pure-mathematics-2-3-pdf

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WORKED EXAMPLE 4.5 b xe−5x e3x
c x2
Differentiate with respect to x. Product rule.
a e2x Quotient rule.

Answer

a d (e2x ) =  2 × e2x  = 2e2x  
dx
ւց
differentiate index original function

Review b d (xe−5x ) = x ×  d ( e−5x )  +  e−5x   ×  d (x)
dx dx dx

= x × (−5e−5x ) + e−5x  × (1)

= −5xe−5x  + e−5x

= e−5x (1 − 5x)

d  e3x    =   x2  ×   d (e3x ) − e3x  ×  d (x2 )
dx  x2  dx dx
c x4

= x2  × 3e3x − e3x  × 2x
x4

= e3x (3x − 2) 89
x3

Review WORKED EXAMPLE 4.6

The equation of a curve is y = e2x − 9ex + 7x. Find the exact value of the x-coordinate
at each of the stationary points and determine the nature of each stationary point.

Answer

y = e2x − 9ex + 7x Differentiate.

  dy = 2e2x − 9ex +7
dx

d2 y = 4e2x − 9ex
dx2

Stationary points occur when dy = 0.
dx

Review

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        2e2x − 9ex + 7 = 0 Factorise.

 (2ex − 7)(ex − 1) = 0

(2ex − 7) = 0 or (ex − 1) = 0

ex = 7  or  ex = 1
2

x = ln  7   or  x = 0
 2 

When x = 0, d2 y is negative ⇒ maximum point.
dx2
d2 y
Review When x = ln  7  , dx2 is positive ⇒ minimum point.
 2 

EXERCISE 4C

1 Differentiate with respect to x. b e−4x c 2e6x
a e5x
d 3e−5x x f e2x−7

g ex2 −3 e 4e 2 i 5 ex − 1
e2x
h 2x + 3e x

90 j 2(e3x − 1) k 3e2x + e−2x l 5(ex2 − 2x)
2

Review 2 a Sketch the graph of the function y = 1 − e2−x.
b Find the equation of the normal to the curve y = 1 − e2−x at the point where y = 0.

M 3 The mass, m grams, of a radioactive substance remaining t years after a given time, is given by the formula
m = 300e−0.00012t. Find the rate at which the mass is decreasing when t = 2000.

4 Differentiate with respect to x.

a xex b x2e3x c 5xe−2x

d 2 x  ex e6x e−2x
e f

x x

g ex − 1 h xe3x + e6x i x2ex − x
ex + 2 2 ex + 2

5 Find the gradient of the curve y = 8 at the point where x = 0.
5 + e2x

Review 6 Find the exact coordinates of the stationary point on the curve y = xex.

7 The curve y = 2e2x + e−x cuts the y-axis at the point P. Find the equation of the tangent to the curve at the
point P and state the coordinates of the point where this tangent cuts the x-axis.

8 Find the exact coordinates of the stationary point on the curve y = (x − 4)ex and determine its nature.

9 Find the exact coordinates of the stationary point on the curve y= e2x and determine its nature.
x2

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10 The equation of a curve is y = x2e−x.

a Find the x-coordinates of the stationary points of the curve and determine the nature of these stationary points.

b Show that the equation of the normal to the curve at the point where x = 1 is e2x + ey = 1 + e2.

11 Find the exact value of the x-coordinates of the points on the curve y = x2e−2x at which d2 y = 0.
dx2
12 Find the coordinates of the stationary point on the curve y = e2x−1 .
x
d
P 13 By writing 2 as eln 2 , prove that dx (2x ) = 2x ln 2.

PS 14 The equation of a curve is y = x(3x ).

Review Find the exact value of the gradient of the tangent to the curve at the point where x = 1.

4.4 Derivatives of natural logarithmic functions

EXPLORE 4.2

y 91
4

3 y = ln x
2
1

–1 O 1 2 3 4 5 6 7 8 9x
–1
Review
–2

–3

–4

Here is a sketch of the graph of y = ln x. Can you sketch a graph of its derivative
(gradient function)? Discuss your sketch with your classmates.

Now sketch the graphs using graphing software. Can you suggest a formula for the
derivative of ln x? (It might be helpful to look at the coordinates of some of the points
on the graph of the derivative.)

The derivative of ln x

In Chapter 2, you learnt that if y = ln x, then  ey = x.

Review In Section 4.3, you learnt that if y = ex, then dy = ex.
dx

Using these two results we can find the rule for differentiating ln x.

   y = ln x

 ey = x Differentiate both sides with respect to y.
Rearrange and replace ey by x.
  e y = dx
dy

dy = 1
dx x

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KEY POINT 4.5

d (ln x) = 1
dx x

The derivative of ln f(x)

Consider the function y = ln f(x).

Let y = ln u where u = f(x)

Review dy = 1 du = f ′(x)
du u dx

Using the chain rule: dy = dy × du
dx du dx

= 1 × f ′(x)
u

= f ′(x)
f(x)

d [ln f(x)] = f ′(x)
dx f(x)

KEY POINT 4.6

92 In particular,

d [ln(ax + b)] = a
dx ax + b

Review WORKED EXAMPLE 4.7

Differentiate with respect to x. b ln(4x − 3) c ln 5 − x
a ln 2x
‘Inside’ differentiated.
Answer ‘Inside’

a d (ln 2x) = 2  
dx 2x

=1
x

b d [ln(4x − 3)] = 4 3  
dx 4x −

Review

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c Method 1:

d 1 (5 − x)− 1 (−1)
dx 2 2

 ln 5 − x  = 5−x ‘Inside’ differentiated.
‘Inside’
= − 1 x)
2(5 −

Method 2: using the rules of logarithms before differentiating.

d  ln 5− x  = d  1    Use ln am = m ln a
dx dx  ln(5 − 
 x)2  ‘Inside’ differentiated.
‘Inside’
Review           = d 1 ln(5 − x ) 
dx  2 

= 1 × d [ln(5 − x)]
2 dx

= 1 × −1
2 5−x

= − 1 x)
2(5 −

WORKED EXAMPLE 4.8

93

Review Differentiate with respect to x. ln 2x
a 2x4 ln 5x b x3

Answer

a d (2x4 ln 5x) =  2x4   ×   d (ln 5x) + ln 5x ×  d (2x4 ) Product rule.
dx dx dx Quotient rule.

=  2x4   ×   5 + ln 5x × 8x3
5x

= 2x3  + 8x3 ln 5x

( )b x3   ×   d ln 2x − ln 2x  ×  d (x3 )
d  ln 2x    =  dx dx
dx x3 (x3 )2

= x3   ×   2   −  ln 2x ×  3x2
2x
x6

Review = x2 − 3x2 ln 2x
x6

= 1− 3 ln 2x
x4

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EXERCISE 4D

1 Differentiate with respect to x. b ln 7x c ln(2x + 1)
a ln 3x
d 5 + ln(x2 + 1) e ln(2x − 1)2 f ln x − 3
g ln(x + 3)5
h 3x + ln  2  i 5x + ln  1 2 
j ln(ln x)  x   − 2x 

k ln ( 2 − x )2 l ln(5x + ln x)

ReviewPS 2 The answers to question 1 parts a and b are the same. Why is this the case? How many different ways can you
justify this?

3 Differentiate with respect to x.

a x ln x b 2x3 ln x c x ln(2x + 1)

d 3x ln 2x e x ln(ln x) f ln 5x
x

g 2 h ln(3x − 2) i ln(2x + 1)
ln x x 4x − 1

4 a Sketch the graph of the function y = ln(2x − 3).

b Find the gradient of the curve y = ln(2x − 3) at the point where x = 5.

94 5 Find the gradient of the curve y = e2x − 5 ln(2x + 1) at the point where x = 0.

Review 6 A curve has equation y = x2 ln 5x.
d2 y
Find the value of dy and dx2 at the point where x = 2.
dx

7 The equation of a curve is y = x2 ln x. Find the exact coordinates of the stationary point on this curve and
determine whether it is a maximum or a minimum point.

8 The equation of a curve is y = ln x . Find the exact coordinates of the stationary point on this curve and
x

determine whether it is a maximum or a minimum point.

9 Find the equation of the tangent to the curve y = ln(5x − 4) at the point where x = 1.

10 Use the laws of logarithms to help differentiate these expressions with respect to x.

a ln 5x − 1 b ln  1 2  c ln [x(x + 1)5 ]
 3x + 

d ln  2x + 3  e ln  1 − 3x  f ln  x(x − 2) 
 x −1   x2   x+4 

Review g ln  3−x  h ln  8  i ln  (x + 2) (2x − 1) 
 4) (x   1)2 (x   x(x + 5) 
(x + − 1) (x + − 2)

11 Find dy , in terms of x, for each of the following.
dx
a ey = 2x2 − 1 b ey = 3x3 + 2x
c ey = (x + 1)(x − 5)

TIP
Take the natural logarithm of both sides of the equation before differentiating.

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PS 12 A curve has equation x = 1  e y(2x−3) + 4 .
5
dy
Find the value of dx when x = 1.

4.5 Derivatives of trigonometric functions

EXPLORE 4.3

Review Graphing software has been used to draw the graphs of y = sin x and y = cos x for

0 ø x ø 2π together with their gradient (derived) functions.

yy y = cos x

1 gradient 1

function gradient

function
O π π 3π 2π x O π π 3π 2π x
22 22
y = sin x

–1 –1

Discuss with your classmates why, for the function y = sin x , you would expect
the graph of the gradient function to have this shape. Do the same for the graph of

y = cos x .

Can you suggest how to complete the following formulae? 95

Review d (sin x) = d (cos x) =
dx dx

The derivative of sin x

Consider the function f(x) = sin x, where x is measured in radians, and two points whose
x-coordinates are x and x + δx where δx is a small increase in x.

y f(x) = sin x
f(x + δx)

f(x)

O x x + δx x

Review dy = lim f(x + δx) − f(x)
dx (x + δx) − x
δx → 0

= lim sin(x + δx) − sin x Expand sin(x + δx)
δx
δx → 0 As δx → 0, cos δx → 1
and sin δx → δx.
= lim sin x cos δx + cos x sin δx − sin x
δx
δx → 0

= cos x

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KEY POINT 4.7

d (sin x) = cos x 
dx

Similarly, we can show that

d (cos x) = −sin x
dx

We can find the derivative of tan x using these two results together with the quotient rule.

Review d (tan x) = d  sin x  Use the quotient rule.
dx dx  cos x 

cos x ×   d (sin x) − sin x  ×  d (cos x)
=  dx dx
(cos x)2

=  cos x × cos x − sin x × (−sin x)
cos2 x

=   cos2 x + sin2 x  Use cos2 x + sin2 x = 1.
cos2 x

=   1  x Use 1 = sec x.
cos2 cos x

96 = sec2 x

KEY POINT 4.8

Review d (tan x) = sec2 x
dx

WORKED EXAMPLE 4.9

Differentiate with respect to x.

a 2 sin x b x cos x c tan x d (3 + 2 sin x)5
x2

Answer

a d (2 sin x) = 2 d (sin x)
dx dx

 = 2 cos x

Review b d (x cos x) =  x ×   d (cos x) + cos x ×   d (x) Product rule.
dx dx dx

= −x sin x + cos x

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d   tan x    =   x2  ×   d (tan x) − tan x ×  d (x2 ) Quotient rule.
dx x2 dx dx Chain rule.
c x4

=  x2  × sec2 x− tan x × 2x
x4

         =   x sec2 x− 2 tan x  
x3

d d [(3 + 2 sin x)5 ] = 5(3 + 2 sin x)4  × 2 cos x
dx

Review = 10 cos x(3 + 2 sin x)4

Derivatives of sin(ax + b), cos(ax + b) and tan(ax + b)

Consider the function y = sin(ax + b) where x is measured in radians.

Let   y = sin u where u = ax + b

dy = cos u du = a
du dx

Using the chain rule: dy = dy × du
dx du dx

= cos u × a 97

= a cos(ax + b)

Review KEY POINT 4.9 TIP

  d [sin(ax + b)] = a cos(ax + b) It is important to
dx remember that, in
calculus, all angles are
Similarly, it can be shown that: measured in radians
unless a question tells
  d [cos(ax + b)] = −a sin(ax + b) you otherwise.
dx

  d [tan(ax + b)] = a sec2 (ax + b)
dx

WORKED EXAMPLE 4.10

Differentiate with respect to x. cos  2x − π 
 4 
a 2 sin 3x b 4x tan 2x c d (3 − 2 cos 5x)4
x2
Review
Answer

a d (2 sin 3x) = 2 d (sin 3x)
dx dx

  = 2 × cos 3x × (3)

  = 6 cos 3x

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b d (4x tan 2x) = 4x  ×  d (tan 2x) + tan 2x ×   d (4x) Product rule.
dx dx dx Quotient rule.

= 4x × sec2 2x (2) + tan 2x × (4) Chain rule.

= 8x sec2 2x + 4 tan 2x

d  cos  2x − π   x2  ×   d  cos  2x − π   − cos  2x − π   ×  d [x2 ]
dx   4   dx   4   4  dx
c   =  
 x2  (x2 )2

 

Review −2x2 sin  2x − π  − 2x cos  2x − π
=   4   4

x4

−2x sin  2x − π  − 2 cos  2x − π   
   =   4   4 

x3

d d  ( 3 − 2 cos 5x )4    = 4(3 − 2 cos 5x )3 × 10 sin 5x
dx  

  = 40 sin 5x(3 − 2 cos 5x)3

EXERCISE 4E

98 1 Differentiate with respect to x.

a 2 + sin x b 2 sin x + 3 cos x c 2 cos x − tan x

Review d 3sin 2x e 4 tan 5x f 2 cos 3x − sin 2x

g tan( 3x + 2 ) h sin  2x + π  i 2 cos  3x − π 
 3   6 

2 Differentiate with respect to x.

a sin3 x b 5 cos2 3x c sin2 x − 2 cos x

d ( 3 − cos x )4 e 2 sin3  2x + π f 3 cos4 x + 2 tan2  2x − π 
 6  4 

3 Differentiate with respect to x.

a x sin x b 5x cos 3x c x2 tan x d x cos3 2x

e 5 f x g tan x h sin x
cos 3x cos x x 2 + cos x

i sin x 1 k 3x l sin x + cos x
3x − 1 j sin3 2x sin 2x sin x − cos x

Review 4 Differentiate with respect to x.

a esin x b ecos 2x c etan 3x d e( sin x−cos x )

e ex cos x f ex sin 2x g ex (2 cos x − sin x) h x3ecos x
i ln(cos x) j x ln(sin x)
k cos 2x l x sin 2x
e2 x +1 e2x

5 Find the gradient of the curve y = 3 sin 2x − 5 tan x at the point where x = 0.

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6 Find the exact value of the gradient of the curve y = 2 sin 3x − 4 cos x at the point  π , −2  .
 3 

7 Given that y = sin2 x for 0 ø x ø π, find the exact values of the x-coordinates of the points on the curve

where the gradient is 3 .
2
5
P 8 Prove that the gradient of the curve y = 2 − tan x is always positive.

9 a By writing sec x as 1 , find d (sec x).
cos x dx

b By writing cosec x as 1 x , find d (cosec x).
sin dx
Review
c By writing cot x as cos x , find d (cot x).
sin x dx

P 10 Prove that the normal to the curve y = x sin x at the point P  π ,  π  intersects the x-axis at the point (π, 0).
 2 2 

11 The equation of a curve is y = 5 sin 3x − 2 cos x . Find the equation of the tangent to the curve at the point
π
 3 , −1 . Give the answer in the form y = mx + c, where the values of m and c are correct to 3 significant figures.


12 A curve has equation y = 3 cos 2x + 4 sin 2x + 1 for 0 ø x ø π. Find the x-coordinates of the stationary
points of the curve, giving your answer correct to 3 significant figures.

13 A curve has equation y = ex cos x for 0øxø π . Find the exact value of the x -coordinate of the stationary
2
point of the curve and determine the nature of this stationary point.
99

14 A curve has equation y= sin 2x for 0øxø π . Find the exact value for the x-coordinate of the stationary
e2x 2
Review point of this curve.

15 A curve has equation y= e3x for 0,x, π . Find the exact value for the x-coordinate of the stationary
sin 3x 2
point of this curve and determine the nature of this point.

PS 16 A curve has equation y = sin 2x − x for 0 ø x ø 2π. Find the x-coordinates of the stationary points of the
curve, and determine the nature of these stationary points.

PS 17 A curve has equation y = tan x cos 2x for 0 ø x , π . Find the x -coordinate of the stationary point on the
2
curve giving your answer correct to 3 significant figures.

Review 4.6 Implicit differentiation

All the functions that we have differentiated so far have been of the form y = f(x). These
are called explicit functions as y is given explicitly in terms of x. However, many functions
cannot be expressed in this form, for example x3 + 5xy + y3 = 16.

When a function is given as an equation connecting x and y, where y is not the subject, it

is called an implicit function.

The chain rule  dy = dy × du  and the product rule  d ( uv ) = u dv + v du  are used
dx du dx dx dx dx

extensively to differentiate implicit functions.

This is illustrated in the following Worked examples.

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WORKED EXAMPLE 4.11

Differentiate each of these expressions with respect to x.

a y3 b 4x2 y5 c x3 + 5xy

Answer

a d ( y3 ) = d (y3) × dy Chain rule.
dx dy dx

= 3y2 dy
dx
Review
b d (4x2 y5 ) = 4x2 d ( y5 ) + y5 d (4x2 ) Product rule.
dx dx dx

 = 4x2 d ( y5 ) + 8xy5   Chain rule: d ( y5 ) = 5y4 dy .
dx dx dx

 = 20x2 y4 dy + 8xy5  
dx

c d (x3 + 5xy) = d (x3 ) + d (5xy) Use the product rule for d (5xy).
dx dx dx dx

= 3x2 + 5x d (y) + y d (5x)
dx dx

= 3x2 + 5x dy + 5y
dx

100

Review WORKED EXAMPLE 4.12

Find dy for the curve x3 + y3 = 4xy. Hence find the gradient of the curve at the point (2, 2).
dx

Answer

x3 + y3 = 4xy Differentiate with respect to x.

d (x3 ) + d ( y3 ) = d (4xy) Rearrange terms.
dx dx dx Factorise.
Rearrange.
3x2 + d (y3) × dy = 4x d (y) + y d (4x)
dy dx dx dx

3x2 + 3y2 dy = 4x dy + 4y
dx dx

3y2 dy − 4x dy = 4y − 3x2
dx dx

Review (3y2 − 4x) dy = 4y − 3x2
dx

dy = 4y − 3x2
dx 3y2 − 4x

When x = 2 and y = 2,

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dy = 4(2) − 3(2)2 = −1
dx 3(2)2 − 4(2)

Gradient of the curve at the point (2, 2) is −1.

WORKED EXAMPLE 4.13

The equation of a curve is 3x2 + 2xy + y2 = 6. Find the coordinates of the two stationary points on the curve.

Review Answer

3x2 + 2xy + y2 = 6 Differentiate with respect to x.

( )d d d y2 d d
dx dx dx dx
dx
(3x2 ) + (2xy) + = (6) Use product rule for (2xy).

( )6x d d d d
+ 2x dx (y) + y dx (2x ) + dx y2 =0 Use chain rule for dx ( y2 ).

6x + 2x dy + 2y + 2y dy = 0 Divide by 2 and rearrange.
dx dx

x dy + y dy = −y − 3x Factorise.
dx dx

(x + y) dy = −y − 3x Rearrange.
dx

dy = −y − 3x 101
dx x+y

Review Stationary points occur when dy = 0.
dx

− y − 3x = 0

y = −3x

Substituting y = −3x into the equation of the curve gives:
3x2 + 2x(−3x) + (−3x)2 = 6

6x2 = 6

x = ±1

The stationary points are (−1, 3) and (1, −3).

EXERCISE 4F

1 Differentiate each expression with respect to x.

a y5 b x3 + 2y2 c 5x2 + ln y
f y2 + xy
Review d 2 + sin y e 6x2 y3 i x3 ln y
l 2xecos y
g x3 − 7xy + y3 h x sin y + y cos x

j x cos 2y k 5y + ex sin y

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2 Find dy for each of these functions.
dx

a x3 + 2xy + y3 = 10 b x2 y + y2 = 5x c 2x2 + 5xy + y2 = 8
e 2ex y + e2x y3 = 10 f ln(xy) = 4 − y2
d x ln y = 2x + 5 h ln x − 2 ln y + 5xy = 3

g xy3 = 2 ln y

3 Find the gradient of the curve x2 + 3xy − 5y + y3 = 22 at the point (1, 3).

4 Find the gradient of the curve 2x3 − 4xy + y3 = 16 at the point where the curve crosses the x-axis.

Review 5 A curve has equation 2x2 + 3y2 − 2x + 4y = 4. Find the equation of the tangent to the curve at the
point (1, −2).

6 The equation of a curve is 4x2 y + 8 ln x + 2 ln y = 4. Find the equation of the normal to the curve at the
point (1, 1).

7 The equation of a curve is 5x2 + 2xy + 2y2 = 45.

a Given that there are two points on the curve where the tangent is parallel to the x-axis, show by
differentiation that, at these points, y = −5x.

b Hence find the coordinates of the two points.

8 The equation of a curve is y2 − 4xy − x2 = 20.

a Find the coordinates of the two points on the curve where x = 4.

102

b Show that at one of these points the tangent to the curve is parallel to the x-axis.

c Find the equation of the tangent to the curve at the other point.
Review
9 The equation of a curve is y3 − 12xy + 16 = 0.
a Show that the curve has no stationary points.
b Find the coordinates of the point on the curve where the tangent is parallel to the y-axis.

10 Find the gradient of the curve 5ex y2 + 2ex y = 88 at the point (0, 4).

11 The equation of a curve is x2 − 4x + 6y + 2y2 = 12. Find the coordinates of the two points on the curve at

which the gradient is 4 .
3

12 The equation of a curve is 2x + y ln x = 4y . Find the equation of the tangent to the curve at the point with

coordinates  1,  1  .
 2 

ReviewPS 13 The equation of a curve is y = xx. Find the exact value of the x-coordinate of the stationary point on this
curve.

PS 14 Find the stationary points on the curve x2 − xy + y2 = 48. By finding d2 y determine the nature of each of
these stationary points. dx2

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4.7 Parametric differentiation

Sometimes variables x and y are given as a function of a third variable t.

For example, x = 1 + 4 sin t and y = 2 cos t.

The variable t is called a parameter and the two equations are called the parametric
equations of the curve.

We can find the curve given by the parametric equations x = 1 + 4 sin t, y = 2 cos t for
0 ø t ø 2π by finding the values of x and y for particular values of t.

t 0 1 π 1 π 3 π π 5 π 3 π 7 π 2π
4 2 4 4 2 4
Review 1
x 1 3.83 5 3.83 1 −1.83 −3 −1.83 2

y 2 1.41 0 −1.41 −2 −1.41 0 1.41

Plotting the coordinates on a grid gives the following graph.

y
3

2

1

–4 –3 –2 –1 O 1 2 3 4 5 6x
–1
103
–2

–3

Review EXPLORE 4.4

Review 1 Use tables of values for t, x and y to sketch each of the following curves.
Use graphing software to check your answers.
a x = t3, y = t2
b x = t2, y = t3 − t
c x = t + 1, y = t − 1
tt

2 Use graphing software to draw the curves given by these parametric equations.

a x = sin t, y = sin 2t
b x = sin 2t, y = sin 3t

c x = sin 4t, y = sin 3t

3 The curves in question 2 are called Lissajous curves. Use the internet to find out
more about these curves. (A Lissajous curve is shown at the start of this chapter.)

When a curve is given in parametric form, in terms of the parameter t, we can use the

chain rule to find dy in terms of t.
dx

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WORKED EXAMPLE 4.14

The parametric equations of a curve are x = t2, y = 5 − 2t.

a Find dy in terms of the parameter t.
dx

b Find the equation of the tangent to the curve at the point where t = 2.

Answer

a x = t2   ⇒     dx = 2t
dt

Review y = 5 − 2t   ⇒      dy = −2
dt

dy = dy  ×  dt Chain rule.
dx dt dx

 = −2 ×   1
2t

 = − 1
t

b When t = 2, gradient = − 21,  x = 4 and y = 1.

Using y − y1 = m(x − x1)

104  y −1 = − 1 (x − 4)
2

   y = − 1 x + 3
2

Review WORKED EXAMPLE 4.15

The parametric equations of a curve are x = 1 + 2 sin2 θ, y = 4 tanθ for π , θ , 3π .
2 2
dy
a Find dx in terms of the parameter θ .

b Find the coordinates of the point on the curve where the tangent is parallel to the y-axis.

Answer

a x = 1 + 2 sin2 θ   ⇒     dx = 4 sinθ cosθ
dθ

y = 4 tanθ    ⇒     dy = 4 sec2 θ = 4
dθ cos2 θ
Review
dy = dy  ×  dθ Chain rule.
dx dθ dx

= 4 θ   ×   4 sin 1 cosθ
cos2 θ

    = sinθ 1
cos3 θ

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b Tangent is parallel to the y-axis when:

  sinθ cos3 θ = 0

 sinθ = 0  or  cosθ = 0 

 sinθ = 0  ⇒  θ = π  ⇒  (1, 0)

There are no solutions to cosθ = 0 in the range π , θ , 3π .
2 2

∴ The point where the tangent is parallel to the y-axis is (1, 0).

Review EXERCISE 4G

1 For each of the following parametric equations, find dy in terms of the given
parameter. dx

a x = 2t3, y = t2 − 5 b x = 2 + sin 2θ , y = 4θ + 2 cos 2θ

c x = 2θ − sin 2θ, y = 2 − cos 2θ d x = 3tanθ , y = 2 sin 2θ

e x = 1 + tanθ , y = cos θ f x = cos 2θ − cosθ , y = sin2 θ

g x = 1 + 2 sin2 θ , y = 4 tanθ h x = 2 + e−t, y = et − e−t

i x = e2t, y = t3et + 1 j x = 2 ln(t + 3) , y = 4et 105
k x = ln(1 − t), y = 5 l x = 1 + t , y = 2 ln t

t

Review 2 The parametric equations of a curve are x = 3t, y = t3 + 4t2 − 3t. Find the two
values of t for which the curve has gradient 0.

3 The parametric equations of a curve are x = 2 sinθ , y = 1 − 3 cos 2θ . Find the

exact gradient of the curve at the point where θ = π .
3
4   for
4 The parametric equations of a curve are x = 2 + ln(t − 1), y =t+ t t . 1.

Find the coordinates of the only point on the curve at which the gradient is equal

to 0.

5 The parametric equations of a curve are x = e2t, y = 1 + 2tet.  Find the equation
of the normal to the curve at the point where t = 0.

6 The parametric equations of a curve are x = e2t, y = t2e−t − 1.

a Show that dy = t(2 − t ) .
dx 2e3t
Review
b Show that the tangent to the curve at the point (1, −1) is parallel to the x-axis
and find the exact coordinates of the other point on the curve at which the

tangent is parallel to the x-axis.

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7 A curve is defined by the parametric equations x = tanθ, y = 2 sin 2θ, for
π
0 , θ , 2 .

a Show that dy = 4 cos2 θ(2 cos2 θ − 1).
dx

b Hence, find the coordinates of the stationary point.

8 The parametric equations of a curve are x = t + 4 ln t, y = t + 9 , for t . 0.
t

a Show that dy = t2 − 9 .
dx t2 + 4t

b The curve has one stationary point. Find the y-coordinate of this point and

Review determine whether it is a maximum or a minimum point.

9 The parametric equations of a curve are x = 1 + 2 sin2 θ, y = 1 + 2 tanθ.  Find

the equation of the normal to the curve at the point where θ = π .
4

10 The parametric equations of a curve are x = 2 sinθ + cos 2θ, y = 1 + cos 2θ, for

0 ø θ ø π .
2

a Show that dy = 2 sinθ .
dx 2 sinθ − 1
b Find the coordinates of the point on the curve where the tangent is parallel to

the x-axis.

106 c Show that the tangent to the curve at the point  3 ,  3  is parallel to the
y -axis.  2 2 

ReviewPS 11 The parametric equations of a curve are x = ln(tan t), y = 2 sin 2t, for

0 , t , π . WEB LINK
2
Try the Parametric
a Show that dy = sin 4t. points resource on
dx the Underground
Mathematics website.
b Hence show that at the point where x = 0 the tangent is parallel to the

x -axis.

Review

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Product rule 107

● d ( uv ) = u dv + v du
dx dx dx

Quotient rule

d  u  = v  du  − u  dv
dx v dx dx
v2
●

Exponential functions

Review ● d (ex ) = ex d  eax+b  = ae ax + b d [ef(x) ] = f ′(x) × e f( x )
dx dx dx

Logarithmic functions

● d (ln x) = 1 d [ln(ax + b)] = a d [ln(f(x))] = f ′(x)
dx x dx ax + b dx f(x)

Trigonometric functions

● d (sin x) = cos x d [sin(ax + b)] = a cos(ax + b)
dx dx

● d (cos x) = −sin x d [cos(ax + b)] = − a sin(ax + b)
dx dx

● d (tan x) = sec2 x d [tan(ax + b)] = a sec2 (ax + b)
dx dx

Review

Review

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END-OF-CHAPTER REVIEW EXERCISE 4

1 The parametric equations of a curve are

x = 1+ ln(t − 2), y = t + 9, for t . 2.
t

i Show that dy = (t2 − 9)(t − 2). [3]
dx t2 [3]

ii Find the coordinates of the only point on the curve at which the gradient is equal to 0.

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q4 November 2010

2 Find the value of dy when x = 4 in each of the following cases:
dx
Review
i y = x ln(x − 3), [4]

ii y = x − 1 . [3]
x + 1

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q5 June 2011

3 The parametric equations of a curve are x = e3t, y = t2et + 3.

i Show that dy = t(t + 2) . [4]
dx 3e2t [2]

ii Show that the tangent to the curve at the point (1, 3) is parallel to the x-axis.

iii Find the exact coordinates of the other point on the curve at which the tangent is parallel [2]
to the x-axis.

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 November 2011
108 4 Find the gradient of each of the following curves at the point for which x = 0.

i y = 3sin x + tan 2x [3]
[3]
Review ii y = 1 6 Cambridge International AS & A Level Mathematics 9709 Paper 21 Q2 June 2014
+ e2x

5 The equation of a curve is 2x2 + 3xy + y2 = 3.

i Find the equation of the tangent to the curve at the point (2, −1), giving your answer in the form [6]
ax + by + c = 0, where a, b and c are integers.

ii Show that the curve has no stationary points. [4]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 June 2014

6 The equation of a curve is y3 + 4xy = 16.

i Show that dy = − 3 4y 4x . [4]
dx y2 + [2]

ii Show that the curve has no stationary points.

iii Find the coordinates of the point on the curve where the tangent is parallel to the y-axis. [4]

Review Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 June 2015

7 The equation of a curve is y = 6 sin x − 2 cos 2x.

Find the equation of the tangent to the curve at the point  1 π, 2  . Give the answer in the
 6 

form y = mx + c, where the values of m and c are correct to 3 significant figures. [5]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q3 June 2015

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8y

A

D B
O x

Review C

The parametric equations of a curve are x = 6 sin2 t, y = 2 sin 2t + 3 cos 2t, for 0 ø t , π. The curve crosses the

x-axis at the points B and D and the stationary points are A and C , as shown in the diagram.

i Show that dy = 2 cot 2t − 1. [5]
dx 3

ii Find the values of t at A and C , giving each answer correct to 3 decimal places. [3]

iii Find the value of the gradient of the curve at B. [3]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 November 2015

9 The equation of a curve is 3x2 + 4xy + y2 = 24. Find the equation of the normal to the curve [8]
at the point (1, 3), giving your answer in the form ax + by + c = 0 where a, b and c are integers.

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q6 November 2016 109

Review 10 y

P

–1 O 1x

The diagram shows the curve y =  1 − x   .
 1 + x 

i By first differentiating 1 − x , obtain an expression for dy in terms of x. Hence show that the gradient of
1 + x dx

the normal to the curve at the point (x,  y) is (1 + x)  (1 − x2 ). [5]

Review ii The gradient of the normal to the curve has its maximum value at the point P shown in the diagram.

Find, by differentiation, the x-coordinate of P. [4]

Cambridge International A Level Mathematics 9709 Paper 31 Q9 June 2010

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11 The parametric equations of a curve are x = 3(1 + sin2 t), y = 2 cos3 t.

Find dy in terms of t, simplifying your answer as far as possible. [5]
dx

Cambridge International A Level Mathematics 9709 Paper 31 Q2 November 2011

12 The equation of a curve is ln(xy) − y3 = 1.

i Show that dy = y − 1) . [4]
dx x(3y3 [4]

ii Find the coordinates of the point where the tangent to the curve is parallel to the y-axis,

Review giving each coordinate correct to 3 significant figures.

Cambridge International A Level Mathematics 9709 Paper 31 Q7 November 2012

13 The curve with equation 6e2x + key + e2y = c, where k and c are constants, passes through the point P with
coordinates (ln 3, ln 2).

i Show that 58 + 2k = c. [2]

ii Given also that the gradient of the curve at P is −6, find the values of k and c. [5]

Cambridge International A Level Mathematics 9709 Paper 31 Q5 June 2011

110

Review

Review

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111

Review Chapter 5
Integration

In this chapter you will learn how to:

■ extend the idea of ‘reverse differentiation’ to include the integration of e ax + b , 1 b ,
sin(ax + b), cos(ax + b) and sec2(ax + b) ax +

use trigonometrical relationships in carrying out integration

■■P2 understand and use the trapezium rule to estimate the value of a definite integral.

Review

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PREREQUISITE KNOWLEDGE

Where it comes from What you should be able to do Check your skills

Chapter 4 Differentiate standard functions. 1 Find dy when:
dx
Review Chapter 4 Use standard trigonometric relationships to
prove trigonometric identities. a y = 3sin 2x − 5 cos x
b y = e5x−2

c y = ln(2x + 1)

2 Prove that
cos 4x + 4 cos 2x ≡ 8 cos4 x − 3.

Review Why do we study integration? REWIND

In this chapter we will learn how to integrate exponential, logarithmic and trigonometrical This chapter builds
functions and how to apply integration to solve problems. on the work we did in
Pure Mathematics 1
It is important that you have a good grasp of the work covered in Chapter 3 on Coursebook,
trigonometry, as you will be expected to know and use the trigonometrical relationships Chapter 9, where we
when solving integration problems. learnt how to integrate
(ax + b)n, where n ≠ −1.
∫b
WEB LINK
P2 Lastly, we will learn how to find an estimate for f(x) dx using a numerical method. This
Explore the Calculus
a of trigonometry and
logarithms station
112 b on the Underground
Mathematics website.
∫is used when we are not able to find the value of f(x) dx using an algebraic method.
a REWIND

5.1 Integration of exponential functions In Chapter 4, we learnt
the following rules
Since integration is the reverse process of differentiation, the rules for integrating exponential for differentiating
functions are: exponential functions:

KEY POINT 5.1 ∫ eax+b dx = 1 eax+b + c d (ex ) = ex
a dx
∫ ex dx = ex + c d (eax+b ) = aeax+b
dx
WORKED EXAMPLE 5.1

Find: ∫b e−2x dx ∫c e5x−4 dx

∫a e3x dx

Review Answer ∫b e−2x dx  = 1 e−2x +c ∫c e5x−4 dx = 1 e5x−4 + c
−2 5
∫a e3x dx = 1 e3x + c
3

= − 1 e−2x + c
2

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∫2 Substitute limits.

Evaluate 9e3x−1 dx.

1

Answer

∫2 9e3x−1 dx =  9 e3x −1 2
1  3 1

= (3e5 ) − (3e2 )

Review   = 3e2 (e3 − 1)

WORKED EXAMPLE 5.3

y y = x2 + e–2x

O 12 x

Find the area of the shaded region.

113

Answer

Review ∫Area = 2 (x2 + e−2x ) dx
1

=  x3 − 1 e−2x 2 Substitute limits.
 3 2 1 Simplify.

=  8 − 1 e−4  −  1 − 1 e−2 
 3 2   3 2 

= 7 + 1 e−2 − 1 e−4
32 2

≈ 2.39 units2

EXERCISE 5A

Review 1 Find: ∫b e−4x dx ∫c 6e3x dx
∫e 2e−x dx ∫f e2x+4 dx
∫a e2x dx ∫h 6e2−3x dx ∫i 2e8x−3 dx

∫d 4e 1 x dx
2

∫g e3x−1 dx

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2 Find: ∫b 5ex (2 + e3x ) dx ∫c (e2x − 1)2  dx

∫a e−x(1 + ex ) dx

∫d 4 + e2x dx ∫e 8 − ex dx ∫f (ex − 2)2 dx
e2x 2e2x e2x

3 Evaluate: 1 ∫ ln 2

∫2 ∫b 2 e4x dx c 5e−2x dx
0
a e3x dx 0
Review ∫e 1 8 dx
0 0  e2 x −1 ∫1

∫d 3 e1+2x  dx f (ex + 1)2  dx
0
0
∫1
∫h 1 2ex − 5 2 dx ∫i2 (ex + 1)2 dx
g (ex + e2x )2 dx 0 ex  0 e2x

0

4 A curve is such that dy = 6e2x + 2e−x. Given that the curve passes through the point (0, 2), find the equation
dx

of the curve.

5 A curve is such that d2 y = 20e−2x.  Given that dy = −8 when x = 0 and that the curve passes through the
dx2 dx

point  1,  5  , find the equation of the curve.
 e2 

114 6 Find the exact area of the region bounded by the curve y = 1 + e2x−5, the x-axis and the lines x = 1 and
x = 3.

Review 7 1x

y y = 2e 2 – 2x + 3

O 3x

Find the exact area of the shaded region.

8 a Show that d (xex − ex ) = xex.
dx

by y = xex

Review O 3x

Use your result from part a to evaluate the area of the shaded region.

∫a

9 a Find (4e−2x + 5e−x ) dx, where a is a positive constant.

0

∫∞

b Hence find the value of (4e−2x + 5e−x ) dx.

0

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PS 10 y

M

Ox

The diagram shows the curve y = 2ex + 8e−x − 7 and its minimum point M.
Find the area of the shaded region.

P PS 11 y y = ex

Review 3
2

O ln 2 ln 3 x

The diagram shows the graph of y = ex. The points (ln 2, 2) and (ln 3, 3) lie on the curve.

∫ ln 3 ∫b 3 = ln 27  .
Hence show that 2 ln y dy 4e 
a Find the value of ex dx.

ln 2

5.2 Integration of 1 b
ax +

Since integration is the reverse process of differentiation, the rules for integration are: REWIND 115

KEY POINT 5.2 In Chapter 4, we learnt

Review the following rules

∫1 = + ∫1 = 1 + + c,  ax +b.0 for differentiating
x ax + a
dx ln x c,  x . 0 b dx ln(ax b) logarithmic functions:

d (ln x) = 1 , x . 0
dx x

WORKED EXAMPLE 5.4 d [ ln(ax + b) ] =

dx ax

Find each of these integrals and state the values of x for which the integral is valid. + = a , ax +b.0
ax + b

∫a 2 dx ∫b 4 1 dx ∫c 2 6 dx It is important to
x 2x + − 3x remember that ln x is
only defined for x . 0.
Answer

2 dx = 2 1 dx
xx
∫ ∫a

 = 2 ln x + c,  x . 0

4 1
2x + 1 2x +
Review ∫ ∫b
dx = 4 1 dx

= 4  1 ln(2x + 1) + c Valid for 2x + 1 . 0.
 2

= 2 ln(2x + 1) + c,  x . − 1 
2

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∫ ∫c 6 1
2 − 3x dx = 6 2 − 3x dx

= 6  1  ln(2 − 3x) + c Valid for 2 − 3x . 0.
−3

= −2 ln(2 − 3x) + c,  x , 2
3

EXPLORE 5.1

Review ∫Fiona is asked to find x−1 dx.

∫She tries to use the formula xn dx = n 1 1 x n +1 + c to obtain her answer.
+

∫Fiona is also asked to find (2x + 3)−1 dx.

∫She tries to use the formula (ax + b)n dx = 1 (ax + b)n+1 +c to obtain her
a(n + 1)
answer.

Discuss with your classmates why Fiona’s methods do not work.

116 EXPLORE 5.2

Review ∫Rafiu and Fausat are asked to find 1 1) dx .
2(3x +

1 1
2(3x + 6x +
∫ ∫Rafiu writes: dx = dx
1) 2

= 1 ln(6x + 2) + c
6

1 1 1
2(3x + 1) 2 3x +
∫ ∫Fausat writes: dx = 1 dx

=  1  1 ln(3x + 1) + c
 2  3

= 1 ln(3x + 1) + c
6

Decide who is correct and discuss the reasons for your decision with your classmates.

ReviewConsider the integrals 3 1 dx and −2 1 dx.  y y= 1
2 x
∫ ∫2 x −3 x
1
From the symmetry properties of the graph of y = 1 we can see that the shaded areas
x –3 –2 –1 O
–1
that represent the two integrals are equal in magnitude. However, one of the areas is –2 1 2 3x

∫ ∫below the x-axis, which suggests that −2 1 dx = − 3 1 dx.
−3 x 2x

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∫Evaluating 3 1 dx gives:
2x

∫3 1 dx =  ln x 23 = ln 3 − ln 2 = ln 3
2 x 2

∫This implies that −2 1 dx = −ln 3 = ln 2 . WEB LINK
−3 x 23 A more in-depth
explanation for
∫If we try using integration to find the value of −2 1 dx we obtain: this formula can be
−3 x found in the Two
for one resource on
∫ −2 1 dx =  ln x −−32 = ln(−2) − ln(−3) = ln 2 the Underground
−3 x 3 Mathematics website.
Review
There is, however, a problem with this calculation in that ln x is only defined for x . 0 so TIP
ln(−2) and ln(−3) do not actually exist. It is normal practice
to only include the
∫Hence for x , 0, we say that 1 dx = ln x + c. modulus sign when
x finding definite
integrals.
In conclusion, the integration formulae are written as:

KEY POINT 5.3 ∫1 dx = 1 ln ax + b +c
ax + b a
∫ 1 dx = ln x + c
x

WORKED EXAMPLE 5.5 117

Review ∫Find the value of 3 2 6 dx.
2 − 3x

Answer

∫3 6 dx =   6  ln 2 − 3x 3 Substitute limits.
2 2 − 3x  −3 2 Simplify.

= (−2 ln 7) − (−2 ln 4)

= −2 ln 7 + 2 ln 4

= 2 ln 4
7

EXERCISE 5B

Review 1 Find: ∫b 1 dx ∫c 1 1 dx
2x 3x +
∫a 6 dx
x

∫d 6 dx ∫e 5 dx ∫f 3 1) dx
2x − 5 2 − 3x 2(5x −

2 Evaluate:

∫a 5 x 1 2 dx ∫b 4 1 1 dx ∫c4 3 5 dx
0 + 1 2x + −1 2x +

∫d 4 2 dx ∫e 2 3 7 dx ∫f3 3 4 dx
2 4x + 1 0 2x − 2 − 2x

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3 Evaluate:

∫a 10  2 + 5 2  dx
4  3x − 

∫b 3  3 − 4  dx
1  x 2x − 1

∫c 3 2x − 1+ 4 3  dx
−1  2x + 

4 a Given that 4x ≡ 2 + A , find the value of the constant A.
2x − 1 2x − 1
Review
∫b Hence show that 5 4x dx = 8 + ln 9.
1 2x − 1

5 a Find the quotient and remainder when 6x2 + 5x is divided by 2x − 5.

∫b Hence show that 1 6x2 + 5x dx = 23 − 25 ln 5  .
0 2x − 5 2 3

6 A curve is such that dy = 2x + x 3 e .
dx +

Given that the curve passes through the point (e, e2 ), find the equation of the curve.

118 PS 7 y

Review O1 kx

The diagram shows part of the curve y = 2 . Given that the shaded region has area 4, find the value of k.
x+ 3

PS 8 The points P(1, −2) and Q(2, k) lie on the curve for which dy = 3 − 2  . The tangents to the curve at the points
dx x

P and Q intersect at the point R. Find the coordinates of R.

5.3 Integration of sin(ax + b), cos(ax + b) and sec2 (ax + b)

In Chapter 4, we learnt how to differentiate some trigonometric functions:

Review d (sin x) = cos x d [sin(ax + b)] = a cos(ax + b)
dx dx
d
d (cos x) = −sin x dx
dx d [cos(ax + b)] = −a sin(ax + b)
dx
d (tan x) = sec2 x
dx dx
d [tan(ax + b)] = a sec2 (ax + b)
dx

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KEY POINT 5.4 TIP

∫Since integration is the reverse process of differentiation, the rules for integrating are: It is important to
remember that
∫ cos x dx = sin x + c ∫ ∫cos(ax + b) dx = 1 sin(ax + b) + c the formulae for
a differentiating and
integrating these
∫ sin x dx = −cos x + c ∫ ∫sin(ax + b) dx = − 1 cos(ax + b) + c trigonometric
a functions only apply
when x is measured in
∫ sec2 x dx = tan x + c ∫ sec2 (ax + b) dx = 1 tan(ax + b) + c radians.
a

Review WORKED EXAMPLE 5.6

Find: ∫b cos 3x dx ∫c 4 sec2 x  dx
∫b cos 3x dx = 1 sin 3x + c 2
∫a sin 2x dx
3 ∫ ∫c 4 sec2 x  dx = 4 sec2 x  dx
Answer 22

∫a sin 2x dx = − 1 cos 2x + c
2

= 4 ×  2 tan x  + c
 2 

 = 8 tan x + c 119
2

Review WORKED EXAMPLE 5.7

π

∫Find the exact value of 4 (3 − 2 sin 2x) dx.
0

Answer

ππ
∫4  2 4
0 (3 − 2 sin 2x) dx =  3x + 2 cos 2x 0 Substitute limits.

  =  3π + cos π  − (0 + cos 0 )
 4 2 

  =  3π + 0  − (0 + 1)
 4 

Review   = 3π − 1
4

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EXERCISE 5C

1 Find: ∫b cos 4x dx ∫c sin x dx
∫e 5 cos 3x dx 2
∫a sin 3x dx ∫h 3sin(2x + 1) dx
∫d 3sin 2x dx ∫f sec2 2x dx
∫g 2 cos(1 − 5x) dx ∫i 2 sec2(5x − 2) dx

Review 2 Evaluate: ∫b 2 π  1  ∫ 1π
3  2 
∫ 1π sin x dx c 6 sec2 2x dx
0
a 6 cos 4x dx 0

0 ∫ 1π 1
∫π
∫ 1π e 6 (cos 2x + sin x) dx f 4 (5 − 2 sec2 x) dx
1
d 2 (sin 2x + cos x) dx 0 −π
4
0

3 a Find d (x sin x + cos x).
dx

∫ 1π

b Hence find the exact value of 3 x cos x dx.

0

120 4 A curve is such that dy = 1 − 3sin 2x.
dx

Review Given that the curve passes through the point  π , 0  , find the equation of the curve.
 4 

5 A curve is such that d2 y = −12 sin 2x − 2 cos x.
dx2

Given that dy = 4 when x= 0 and that the curve passes through the point  π , −3  , find the equation of
dx  2 

the curve.

6 The point  π , 5    lies on the curve for which dy = 4 sin  2x − π .
 2  dx  2

a Find the equation of the curve.

b Find the equation of the normal to the curve at the point where x = π .
3

PS 7 y

Review Ox

The diagram shows part of the curve y = 1 + 3 sin 2x + cos 2x. Find the exact value of the area of the shaded
region.

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PS 8 y M

O πx

The diagram shows part of the curve y = 3sin 2x + 6 sin x and its maximum point M.
Find the exact area of the shaded region.

P PS 9y

Review √3 y = sin x
2

1
2

O π πx

63

The diagram shows part of the graph of y = sin x. The points  π ,  1  and  π ,  3    lie on the curve.
 6 2   3 2 

∫a π

Find the exact value of 3 sin x dx.
π

6

3 121
∫ ( )b 3 −1.
Hence show that 2 (sin−1 y) d y = π 2 3 −1 − 2
1 12

2

ReviewP PS 10 y

√2 y = cos x

2
1
2

O ππ x

43

The diagram shows part of the graph of y = cos x. The points  π ,  2 and  π ,  1    lie on the curve.
 4 2   3 2 

∫a π

Find the exact value of 3 cos x dx.
π

4

Review ∫b 2 (cos−1 y) d y = π (3 2 − 4) + 3− 2.
Hence show that 2 24 2
1

2

5.4 Further integration of trigonometric functions

Sometimes, when it is not immediately obvious how to integrate a trigonometric function,
we can use trigonometric identities to rewrite the function in a form that we can then
integrate.

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To integrate sin2 x and cos2 x we must use the double angle identities:

cos 2x ≡ 1 − 2 sin2 x and cos 2x ≡ 2 cos2 x − 1
Rearranging these two identities gives:

sin2 x ≡ 1 (1 − cos 2x) and cos2 x ≡ 1 (1 + cos 2x)
22

WORKED EXAMPLE 5.8

Review Find: ∫b 3 cos2 2x dx

∫a sin2 x dx Use sin2 x ≡ 1 (1 − cos 2x).
2
Answer
Use cos2 2x ≡ 1 (1 + cos 4x).
∫ ∫a sin2 x dx = 1 (1 − cos 2x) dx 2
2

 = 1  x − 1 sin 2x  +c
2  2 

 = 1 x − 1 sin 2x + c
24

∫ ∫b 3 cos2 2x dx = 3 (1 + cos 4x) dx
2

= 3  x + 1 sin 4x  + c
2  4 
122
= 3 x + 3 sin 4x + c
28

Review WORKED EXAMPLE 5.9

∫Find sin4 x dx.

Answer

sin4 x = (sin2 x)2 Use sin2 x ≡ 1 (1 − cos 2x).
2
=  1 (1 − cos 2x ) 2
 2  Use cos2 2x ≡ 1 (1 + cos 4x).
2
= 1 (1 − 2 cos 2x + cos2 2x)
4

= 1  1 − 2 cos 2x + 1 + 1 cos 4x 
4  2 2 
Review
= 3 − 1 cos 2x + 1 cos 4x
82 8

∫ ∫sin4 x dx =  3 − 1 cos 2x + 1 cos 4x  dx
 8 2 8 

     = 3 x − 1 sin 2x + 1 sin 4x + c
84 32

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To integrate tan2 x we use the identity 1 + tan2 x ≡ sec2 x.
Rearranging this identity gives tan2 x ≡ sec2 x − 1.

WORKED EXAMPLE 5.10

π

∫Find the exact value of 4 5 tan2 x dx.
0

Answer

Review ππ Use tan2 x ≡ sec2 x − 1.
Substitute limits.
∫ ∫4 5  tan2 x dx = 4 (5 sec2 x − 5) dx
00

 = − π

 5 tan x 5x  4
0

 =  5 − 5π  − (0 − 0)
 4 

 = 5 − 5π
4

Worked example 5.11 uses the principle that integration is the reverse process of differentiation.

WORKED EXAMPLE 5.11

123

Review a Show that if y = sec x then dy = tan x sec x.
dx

∫ 1π

b Hence find the exact value of 4 (cos 2x + 5 tan x sec x) dx.

0

Answer

a y = 1 Use the quotient rule.
cos x
∫Use tan x sec x dx = sec x.
dy = (cos x) (0) − (1) (−sin x)
dx cos2 x Substitute limits.

 = sin x
cos2 x

 = tan x sec x

∫ 1π

b 4 (cos 2x + 5 tan x sec x) dx

0

Review =  1 sin 2x + 5 sec x  1π
 2  4
0

=  1 + 5 2  − (0 + 5)
 2 

=5 2 − 9
2

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WORKED EXAMPLE 5.12

a Prove that 2 cosec 2x tan x ≡ sec2 x.

1
∫π
b Hence find the exact value of 3 (5 + cosec 2x tan x) dx.
1
−π
3

Answer

a 2 cosec 2x tan x ≡ 2 tan x Use tan x = sin x   and
sin 2x cos x

Review   ≡ 2 sin x sin 2x = 2 sin x cos x.
(2 sin x cos x) cos x Simplify.

           ≡ 1 Use part a.
cos2 x Simplify.

  ≡ sec2 x

1π 1π  1 
∫ ∫b 3 + = 3  + 2 sec2 
−1π (5 cosec 2x tan x) dx −1π 5 x dx

33

=  5x + 1 tan x 1 π
 2 1
3 π
− 3

=  5π + 3  −  − 5π − 3
 3 2   3 2 
124
  = 10π + 3
3

Review WORKED EXAMPLE 5.13

a Use the expansions of cos(3x − x) and cos(3x + x) to show that:

cos 2x − cos 4x ≡ 2 sin 3x sin x

∫b Hence show that 1π sin 3x sin x dx = 1.
4

04

Answer

a cos 2x − cos 4x ≡ cos(3x − x) − cos(3x + x)

≡ (cos 3x cos x + sin 3x sin x) − (cos 3x cos x − sin 3x sin x)

≡ 2 sin 3x sin x

1π 1 1π

4 sin 3x 4 (cos 2x − cos 4x) dx
∫ ∫b =
Review sin x dx Use part a.
0 20 Simplify.

= 1  1 − 1 1 π
2  2 4
sin 2x sin 4x 4
0

= 1  1 − 0  − (0 − 
2   2  0) 

=1
4

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EXERCISE 5D ∫b 4 cos2  x  dx ∫c sin2 3x dx
2 ∫f cos4 x dx
1 Find:
∫e 6 tan2 (3x) dx
∫a 3 cos2 x dx
∫d 2 tan2 x dx

Review 2 Find the value of: b ∫ 1π3 3 tan2 x dx ∫ 1π
1π
∫ 1π 6 c 3 cos2 x dx

a 3 sin2 x dx 0

0 ∫ 1π

∫ 1π ∫ 1π f 6 tan2 2x dx

d 8 cos2 2x dx e 6 4 sin2 x dx 0

0 0

3 Find the value of:

∫a 1 π  cos2 x − 1  ∫ 1π ∫ 1π
6  cos2  dx
x b 6 (cos2 x + sin 2x) dx c 4 (2 sin x + cos x)2 dx
0
0 0

∫ 1π ∫e 1 π 1 + cos4 x ∫f 1 π 5
6 cos2 x 6 cos
d 3 (cos x − 3 sin x)2 dx dx 1 + 4x dx
0 0
0

4 a By differentiating cos x , show that if y = cot x then dy = −cosec2 x. 125
sin x dx

∫b Hence, show that 1π = 3.
Review 4 cosec2 2x dx 6
1π
6

5 a Show that tan x + cot x ≡ 2 .
sin 2x

1
∫b π 8
Hence show that 2 tan x + cot x dx = 1.
1
π
3

6 a Use the expansions of sin(5x + 2x) and sin(5x − 2x) to show that:

sin 7x + sin 3x ≡ 2 sin 5x cos 2x

∫b Hence show that 1 2 sin 5x cos 2x dx = 11 − 3 3.
π 42

3
1

π
6

Review 7 a Show that sin 3x ≡ 3sin x − 4 sin3 x.

∫b Hence show that 1π 5 .

3 2 sin3 x dx =
0 12

8 a Show that cos 3x ≡ 4 cos3 x − 3 cos x.

1
∫b Hence show that π 17
6 (4 cos3 x + 2 cos x) dx = .

06

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PS 9 The diagram shows part of the curve y = cos x + 2 sin x. Find the exact volume y
of the solid formed when the shaded region is rotated through 360° about the

x-axis.

PS 10 a Show that sec2 x + sec x tan x ≡ 1. O 1π x
sin x 2

1−

b Show that if y = sec x then dy = sec x tan x.
dx

∫c 1π 4 dx = 2(1 + 3 ).
Review Hence show that 6 1 − sin 2x
0

P2 5.5 The trapezium rule y
y = f(x)
We already know that the area, A, under the curve y = f(x) between x = a and x = b
A
∫b
Oa bx
can be found by evaluating f(x) dx. Sometimes we might not be able to find the value of

a

∫b
f(x) dx by the integration methods that we have covered so far or maybe the function
a
cannot be integrated algebraically.

In these situations an approximate answer can be found using a numerical method called
the trapezium rule.

126

This numerical method involves splitting the area under the curve y = f(x) between
x = a and x = b into equal width strips.

y
Review
y = f(x)
y0 y1 y2 y3 y4 y5 y6 x

hhhhhh

Oa b

The area of each trapezium-shaped strip can be found using the formula:

area = 1 (sum of parallel sides) × width
2

ReviewThe sum of the areas of all the strips gives an approximate value for the area under the

b

∫curve and hence an approximate value for f(x) dx.
a

For the previous diagram, there are 6 strips, each of width h, where h = b − a and the
6

length of the vertical edges (ordinates) of the strips are y0,   y1,   y2,   y3,   y4,   y5   and   y6.

The sum of the areas of the trapezium-shaped strips is:

h ( y0 + y1 ) + h ( y1 + y2 ) + h ( y2 + y3 ) + h ( y3 + y4 ) + h ( y4 + y5 ) + h ( y5 + y6 )
2 2 2 2 2 2

= h [ y0 + y6 + 2( y1 + y2 + y3 +  y4 + y5 )]
2

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When there are n strips and the ordinates are y0,   y1,   y2,   y3,  … ,   yn−1,   yn then the sum
of the areas of the strips is:

h ( y0 + y1 ) + h ( y1 + y2 ) + h ( y2 + y3 ) + … + h ( yn−2 + yn −1 ) + h ( yn−1 + yn )
2 2 2 2 2

b

∫Hence we can find an approximate value for f (x) dx using:
a

KEY POINT 5.5 TIP

∫b f(x) dx ≈ h [ y0 + yn + 2( y1 + y2 + y3 + … + yn −1 )] where h= b − a. • The width of the
a 2 n strip is the interval
along the x-axis.
Review An easy way to remember this rule in terms of the ordinates is:
• The number of
∫b ordinates is one
f (x) dx ≈ half width of strip × (first + last + twice the sum of all the others). more than the
a number of strips.

WORKED EXAMPLE 5.14 • You should not use
a numerical method
The diagram shows the curve y = x − 2 ln x. y when an algebraic
method is available
Use the trapezium rule with 2 intervals to estimate the to you, unless you
are specifically
shaded area, giving your answer correct to 2 decimal asked to do so.

places. State, with a reason, whether the trapezium rule

gives an under-estimate or an over-estimate of the true 127

value. O 0.5 3.5 x
Answer
Review
a = 0.5, b = 3.5 and h = 1.5

y

O 0.5 2 3.5 x

x 0.5 2 3.5
y 1.8863 0.6137 0.9945

y0 y1 y2

Review Area ≈ h [ y0 + y2 + 2 y1]
2

≈ 1.5 [1.8863 + 0.9945 + 2 × 0.6137]
2

≈ 3.08115 …

≈ 3.08 (to 2 decimal places)

It can be seen from the diagram that this is an over-estimate since the top edges of the
strips all lie above the curve.

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WORKED EXAMPLE 5.15

The diagram shows the curve y = 2 4x − x2 .  y

Use the trapezium rule with 4 intervals to estimate the O 4x

∫4

value of 2 4x − x2 dx giving your answer correct

0

to 2 decimal places. State, with a reason, whether the

trapezium rule gives an under-estimate or an over-

estimate of the true value.

Review Answer y
a = 0, b = 4 and h = 1

x0 1 23 4
4 23 0
y 0 23 y2 y3 y4

y0 y1 O 1 2 3 4x

∫4 4x − x2   dx ≈ h [ y0 + y4 + 2( y1 + y2 + y3 )]
2 2
0

( )≈1
2  0 + 0+2 2 3 +4+2 3 

≈4 3+4

128 ≈ 10.93 (to 2 decimal places)

Review It can be seen from the diagram that this is an under-estimate since the top edges of
the strips all lie below the curve.

Worked example 5.14 involves a ‘concave’ curve and the diagram shows that the trapezium DID YOU KNOW?
rule gives an over-estimate for the area.
If you double the
Worked example 5.15 involves a ‘convex’ curve and the diagram shows that the trapezium number of strips in
rule gives an under-estimate for the area. a trapezium rule
approximation, the
If a curve is partially convex and partially concave over the required interval it is not so error reduces to
easy to predict whether the trapezium rule will give an under-estimate or an over-estimate approximately one
for the area. quarter the previous
error.
The more strips that are used, the more accurate the estimate will be.

EXERCISE 5E

Review 1 Use the trapezium rule with 2 intervals to estimate the value of each of these
definite integrals. Give your answers correct to 2 decimal places.

∫4 ∫2 ∫c4 3 2 dx
0 + ex
a x2 − 2 dx b ex − 12 dx

2 0

∫d 7 2 5) dx ∫e 1π 2 + 1 x dx ∫ 12
1 ln(x + 3 tan
0 f log10 x dx

2

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2y

Ox

The diagram shows part of the curve y = x2e−x.  Use the trapezium rule with

∫5

4 intervals to estimate the value of x2e−x dx, giving your answer correct to

1

2 decimal places.

3 a Use the trapezium rule with 6 intervals to estimate the value of

Review ∫7π cosec  1 x  dx, giving your answer correct to 2 decimal places.
4  2 
1π
4

b Use a sketch of the graph of y = cosec  1 x  for 0 ø x ø 2π, to explain
 2 

whether the trapezium rule gives an under-estimate or an over-estimate of the

∫true value of 7π cosec  1 x  dx.
4  2 
1π
4

4y

129

Review O 1 π x
2

The diagram shows the curve y = ex cos x for 0 ø x ø 1 π.  Use the trapezium

∫1π 2

rule with 3 intervals to estimate the value of 2 ex cos x dx, giving your answer

0

correct to 2 decimal places. State, with a reason, whether the trapezium rule

1π

∫gives an under-estimate or an over-estimate of the true value of 2 ex cos x dx.
0

5y

Review O 1 2 3x WEB LINK

The diagram shows part of the curve y = ex . Use the trapezium rule with 4 Try the When does the
2x trapezium rule give
the exact answer here?
intervals to estimate the area of the shaded region, giving your answer correct to resource on the
2 decimal places. Underground
Mathematics website.
State, with a reason, whether the trapezium rule gives an under-estimate or an
over-estimate of the true value of the shaded area.

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Checklist of learning and understanding

Integration formulae ∫

∫● ex dx = ex + c ∫ ∫eax+b dx = 1 eax+b + c
a

∫● 1 dx = ln x + c ∫ ∫1 b dx = 1 ln ax + b +c
x ax + a
∫1 1
● cos x dx = sin x + c ∫ cos(ax + b) dx = 1 sin(ax + b) + c
a

Review ∫● sin x dx = −cos x + c ∫ ∫sin(ax + b) dx = − 1 cos(ax + b) + c
∫● sec2 x dx = tan x + c a

∫ sec2(ax + b) dx = 1 tan(ax + b) + c
a

P2 The trapezium rule

b

∫● The trapezium rule can be used to find an approximate value for f(x) dx. If the region under
a

the curve is divided into n strips each of width h, then:

b b−
n
∫ f(x) dx ≈ h [ y0 + yn + 2( y1 + y2 + y3 + … + yn −1 )] where h = a .
2

a

130

Review

Review

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END-OF-CHAPTER REVIEW EXERCISE 5

∫1 1π (1 + 2 tan2 x) dx 43 − π.
Show that 3 = 3 6 [4]
1π
6

∫2 −2  4 − 2x  dx, giving your answer in the form + ln b  , where a, b and
Find the exact value of −4  1 − 2x  a c 

c are integers. [4]

P2 3 y

Review R

O 1x [4]

The diagram shows part of the curve y = 2 − x2 ln(x + 1). The shaded region R is bounded by the curve
and by the lines x =  0, x = 1 and y = 0. Use the trapezium rule with 4 intervals to estimate the area
of R giving your answer correct to 2 decimal places. State, with a reason, whether the trapezium rule
gives an under-estimate or an over-estimate of the true value of the area of R.

∫ 1π [5]
[5]
4 Find the exact value of 6 (cos 3x − sin 2x) dx.

0

∫5 1 9 23 . 131

Show that (3ex − 2)2 dx = e2 − 12e +
0 22

Review ∫6 a Find k (5e−2x + 2e−3x ) dx, where k is a positive constant. [4]
0 [1]
[5]
∫∞

b Hence find the exact value of (5e−2x + 2e−3x ) dx.

0

∫7 4 3 = 3 ln 21 .
Show that 2 5x + 1 dx 5 11

8 a Find the value of the constant A such that 8x 5 ≡ 4 + A 5 . [2]
2x + 2x +

∫b 3 8x = − 11.
Hence show that 1 2x + 5 dx 8 10 ln 7 [5]
[3]
9 i Show that 12 sin2 x cos2 x ≡ 3 (1 − cos 4x).
2

∫ii 1π π 33
3 12 sin2 cos2 x dx = 8 + 16
Hence show that 1π x . [3]

Review 4

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q3 June 2013

∫10 a Find 4ex (3 + e2x ) dx. [3]

∫b 1π 1
4 2
Show that −1 (3 + 2 tan2 θ ) dθ = (8 + π). [4]

4 π

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q6 June 2011

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11 i Show that (2 sin x + cos x)2 can be written in the form 5 + 2 sin 2x − 3 cos 2x. [5]
ii 22 [4]

∫ 1π

Hence find the exact value of 4 (2 sin x + cos x)2 dx .

0

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 June 2012

12 i By differentiating cos x , show that if y = cot x then dy = −cosec2 x. [3]
sin x dx

Review ii By expressing cot2 x in terms of cosec2 x and using the result of part i, show

1
∫that π 1
2 cot2 x dx = 1− 4 π. [4]
1 [3]
π
4
iii 1 can be expressed as 1 cosec2 x.
Express cos 2x in terms of sin2 x and hence show that 1 − cos 2x 2

∫Hence, using the result of part i, find 1− 1 dx.
cos 2x

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q8 June 2010

∫13 i Find 1 + cos4 2x dx. [3]
cos2 2x

∫132 14  +  6  giving your answer in the form
ii Without using a calculator, find the exact value of 4  2 3x − 2  dx,

ln(aeb ), where a and b are integers. [5]

Review Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 June 2016

14 y

Q

O πx

The diagram shows the curve y = x sin x, for 0 ø x ø π. The point Q  1 π, 1 π  lies on the curve.
 2 2 

Review i Show that the normal to the curve at Q passes through the point (π, 0). [5]
[2]
ii Find d (sin x − x cos x).
dx [3]

∫ 1π

iii Hence evaluate 2 x sin x dx.

0

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q8 November 2010

15 i Using the expansion of cos(3x − x) and cos(3x + x), prove that 1 (cos 2x − cos 4x) ≡ sin 3x sin x. [3]
ii 2 [3]

∫Hence show that 1π sin 3x sin x dx = 1 3.
3 8
1π
6

Cambridge International A Level Mathematics 9709 Paper 31 Q4 June 2010

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133

Review Chapter 6
Numerical solutions of equations

In this chapter you will learn how to:

■ locate approximately a root of an equation, by means of graphical considerations and/or
searching for a sign change

■ understand the idea of, and use the notation for, a sequence of approximations that converges to
a root of an equation

■ understand how a given simple iterative formula of the form xn+1 = F(xn ) relates to the
equation being solved, and use a given iteration, or an iteration based on a given rearrangement
of an equation, to determine a root to a prescribed degree of accuracy.

Review

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PREREQUISITE KNOWLEDGE

Where it comes from What you should be able to do Check your skills
IGCSE / O Level Mathematics
Substitute numbers for letters 1 Evaluate each of the following
in complicated formulae and
algebraic expressions, including expressions when x = 1.5.
those involving logarithmic
functions, exponential a 2  x + 1 
functions and major and minor 3  x2 
trigonometric functions.
b 100 sin x − 99x, (x in radians)

c e−x

Review d ln x + 5

IGCSE / O Level Mathematics Rearrange complicated formulae 2 Rearrange each formula so that
and equations. x is the subject.

a y = 4x2 + 13, given x > 0

b 0 = y − 3x + 7
w

c 3 = 3 1+ y
x5

Pure Mathematics 1 Coursebook, Understand the relationship 3 Show that the x-coordinates of

Chapters 1 and 2 between a graph and its associated the solutions to the simultaneous

algebraic equation, and use the equations:

134 relationship between points y = x2 − 6x + 10
of intersection of graphs and y=x
solutions of equations.

Review also satisfy the equation

x2 − 7x + 10 = 0.

Pure Mathematics 1 Coursebook, Know that the values of x that 4 Find the roots of the equation
Chapter 1 make both sides of an equation x2 − 7x + 10 = 0.
equal are called the roots of the
equation.

IGCSE / O Level Mathematics Also, the roots of f(x) = 0 are the 5 Sketch the graph, y = f(x), of
Pure Mathematics 1 Coursebook, x-intercepts on the graph of each function.
Chapter 5 y = f(x). a f(x) = x3 + 1

Sketch graphs of quadratic b f(x) = cos 2x − 3
or cubic functions or the
trigonometric functions sine,
cosine, tangent.

c f(x) = 4x2 − 16x

Review Chapters 2 and 3 Sketch graphs of exponential 6 Sketch the graph, y = g(x), of
or logarithmic functions or the each function.
trigonometric functions cosecant,
secant, cotangent. a g(x) = ex
2

b g(x) = ln(x − 1)

c g(x) = cosec x

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Review Why solve equations numerically?

You are used to solving equations using direct, algebraic methods such as factorising or 135
using the quadratic formula.
Review
You might be surprised to learn that not all equations can be solved using a direct,
algebraic process.

Numerical methods are ways of calculating approximate solutions to equations. They are
extremely powerful problem-solving tools and many are available. They are widely used in
engineering, computing, finance and many other applications.

A well-programmed computer is quickly able to solve a complicated equation using a
numerical method. So why do we need to understand how the method works?

Knowing how a numerical method works helps us to:

● understand when the results are likely to be reasonable
● understand how to use available software correctly
● select an appropriate method when choices are available
● write our own programs when we need to do so, if we have the programming skills!

To be a powerful problem-solving tool, the method must be used correctly. This might
mean the method needs to work quickly or work to obtain a particular level of accuracy.
When used incorrectly, numerical methods might be very slow or even fail to work, as we
will see later in this chapter.

In this chapter we will focus on one numerical method, the numerical solution of an
equation using an iterative formula.

6.1 Finding a starting point

As we will see throughout this chapter, a good starting point is very important.

We need a method to find an approximate solution to an equation first.

We might find this using, for example, a graphical approach or the change of sign approach.

For suitable functions, each of these approaches will produce two values: one above and
one below the solution that we are finding.

WORKED EXAMPLE 6.1

By sketching graphs of y = x3 and y = 4 − x, show that the equation x3 + x − 4 = 0
has a root α between 1 and 2.

Answer

y y = x3
4
Review y =4–x

O 12 4x TIP

Make sure your sketch
shows all the key
features of each graph,
such as the coordinates
of turning points and
the coordinates of any
intercepts.

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There is one point of intersection and its x-coordinate is between 1 and 2.
At this point x3 = 4 − x.
This means that the equation x3 = 4 − x has one root between 1 and 2.
Rearranging this equation gives x3 + x − 4 = 0.
Hence x3 + x − 4 = 0 has a root, α , between 1 and 2.

WORKED EXAMPLE 6.2

Review Show, by calculation, that the equation f(x) = x5 + x − 1 = 0 has a root α between 0 and 1.

Answer Why? Think about it and check the
Find the value of α such that f(α ) = 0. explanation that follows.

f(0) = 05 + 0 − 1 = −1 As α is between 0 and 1, then f(0) might have
f(1) = 15 + 1 − 1 = 1 the opposite sign to f(1).

The change of sign indicates the presence of a root, This conclusion is important as it completes
136 so 0 , α , 1. the argument.

Review To see why the change of sign method works in this case, look at the graph of y = f(x).

y
f(1) is positive

O 1x

There are no breaks
f(0) is negative in the graph:

it is continuous.

Since it starts below the x-axis at x = 0 and finishes above the x-axis at x = 1, it must cross the x-axis somewhere
in between, say at x = α, where 0 , α , 1. So α is a root of the equation, that is, f(α ) = 0.

Review

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WORKED EXAMPLE 6.3

a By sketching a suitable pair of graphs, show that the equation cos x = 2x − 1 (where x is in radians) has

only one root for 0 ø x ø 1 π.
2

b Verify by calculation that this root lies between x = 0.8 and x = 0.9.

Answer Why are these graphs suitable?

Review a Draw the graphs of y = cos x and y = 2x − 1.

y
y = 2x – 1

1

y = cos x

O 0.5 πx
2

–1

The graphs intersect once only and so the equation

cos x = 2x −1 has only one root for 0øxø 1 π. 137
2

b cos x = 2x − 1 so cos x − 2x + 1 = 0

Review Let f(x) = cos x − 2x + 1 and so f(x) = 0 then
f(0.8) = cos 0.8 − 2(0.8) + 1 = 0.0967…
f(0.9) = cos 0.9 − 2(0.9) + 1 = −0.1783…

Change of sign indicates the presence of a root.

Review EXERCISE 6A

1 a On the same axes, sketch the graphs of y = 1 + x and y = x2.
b Using your answer to part a, explain why the equation x2 − 1 + x = 0 has two
roots.
c Show, by calculation, that the smaller of these two roots lies between x = −1
and x = 0.

2 a By sketching graphs of y = x3 + 5x2 and y = 5 − 2x, determine the number
of real roots of the equation x3 + 5x2 + 2x − 5 = 0.

b Verify by calculation that the largest root of x3 + 5x2 + 2x − 5 = 0 lies
between x = 0 and x = 2.

3 The curve y = x3 + 5x − 1 cuts the x-axis at the point (α, 0).
a By sketching the graph of y = x3 and one other suitable graph, deduce that
this is the only point where the curve y = x3 + 5x − 1 cuts the x-axis.
b Using calculations, show that 0.1 , α , 0.5.

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Review 4 a On the same axes, sketch the graphs of y = ln(x + 1) and y = 3x − 4.
b Hence deduce the number of roots of the equation ln(x + 1) − 3x + 4 = 0.

5 a By sketching graphs of y = ex and y = x + 6, determine the number of roots
of the equation ex − x − 6 = 0.

b Show, by calculation, that ex − x − 6 = 0 has a root that lies between
x = 2.0 and x = 2.1.

6 a Show, by calculation, that (x + 2)e5x = 1 has a root between x = 0 and
x = −0.2.

b Show, by sketching the graph of y = e5x and one other suitable graph, that
this is the only root of this equation.

7 a Show by calculation that the equation cos−1 2x = 1 − x has a root α , where
0.4 ø α ø 0.5.

b The diagram shows the graph of y = cos−1 2x where −0.5 ø x ø 0.5.

y
π

3π
4

π
2

π
138 4

Review –1 –0.5 O 0.5 1 x

By drawing a suitable graph on a copy of the diagram, show that α is the only
root of cos−1 2x = 1 − x for −0.5 ø x ø 0.5.

8 The curve with equation y = sin x where x,− π intersects the line y = 1 at
2x + 3 2
the point (α, 1).

a By sketching the graph of y = sin x for −2π , x , − π radians and one
2
other suitable graph, deduce that this is the only point where the curve

y = sin x intersects the line y = 1.
2x + 3

b Using calculations, show that −2 , α , − 1.9.

ReviewPS 9 a By sketching a suitable pair of graphs, show that the equation
x3 + 4x = 7x + 4 has only one root for 0 ø x ø 5.

b Verify by calculation that this root lies between x = 2 and x = 3.

PS 10 a Show graphically that the equation 2x = x + 4 has exactly two roots. REWIND
b Show, by calculation, that the larger of the two roots is between 2.7 and 2.8.
You might need to look
PS 11 a By sketching two suitable graphs on the same diagram, show that the back at Chapter 3
before attempting
equation cot x = x2 has one root between 0 and π radians. Question 11.
2

b Show, by calculation, that this root, α , lies between 0.8 and 1.

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