( 2 x + 3 )3
∫ ( 2 x + 3 ) dx = 3 × 2 + c
( 2 x + 3 )3
=
+ c.
6
2
⌠
1
dx where x ≠ − 95 .
3
⌡ 5x + 9
3. Find 
⌠

⌡
3
⌠
1
1
dx = 
1 dx
5x + 9
⌡ (5x + 9 ) 3
1
= ∫ ( 5 x + 9 )− 3 dx
2
=
(5x + 9 ) 3
2 ×5
3
3
=
5x + 9
10
3
+c
2
+c
2
3 3 5x + 9 + c .
= 10
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Higher Mathematics
4. Evaluate
3
∫0
3
∫0
Unit 3 – Further Calculus
3 x + 4 dx where x ≥ − 34 .
1
3
3 x + 4 dx = ∫0 ( 3 x + 4 ) 2 dx
3
 (3x + 4 ) 2 
= 3

 2 × 3  0
3
=  29 ( 3 x + 4 )3 

3
0
3
3
= 29 ( 3 × 3 + 4 ) − 29 ( 3 × 0 + 4 )
=
3
2
9
= 29
13 −
(
2
9
4
)
3
13 − 8
3
(or 8.638 to 3 d.p.).
Note
Changing powers back
into roots here makes it
easier to evaluate the
two brackets.
Remember
To evaluate 43 , it is
easier to work out 4
first.
Warning
Make sure you don’t confuse differentiation and integration – this could
lose you a lot of marks in the exam.
Remember the following rules for differentiating and integrating expressions
n
of the form ( ax + b ) :
d  ax + b )n  = an ( ax + b )n −1 ,

dx (
n
∫ ( ax + b )
dx =
( ax + b )n+1
a ( n + 1)
+ c.
These rules will not be given in the exam.
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Higher Mathematics
Unit 3 – Further Calculus
Using Differentiation to Integrate
Recall that integration is just the process of undoing differentiation. So if we
differentiate f ( x ) to get g ( x ) then we know that
∫ g ( x ) dx = f ( x ) + c .
EXAMPLES
5
with respect to x.
( 3 x − 1)4
1
⌠
dx .
(b) Hence, or otherwise, find 
5
⌡ ( 3 x − 1)
5. (a) Differentiate y =
(a)
y=
5
= 5 ( 3 x − 1)−4
4
( 3 x − 1)
dy
= 5 × 3 × ( −4 ) ( 3 x − 1)−5
dx
60
=−
.
( 3 x − 1)5
⌠
60
5
dx
=
+ c . So:
5
( 3 x − 1)4
⌡ ( 3 x − 1)
(b) From part (a) we know  −
⌠
1
5
dx =
+c
5
( 3 x − 1)4
⌡ ( 3 x − 1)
−60
Note
We could also have used
the special integral to
obtain this answer.

1
5
⌠
1 
dx
=
−
+
c


5
60  ( 3 x − 1)4
⌡ ( 3 x − 1)


=−
6. (a) Differentiate y =
1
(x
⌠
(b) Hence, find 

3
5
− 1)
x2
(
3
)
⌡ x −1
(a)
y=
1
5
( x 3 − 1)
1
+ c1
12 ( 3 x − 1)4
6
where c1 is some constant.
with respect to x.
dx .
= ( x 3 − 1)
−5
−6
dy
= −5 ( x 3 − 1) × 3 x 2
dx
15 x 2
=− 3
6 .
x
−
1
(
)
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HSN23200
Higher Mathematics
Unit 3 – Further Calculus
⌠
(b) From part (a) we know 
−

⌡
⌠
x2
−15
 x3 −1
⌡
(
)
⌠
x2
 3
 x −1
⌡
(
6
6
dx =
6
(x
1
(x
3
5
− 1)

)
15 x 2
3
− 1)
6
dx =
1
(x
3
5
− 1)
+ c . So:
+c
Note
In this case, the special
integral cannot be used.

5 +c
3

x
−
1
(
)


1
=−
where c1 is some constant.
5 + c1
15 ( x 3 − 1)
1
dx = − 15

1
Integrating sin(ax + b) and cos(ax + b)
Since we know the derivatives of sin ( ax + b ) and cos ( ax + b ) , it follows
that their integrals are:
∫ cos ( ax + b ) dx = 1a sin ( ax + b ) + c ,
∫ sin ( ax + b ) dx = − 1a cos ( ax + b ) + c .
These are given in the exam.
EXAMPLES
1. Find ∫ sin ( 4 x + 1) dx .
∫ sin ( 4 x + 1) dx = − 14 cos ( 4 x + 1) + c .
(
)
2. Find ∫ cos 32 x + π5 dx .
∫ cos ( 2 x + π5 ) dx = 23 sin ( 2 x + π5 ) + c .
3
3. Find the value of
1
∫0
3
1
∫0 cos ( 2 x − 5) dx .
1
cos ( 2 x − 5 ) dx =  12 sin ( 2 x − 5 ) 
0
= 12 sin ( −3 ) − 12 sin ( −5 )
= 12 ( −0.141 − 0.959 )
= −0.55 (to 2 d.p.).
hsn.uk.net
Page 157
Remember
We must use radians
when integrating or
differentiating trig.
functions.
HSN23200
Higher Mathematics
Unit 3 – Further Calculus
(
)
4. Find the area enclosed by the graph of y = sin 3 x + π6 , the x-axis and
the lines x = 0 and x = π6 .
y
(
y = sin 3 x + π6
O
π
)
x
π
6
π
⌠ 6 sin 3 x + π dx =  − 1 cos 3 x + π  6
6
6 0
⌡0
 3
(
)
(
)
( ( ) )) − ( − 31 cos (3 ( 0) + π6 ))
= ( ( − 13 ) × ( − 12 ) ) + ( 13 × 2 )
(
= − 13 cos 3 π6 + π6
3
= 16 + 63
1+ 3
=
.
6
So the area is
1+ 3
square units.
6
5. Find ∫ 2cos ( 12 x − 3 ) dx .
∫ 2 cos ( 12 x − 3 ) dx =
(
)
= 4 sin ( 12 x − 3 ) + c .
6. Find
2 sin 1 x − 3 + c
1
2
2
∫ (5cos ( 2 x ) + sin ( x −
∫ (5cos ( 2 x ) + sin ( x −
hsn.uk.net
)
3 ) dx .
)
3 ) dx = 52 sin ( 2 x ) − cos ( x − 3 ) + c .
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Higher Mathematics
7. (a) Differentiate
Unit 3 – Further Calculus
1
with respect to x.
cos x
tan x
(b) Hence find ⌠
dx .

⌡ cos x
1
d ( cos x )−1 = −1( cos x )−2 × − sin x
(a)
= ( cos x )−1 , and dx
cos x
sin x
.
=
cos2 x
(b)
sin x
tan x cos
sin x
x
=
=
.
cos x cos x cos 2 x
sin x
1
From part (a) we know ⌠
dx
=
+c .

⌡ cos 2 x
cos x
tan x
1
Therefore ⌠
dx =
+ c.

⌡ cos x
cos x
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