PAGE PROOFS Integral calculus - Wiley · 8 Integral calculus 8.1 Kick off with CAS 8.2 Areas under...

8Integral calculus

8.1 Kick off with CAS

8.2 Areas under and between curves

8.3 Linear substitutions

8.4 Other substitutions

8.5 Integrals of powers of trigonometric functions

8.6 Integrals involving inverse trigonometric functions

8.7 Integrals involving partial fractions

8.8 Review

c08IntegralCalculus.indd 378 14/08/15 4:31 PM

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ROOFS

8.1 Kick off with CASExploring integration with CAS

In this topic, we will be integrating various types of functions.

1 Use CAS to � nd each of the following.

a 3(3x + 4)5dx b 31

(3x + 4)2 dx c 3

1

"3x + 4 dx d 3

13x + 4

dx

Can you predict 3(ax + b)ndx? What happens when n = –1?


a 31

"4 − 9x2 dx b 3

1

"9 − 16x2 dx c 3

1

"16 − 25x2 dx

Can you predict 31

"a2 − b2x2 dx, where a and b are positive constants?


a 31

4 + 9x2 dx b 3

19 + 16x2

dx c 31

16 + 25x2 dx

Can you predict 31

a2 + b2x2 dx, where a and b are positive constants?


a 31

4 − 9x2 dx b 3

19 − 16x2

dx c 31

16 − 25x2 dx

Can you predict 31

a2 − b2x2 dx, where a and b are positive constants?

Please refer to the Resources tab in the Prelims section of your eBookPLUS for a comprehensive step-by-step guide on how to use your CAS technology.


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ROOFS

Areas under and between curvesArea bounded by a curve and the x-axisBasic integration techniques and evaluating areas bounded by curves and the x-axis, have been covered in the Mathematical Methods course. The examples and theory presented here are a review of this material.

Recall that the de� nite integral, A = 3

b

a

f(x)dx, gives a

measure of the area A bounded by the curve y = f(x), the x-axis and the lines x = a and x = b. This result is known as the Fundamental Theorem of Calculus.

a b0

yy = f(x)

x

8.2

Find the area bounded by the line y = 2x + 3, the x-axis and the lines x = 2 and x = 6.

THINK WRITE/DRAW

1 Draw a diagram to identify the required area and shade this area.

68

42

–2–4–6

2 4 121086

1012141618

–2–4–6 0

y

y = 2x + 3

x

2 The required area is given by a de� nite integral.

A = 3

6

2

(2x + 3)dx

3 Perform the integration using square bracket notation.

A = cx2 + 3x d6

2

4 Evaluate. A = (62 + 3 × 6) − (22 + 3 × 2)

= (36 + 18) − (4 + 6)

= 44

5 State the value of the required area region. The area is 44 square units.

6 The shaded area is a trapezium. As a check on the result, � nd the area using the formula for a trapezium.

The width of the trapezium is h = 6 − 2 = 4,and since y = 2x + 3:when x = 2, y = 7 and when x = 6, y = 15.

WORKED EXAMPLE 111

380 MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4


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Using symmetrySometimes symmetry can be used to simplify the area calculation.

68

42

–2–4–6

2 44

121086

1012141618

–2–4–6 0

y

y = 2x + 3

x

77

15

The area of a trapezium is h2

(a + b).

The area is 42(7 + 15) = 44 square units

Find the area bounded by the curve y = 16 − x2, the x-axis and the lines x = ±3.

THINK WRITE/DRAW

1 Factorise the quadratic to � nd the x-intercepts.

y = 16 − x2

y = (4 − x)(4 + x)The graph crosses the x-axis at x = ±4 and crosses the y-axis at y = 16.


68

42

–2–4

1 2 6543

1012141618

–5–6 –1–2–4 –3 0

y

x

y = 16 – x2

3 The required area is given by a de� nite integral; however, we can use symmetry.

A = 3

3

−3

(16 − x2)dx

= 3

0

−3

(16 − x2)dx + 3

3

0

(16 − x2)dx

WORKED EXAMPLE 222

Topic 8 INTEGRAL CALCULUS 381


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ROOFS

Areas involving basic trigonometric functionsFor integrals and area calculations involving the basic trigonometric functions,

we use the results 3cos(kx)dx = 1k

sin(kx) + c and 3sin(kx)dx = −1k

cos(kx) + c,

where x ∈ R, k ≠ 0, and k and c are constants.

Find the area under one arch of the sine curve y = 5 sin(3x).

THINK WRITE/DRAW


y = 5 sin(3x) has an amplitude of 5 and a period

of 2π3

.

34

21

–1–2–3–4–5–6

56

0

y

xπ–3

2π––3

π

y = 5 sin(3x)

2 One arch is de� ned to be the area under one half-cycle of the sine wave.

The graph crosses the x-axis at sin(3x) = 0,

when x = 0, π3

and 2π3

. The required area is

A = 3

π3

0

5 sin(3x)dx

WORKED EXAMPLE 333

However, 3

0

−3

(16 − x2)dx = 3

3

0

(16 − x2)dx

A = 23

3

0

(16 − x2)dx


A = 2 c16x − 13x3 d

3

0

5 Evaluate. A = 2 c a16 × 3 − 13(3)3b − 0 d

A = 2(48 − 9)

6 State the value of the required area. The area is 78 square units.



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ROOFS


A = 5 c−13

cos(3x) dπ3

0

4 Evaluate, taking the constant factors outside the brackets.

A = −53[cos(π) − cos(0)]

= −53[−1 − 1]

5 State the value of the required area. The area is 103

square units.

Areas involving basic exponential functionsFor integrals and area calculations involving the basic exponential functions, we use

the result 3ekxdx = 1k

ekx + c, where x ∈ R, k ≠ 0, and k and c are constants.

Find the area bounded by the coordinate axes, the graph of y = 4e−2x and the line x = 1.

THINK WRITE/DRAW


2

1

–1

1 32

3

4

5

–1 0

y

x

y = 4e–2x

2 The required area is given by a de� nite integral.

A = 3

1

0

4e−2xdx


A = 4 c−12e−2x d

1

0

= −2 ce−2x d1

0

4 Evaluate. A = −2[e−2 − e0] = −2(e−2 − 1)

5 State the value of the required area. The exact area is 2(1 − e−2) square units.

WORKED EXAMPLE 444



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Areas involving signed areasWhen evaluating a de� nite integral, the result is a number; this number can be positive or negative. A de� nite integral which represents an area is a signed area; that is, it may also be positive or negative. However, areas cannot be negative.

Areas above the x-axisWhen a function is such that f(x) ≥ 0 for a ≤ x ≤ b, where b > a, that is, the function lies above the x-axis, then the de� nite integral that represents the area A is positive:

A = 3

b

a

f(x)dx > 0.

Areas below the x-axisWhen a function is such that f(x) ≤ 0 for a ≤ x ≤ b, where b > a, that is, the function lies below the x-axis, then the de� nite integral that represents the area A is negative:

3

b

a

f(x)dx < 0.

So, when an area is determined that is bounded by a curve that is entirely below the x-axis, the result will be a negative number. Because areas cannot be negative, the absolute value of the integral must be used.

A = 33

b

a

f(x)dx 3 = −3

b

a

f(x)dx = 3

a

b

f(x)dx

a b0

y

x

y = f(x)

a b0

y

x

y = f(x)

Find the area bounded by the curve y = x2 − 4x + 3 and the x-axis.

THINK WRITE/DRAW


y = x2 − 4x + 3y = (x − 3)(x − 1)

The graph crosses the x-axis at x = 1 and x = 3 and crosses the y-axis at y = 3.

2 Sketch the graph, shading the required area.

2

1

–1

1 3 4 52

3

4

5

–1 0

y

x

WORKED EXAMPLE 555



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Areas both above and below the x-axisWhen dealing with areas that are both above and below the x-axis, each area must be evaluated separately.

Since A1 = 3

b

a

f(x)dx > 0 and

A2 = 3

c

b

f(x)dx < 0, the required area is

A = A1 + ∣A2∣ = 3

b

a

f(x)dx + 33

c

b

f(x)dx 3

= A1 − A2 = 3

b

a

f(x)dx − 3

c

b

f(x)dx = 3

b

a

f(x)dx + 3

b

c

f(x)dx

b ca0

A1

A2

y

x

y = f(x)

3 The required area is below the x-axis and will evaluate to a negative number. The area must be given by the absolute value or the negative of this de� nite integral.

A = 33

3

1

(x2 − 4x + 3)dx 3

= −3

3

1

(x2 − 4x + 3)dx


A = − c13x3 − 2x2 + 3x d

3

1

5 Evaluate the de� nite integral. A = − c a13

× 33 − 2 × 32 + 3 × 3b − a13

× 13 − 2 × 12 + 3 × 1b d = 4

3

6 State the value of the required area.

The area is 43 square units.

Find the area bounded by the curve y = x3 − 9x and the x-axis.

THINK WRITE/DRAW

1 Factorise the cubic to � nd the x-intercepts. y = x3 − 9xy = x(x2 − 9)y = x(x + 3)(x − 3)

The graph crosses the x-axis at x = 0 and x = ±3.

WORKED EXAMPLE 666



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Area between curvesIf y1 = f(x) and y2 = g(x) are two continuous curves that do not intersect between x = a and x = b, then the area between the curves is obtained by simple subtraction.

2 Sketch the graph, shading the required area.

68

42

–2–4–6–8

–10–12

1 2 543

1012

–5 –1–2–4 –3 0

y

x

y = x3– 9x

3 The required area is given by a definite integral.

If we work out A = 3

3

−3

(x3 − 9x)dx it comes to

zero, as the positive and negative area have cancelled out.

Let A1 = 3

0

−3

(x3 − 9x)dx and A2 = 3

3

0

(x3 − 9x)dx,

so that A1 > 0 and A2 < 0, but A1 = ∣A2 ∣ by symmetry.

4 Perform the integration, using square bracket notation.

A1 = 3

0

−3

(x3 − 9x)dx

A1 = c14x

4 − 92x

2 d−3

0

5 Evaluate the definite integral. A1 = c a0 − 14

× (−3)4 − 92

× (−3)2b d

A1 = 814

= 2014

A2 = 3

3

0

(x3 − 9x)dx = −814

= −2014

A1 + ∣A2 ∣ = 2 × 814

= 812

= 4012

6 State the value of the required area. The area is 4012 square units.

a b0

y

x

y1 = f(x)

y2 = g(x)

386 MAths Quest 12 sPeCIALIst MAtheMAtICs VCe units 3 and 4


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ROOFS

The area A1 is the entire shaded area bounded by the curve y2 = g(x), the x-axis and

the lines x = a and x = b, so A1 = 3

b

a

g(x)dx. The red area, A2, is the area bounded by

the curve y1 = f(x), the x-axis and the lines x = a and x = b, so A2 = 3

b

a

f(x)dx.

The required area is the blue area, which is the area between the curves.

A = A1 − A2 = 3

b

a

g(x)dx − 3

b

a

f(x)dx, and by the properties of de� nite integrals

A = A1 − A2 = 3

b

a

(g(x) − f(x))dx = 3

b

a

(y2 − y1)dx.

Note that when � nding areas between curves, it does not matter if some of the area is above or below the x-axis.

We can translate both curves up k units parallel to the y-axis so that the area between the curves lies entirely above the x-axis as shown below right.

A = 3

b

a

(g(x) + k)dx − 3

b

a

( f(x) + k)dx

A = 3

b

a

(g(x) − f(x))dx

A = 3

b

a

( y2 − y1)dx

Provided that y2 ≥ y1 for a < x < b, it does not matter if some or all of the area is above or below the x-axis, as the required area between the curves will be a positive number. Note that only one de� nite integral is required, that is y2 − y1 = g(x) − f(x). Evaluate this as a one de� nite integral.

ba

0

y

x

y1 = f(x)

y2 = g(x)

ba 0

y

x

y1 = f(x) + k

y2 = g(x) + k

Find the area between the parabola y = x2 − 2x − 15 and the straight line y = 2x − 3.

THINK WRITE/DRAW


y = x2 − 2x − 15y = (x − 5)(x + 3)The parabola crosses the x-axis at x = 5 and x = −3 and crosses the y-axis at y = −15.The straight line crosses the x-axis at x = 3

2 and

crosses the y-axis at y = −3.

WORKED EXAMPLE 777

AOS 3

Topic 2

Concept 9

Areas of bounded regionsConcept summaryPractice questions



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2 Find the x-values of the points of intersection between the parabola and the straight line.

Let y1 = x2 − 2x − 15 and y2 = 2x − 3. To find the points of intersection, solve y1 = y2.

x2 − 2x − 15 = 2x − 3 x2 − 4x − 12 = 0(x − 6)(x + 2) = 0 x = 6, −2

3 Sketch the graph of the parabola and the straight line on one set of axes, shading the required area. 6

8

42

–2–4–6–8

–10–12–14–16–18

1 2 6 7543

10

–5 –1–2–4 –3 0

y

x

y = x 2– 2x – 15y = 2x – 3

4 The required area is given by a definite integral.

A = 3

b

a

(y2 − y1)dx with a = −2, b = 6,

y1 = x2 − 2x − 15 and y2 = 2x − 3.

y2 − y1 = −x2 + 4x + 12

A = 3

6

−2

(−x2 + 4x + 12)dx


A = c−13x3 + 2x2 + 12x d

6

−2

6 Evaluate the definite integral. A = c a−13

× 63 + 2 × 62 + 12 × 6b

− aa−13

× (−2)3 + 2 × (−2)2 + 12 × (−2)bb dA = c a−13

× 63 + 2 × 62 + 12 × 6b − a a−13

× (−2)3 + 2 × (−2)2 + 12 × (−2)b b d

A = 8513

7 State the value of the required area between the parabola and the straight line.

The area between the straight line and the parabola is 851

3 square units.

388 Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4


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ROOFS

Areas under and between curves1 WE1 Find the area bounded by the line y = 4x + 5, the x-axis, and the lines

x = 1 and x = 3. Check your answer algebraically.

2 Find the area bounded by the line y = 4 − 3x2

and the coordinate axes. Check your answer algebraically.

3 WE2 Find the area bounded by the curve y = 9 − x2, the x-axis and the lines x = ±2.

4 The area bounded by the curve y = b − 3x2, the x-axis and the lines x = ±1 is equal to 16. Given that b > 3, find the value of b.

5 WE3 Find the area under one arch of the sine curve y = 4 sin(2x).

6 Find the area under one arch of the curve y = 3 cosax2b.

7 WE4 Find the area bounded by the coordinate axes, the graph of y = 6e3x and the line x = 2.

8 Find the area bounded by the graph of y = 6(e−2x + e−2x), the x-axis and x = ±1.

9 WE5 Find the area bounded by the curve y = x2 − 5x + 6 and the x-axis.

10 Find the area bounded by the curve y = 3x2 − 10x − 8 and the x-axis.

11 WE6 Find the area bounded by the curve y = x3 − 4x and the x-axis.

12 Find the area bounded by the curve y = 16x − x3, the x-axis, x = −2 and x = 4.

13 WE7 Find the area between the parabola y = x2 − 3x − 18 and the straight line y = 4x − 10.

14 Find the area corresponding to the region {y ≥ x2 − 2x − 8} ∩ {y ≤ 1 − 2x}.

15 a Find the area between the line y = 6 − 2x and the coordinate axis.Check your answer algebraically.

b Find the area between the line y = 3x + 5, the x-axis, x = 1 and x = 4.Check your answer algebraically.

16 a Calculate the area bounded by:i the curve y = 12 − 3x2 and the x-axisii the curve y = 12 − 3x2, the x-axis and the lines x = ±1.

b Determine the area bounded by:i the graph of y = x2 − 25 and the x-axisii the graph of y = x2 − 25, the x-axis and the lines x = ±3.

17 a Find the area under one arch of the sine curve y = 6 sinaπx3b.

b Find the area under one arch of the curve y = 4 cosaπx2b.

c Find the area under one arch of the sine curve y = a sin(nx).

18 a Find the area under the graph of y = 1x between the x-axis and:

i x = 1 and x = 4 ii x = 1 and x = eiii x = 1 and x = a, where a > e.

ExErcisE 8.2

PractisE

consolidatE

Topic 8 InTegral calculus 389


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ROOFS

b Find the area under the graph of y = 1x2

between the x-axis and:

i x = 1 and x = 3ii x = 1 and x = a, where a > 1.

c Find the area bounded by the curve y = x2 − 2x − 15 and the x-axis.

19 a Find the area bounded by the curve y = x2 + 3x − 18, the x-axis and the lines x = −3 and x = 6.

b Find the area bounded by the curve y = x2 − 2x − 24, the x-axis and the lines x = 2 and x = 8.

c Find the area bounded by:i the curve y = x3 − 36x and the x-axisii the curve y = x3 − 36x, the x-axis and the lines x = −3 and x = 6.

20 a If a is a constant, find the area bounded by the curve y = x2 − a2 and the x-axis.

b If a is a constant, find the area bounded by the curve y = x3 − a2x and the x-axis.

21 Find the area between the curves:

a y = x2 and y = xb y = x3 and y = xc y = x4 and y = xd y = x5 and y = x.

22 a i Find the area between the parabola y = x2 − 2x − 35 and the x-axis.

ii Find the area between the parabola y = x2 − 2x − 35 and the straight line y = 4x − 8.

b i Find the area between the parabola y = x2 + 5x − 14 and the x-axis.ii Find the area corresponding to the region

{y ≥ x2 + 5x − 14} ∩ {y ≤ 2x + 4}.

23 a Find the area between the line 2y + x − 5 = 0 and the hyperbola y = 2x.

b Find the area between the line 9y + 3x − 10 = 0 and the hyperbola y = 13x

.

24 a Prove using calculus methods that the area of a right-angled triangle of base length a and height b is given by 1

2ab.

b Prove using calculus that the area of a trapezium of side lengths a and b, and

width h is equal to h2

(a + b).

25 a Consider the graphs of y = x2

3 and y = 4 sinax

2b.

i Find the coordinates of the point of intersection between the graphs.

ii Determine the area between the graphs, the origin and this point of intersection, giving your answer correct to 4 decimal places.

b Consider the graphs of y = 5e−

x

4 and y = x2

.


ii Determine the area between the curves, the y-axis and this point of intersection, giving your answer correct to 4 decimal places.

Master



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ROOFS

26 a Consider the graphs of y = 23e

x

2 and y = 45 sina2x

3b + 42 for x ≥ 0.


ii Determine the area between the curves, the y-axis and this point of intersection, giving your answer correct to 4 decimal places.

b Consider the graphs of y = 190x2

− 5 and y = −32 cosax5b for x ≥ 0.

i Find the coordinates of the � rst two points of intersection.

ii Determine the area between the curves and these � rst two points of intersection, giving your answer correct to 4 decimal places.

Linear substitutionsA linear substitution is of the form u = ax + b.

Finding integrals of the form 3(ax + b)ndx where n ∈ Z

Integrals of the form 3(ax + b)ndx, where a and b are non-zero real numbers and n is

a positive integer, can be performed using a linear substitution with u = ax + b.

The derivative dudx

= a is a constant, and this constant factor can be taken outside the

integral sign by the properties of inde� nite and de� nite integrals. The integration process can then be completed in terms of u. Note that since u has been introduced in this solution process, the � nal answer must be given back in terms of the original variable, x.

8.3AOS 3

Topic 2

Concept 5

Antiderivatives using linear substitutionConcept summaryPractice questions

Find 3(2x − 5)4dx.

THINK WRITE

1 Although we could expand and integrate term by term, it is preferable and easier to use a linear substitution.

Let u = 2x − 5.

3(2x − 5)4dx = 3u4dx

2 Differentiate u with respect to x. u = 2x − 5

dudx

= 2

3 Express dx in terms of du by inverting both sides. dxdu

= 12

dx = 12

du

4 Substitute for dx. 3u4dx = 3u4 12du

5 Use the properties of inde� nite integrals to transfer the constant factor outside the front of the integral sign.

= 123u4du

WORKED EXAMPLE 888



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ROOFS

Finding particular integrals of the form 3(ax + b)ndx where n ∈ Q

Integrals of the form 3 (ax + b)ndx, where a and b are non-zero real numbers,

and n is a rational number, can also be performed using a linear substitution with u = ax + b. First express the integrand (the function being integrated) as a power using the index laws.

6 Perform the integration using 3undu = 1n + 1

un+1

with n = 4 so that n + 1 = 5, and add in the constant +c.

12

× 15

u5 + c

= 110

u5 + c

7 Substitute back for u = 2x − 5 and express the � nal answer in terms of x only and an arbitrary constant +c.

3(2x − 5)4dx = 110

(2x − 5)5 + c

The gradient of a curve is given by 1!4x + 9

. Find the particular curve that passes through the origin.

THINK WRITE

1 Recognise that the gradient of a curve is given by dy

dx.

dy

dx= 1

!4x + 9

2 Integrate both sides to give an expression for y. y = 31

!4x + 9 dx

3 Use index laws to express the integrand as a function to a power and use a linear substitution.

Let u = 4x + 9.

y = 3(4x + 9)−1

2dx

y = 3u−1

2dx

4 The integral cannot be done in this form, so differentiate.

u = 4x + 9dudx

= 4


= 14

dx = 14

du

6 Substitute for dx. y = 3 u−

12 1

4du


y = 143u

−12du

WORKED EXAMPLE 999



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ROOFS

Finding integrals of the form 3(ax + b)ndx when n = −1

Integrals of the form 3(ax + b)ndx, when n = −1, a ≠ 0 and b ∈ R, involve the

logarithm function, since 31x

dx = loge a ∣x ∣ b + c.

8 Perform the integration process using

3undu = 1n + 1

un+1 with n = −12 so that n + 1 = 1

2,

and add in the constant +c.

y = 14 × 1

12

u12 + c

y = 24u

12 + c

y = 12!u + c

9 Simplify and substitute for u = 4x + 9 to express the answer in terms of x and an arbitrary constant +c.

y = 12!4x + 9 + c

10 The arbitrary constant +c in this particular case can be found using the given condition that the curve passes through the origin.

Substitute y = 0 and x = 0 to � nd c:

0 = 12!0 + 9 + c

c = −32

11 Substitute back for c. y = 12!4x + 9 − 3

2

12 State the equation of the particular curve in a factorised form.

y = 12(!4x + 9 − 3)

Antidifferentiate 15x + 4

.

THINK WRITE

1 Write the required integral. 31

5x + 4 dx

2 Use a linear substitution. Let u = 5x + 4.

31

5x + 4 dx = 3

1u

dx

3 The integral cannot be done in this form, so differentiate. u = 5x + 4dudx

= 5


= 15

dx = 15

du

5 Substitute for dx. 31

5x + 4 dx = 3

1u

× 15

du


= 153

1u du

WORKED EXAMPLE 101010



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ROOFS

Finding integrals of the form 3(ax + b)ndx where n ∈ Q

We can generalise the results from the last three examples to state:

3(ax + b)ndx = •1

a(n + 1)(ax + b)n+1 + c n ≠ −1

1a

loge a ∣ax + b ∣b + c n = −1

Evaluating definite integrals using a linear substitutionWhen we evaluate a de� nite integral, the result is a number. This number is also independent of the original variable used. When using a substitution, change the terminals to the new variable and evaluate this de� nite integral in terms of the new variable with new terminals. The following worked example clari� es this process.

AOS 3

Topic 2

Concept 2

Antiderivates involving logarithmsConcept summaryPractice questions

7 Perform the integration process using 31u

du = loge ∣u∣ + c. 31

5x + 4 dx = 1

5 loge a ∣u ∣ b + c

8 Simplify and substitute for u = 5x + 4 to express the � nal answer in terms of x only and an arbitrary constant +c, in simplest form.

31

5x + 4 dx = 1

5 loge a∣5x + 4∣b + c

Evaluate 3

1

0

4(3x + 2)2

dx.

THINK WRITE

1 Write the integrand as a power using index laws and transfer the constant factor outside the front of the integral sign.

3

1

0

4(3x + 2)2

dx

= 43

1

0

(3x + 2)−2dx

2 Use a linear substitution. Note that the terminals in the de� nite integral refer to x-values.

Let u = 3x + 2.

43

1

0

(3x + 2)−2dx = 4 3

x=1

x=0

u−2dx

3 The integral cannot be done in this form, so differentiate. Express dx in terms of du by inverting both sides.

u = 3x + 2

dudx

= 3

dxdu

= 13

dx = 13

du




UNCORRECTED PAGE P

ROOFS

4 Change the terminals to the new variable. When x = 0 ⇒ u = 2 and when x = 1 ⇒ u = 5.

5 Substitute for dx and the new terminals. 43

u=5

u=2

u−2

13

du

6 Transfer the constant factor outside the front of the integral sign.

= 433

5

2

u−2du

7 The value of this de� nite integral has the same value as the original de� nite integral. There is no need to substitute back for x, and there is no need for the arbitrary constant when evaluating a de� nite integral.

= 43c−1

u d 5

2

8 Evaluate this de� nite integral. = 43c−1

5− a−1

2b d

= 43a1

2− 1

5b

= 43a5 − 2

10b

9 State the � nal result. 3

1

0

4(3x + 2)2

dx = 25

Finding integrals using a back substitution

Integrals of the form 3x(ax + b)ndx can also be performed using a linear substitution

with u = ax + b. Since the derivative dudx

= a is a constant, this constant can be taken

outside the integral sign. However, we must express the integrand in terms of u only

before integrating. We can do this by expressing x in terms of u; that is, x = 1a

(u − b).

However, the � nal result for an inde� nite integral must be given in terms of the original variable, x.

Find:

a 3x(2x − 5)4dx b 36x − 5

4x2 − 12x + 9 dx.

THINK WRITE

a 1 Use a linear substitution. a Let u = 2x − 5.

3x(2x − 5)4dx = 3x u4dx




UNCORRECTED PAGE P

ROOFS

2 The integral cannot be done in this form, so differentiate and express dx in terms of du by inverting both sides.

u = 2x − 5

dudx

= 2

dxdu

= 12

dx = 12

du

3 Express x in terms of u. 2x = u + 5

x = 12(u + 5)

4 Substitute for x and dx. 3x(2x − 5)4dx = 312(u + 5)u4 1

2

du

5 Use the properties of indefinite integrals to transfer the constant factors outside the front of the integral sign and expand the integrand.

= 143(u5 + 5u4)du

6 Perform the integration, integrating term by term and adding in the constant.

= 14

× a16u6 + u5b + c

7 Simplify the result by expanding. = 124

u6 + 14u5 + c

8 Substitute u = 2x − 5 and express the final answer in terms of x only and an arbitrary constant +c.

3x(2x − 5)4dx = 124

(2x − 5)6 + 14(2x − 5)5 + c

9 Alternatively, the result can be expressed in a simplified form by taking out the common factors.

124

u6 + 14

u5 + c = u5

24(u + 6) + c

10 Substitute back for u = 2x − 5 and simplify.

=(2x − 5)5

24 (2x − 5 + 6) + c

11 Express the final answer in terms of x only and an arbitrary constant +c.

3x(2x − 5)4dx = 124

(2x − 5)5(2x + 1) + c

b 1 Factorise the denominator as a perfect square.

b 36x − 5

4x2 − 12x + 9 dx = 3

6x − 5(2x − 3)2

dx

2 Use a linear substitution.

Differentiate and express dx in terms of du by inverting both sides.

Let u = 2x − 3.dudx

= 2

dxdu

= 12

dx = 12

du

3 Express the numerator 6x − 5 in terms of u.

2x = u + 3 6x = 3(u + 3) 6x = 3u + 96x − 5 = 3u + 4



UNCORRECTED PAGE P

ROOFS

4 Substitute for 6x − 5, u and dx. 36x − 5

(2x − 3)2 dx = 3

3u + 4u2

× 12

du


= 123a

3u + 4u2

bdu

6 Simplify the integrand. = 123a

3u

+ 4u2bdu

7 Write in index form. = 123a

3u + 4u−2bdu

8 Perform the integration, adding in the constant. The � rst term is a log, but in the

second term, we use 3undu = 1n + 1

un+1

with n = −2, so that n + 1 = −1.

= 12a3 loge a∣u∣b − 4u−1b + c

= 12a3 loge a∣u∣b − 4

ub + c

9 Substitute u = 2x − 3 and express the � nal answer in terms of x only and an arbitrary constant +c, as before.

36x − 5

4x2 − 12x + 9 dx = 3

2 loge a∣2x − 3∣b − 2

2x − 3+ c

Evaluate 3

8

0

x!2x + 9

dx.

THINK WRITE

1 Write the integrand as a power, using index laws. 3

8

0

x!2x + 9

dx = 3

8

0

x(2x + 9) −12dx

2 Use a linear substitution, Let u = 2x + 9.

3

8

0

x(2x + 9)−1

2dx = 3

x=8

x=0

xu−1

2dx


dudx

= 2

dxdu

= 12

dx = 12

du

4 Express x back in terms of u. u = 2x + 92x = u − 9 x = 1

2(u − 9)

5 Change the terminals to the new variable. When x = 0, u = 9, and when x = 8, u = 25.


Defi nite integrals using a back substitution



UNCORRECTED PAGE P

ROOFS

Further examples involving back substitutions with logarithms

6 Substitute for dx, x and the new terminals. 3

x=8

x=0

xu−1

2dx

= 3

u=25

u=9

12(u − 9)u

−12

12du

7 Transfer the constant factors outside the front of the integral sign.

= 143

25

9

(u − 9)u−1

2du

8 Expand the integrand. = 143

25

9

au12 − 9u

−12bdu

9 Perform the integration. = 14c2

3u

32 − 18u

12 d

25

9

10 Evaluate the de� nite integral. = 14c a2

3(25)

32 − 18(25)

12 b − a2

3(9)

32 − 18(9)

12 b d

= 14c a2

3× 125 − 18 × 5b − a2

3× 27 − 18 × 3b d


8

0

x!2x + 9

dx = 223

Find 32x

4x − 3 dx.

THINK WRITE

Method 1

1 Use a linear substitution. Differentiate and express dx in terms of du by inverting both sides.

u = 4x − 3

dudx

= 4

dxdu

= 14

dx = 14

du

2 Express the numerator 2x in terms of u. 4x = u + 3

2x = 12(u + 3)

3 Substitute for 2x, u and dx. 32x

4x − 3 dx = 3

12(u + 3)

u × 14

du

4 Use the properties of inde� nite integrals to transfer the constant factors outside the front of the integral sign.

= 183a

u + 3u bdu




UNCORRECTED PAGE P

ROOFS

The situation above can often happen when evaluating indefinite integrals. Answers may not appear to be identical, but after some algebraic or trigonometric simplification, they are revealed to be equivalent and may differ by a constant only.

Linear substitutions1 WE8 Find 3(5x − 9)6 dx.

2 Find 3(3x + 4)7dx.

3 WE9 A particular curve has a gradient equal to 1!16x + 25

. Find the particular curve that passes through the origin.

4 Given that f ′(x) = 1(3x − 7)2

and f(2) = 3, find the value of f(1).

5 WE10 Antidifferentiate 13x − 5

.

ExErcisE 8.3

PractisE

5 Simplify the integrand. = 183a1 + 3

ubdu

6 Perform the integration, adding in the constant. = 18

au + 3 loge a∣u∣bb + c

= u8

+ 38

loge a∣u∣b + c

7 Substitute u = 4x − 3 and express the final answer in terms of x only and an arbitrary constant +c.

32x

4x − 3 dx = 4x − 3

8+ 3

8 loge a∣4x − 3∣b + c

Method 2

1 Express the numerator as a multiple of the denominator (in effect, use long division to divide the denominator into the numerator).

32x

4x − 3 dx = 1

234x

4x − 3 dx

= 123

(4x − 3) + 34x − 3

dx

2 Simplify the integrand. = 123a1 + 3

4x − 3bdx

3 Perform the integration, adding in the constant. = 12ax + 3

4 loge a∣4x − 3∣b b + c

4 State the final answer. 32x

4x − 3 dx = x

2+ 3

8 loge a∣4x − 3∣b + c

5 Although the two answers do not appear to be the same, the log terms are identical.

However, since 4x − 38

= x2

− 38

, the two

answers are equivalent in x and differ in the

constant only, c1 = −38

+ c.

= 4x − 38

+ 38

loge a∣4x − 3∣b + c

= x2

+ 38

loge a∣4x − 3∣b + c1



UNCORRECTED PAGE P

ROOFS

6 Find an antiderivative of 17 − 2x

.

7 WE11 Evaluate 3

0

−1

9(2x + 3)3

dx.

8 Find the area bounded by the graph of y = 6

!3x + 4, the coordinate

axes and x = 4.

9 WE12 Find:

a 3x(5x − 9)5dx b 32x − 1

9x2 − 24x + 16 dx.

10 Find:

a 3x

(2x + 7)3 dx b 3

x

!6x + 5 dx.

11 WE13 Evaluate 3

5

0

x

!3x + 1 dx.

12 Evaluate 3

1

−1

15x

(3x + 2)2 dx.

13 WE14 Find 36x

3x + 4 dx.

14 Evaluate 3

1

0

4x2x − 5

dx.

15 Integrate each of the following with respect to x.

a (3x + 5)6 b 1(3x + 5)2

c 1(3x + 5)3

d 1

"3 3x + 5

16 Find each of the following.

a 3(6x + 7)8dx b 31

!6x + 7 dx c 3

16x + 7

dx d 31

(6x + 7)2 dx


a x(3x + 5)6 b x(3x + 5)2

c x(3x + 5)3

d x

"3 3x + 5


a 3x(6x + 7)8dx b 3x

!6x + 7 dx

c 3x

6x + 7 dx d 3

x(6x + 7)2

dx

ConsolidatE



UNCORRECTED PAGE P

ROOFS

19 a Given that dxdt

= 1(2 − 5t)2

and x(0) = 0, express x in terms of t.

b A certain curve has its gradient given by 1!3 − 2x

for x < 32. If the point

a−12 , −2b lies on the curve, find the equation of the curve.

c Given that f ′(x) = 33 − 2x

and that f(0) = 0, find f(1).

d A certain curve has a gradient of x

!2x + 9. Find the particular curve that passes

through the origin.

20 Evaluate each of the following.

a 3

2

1

(3x − 4)5dx b 3

2

1

x(3x − 4)5dx c 3

13

0

1

"3 2t + 1 dt d 3

13

0

t

"3 2t + 1 dt

21 a Sketch the graph of y = !2x + 1. Find the area between the curve, the coordinates axes and the line x = 4.

b Sketch the graph of y = 13x + 5

. Determine the area bounded by the curve, the

coordinate axes and x = 3.

c Sketch the curve y = 1(2x + 3)2

. Find the area bounded by this curve and the

x-axis between x = 1 and x = 2.

d Find the area bounded by the curve y = x

!16 − 3x, the coordinate

axes and x = 5.

22 a Find the area of the region enclosed by the curves with the equations:

i y = 4!x − 1 and y = 4!3 − xii y2 = 16(x − 1) and y2 = 16(3 − x).

b Find the area between the curve y2 = 4 − x and the y-axisc Determine the area of the loop with the equation y2 = x2(4 − x).d i Find the area between the curve y2 = a − x where a > 0 and the y-axis

ii Determine the area of the loop with the equation y2 = x2(a − x), where a > 0.

23 Given that a and b are non-zero real constants, find each of the following.

a 3!ax + b dx b 3x!ax + b dx c 31

ax + b dx d 3

xax + b

dx


a 31

!ax + b dx b 3

x

!ax + b dx c 3

1(ax + b)2

dx d 3x

(ax + b)2 dx

25 Given that a, b, c and d are non-zero real constants, find each of the following.

a 3cx + dax + b

dx b 3cx + d

(ax + b)2 dx c 3

cx2 + d

(ax + b)2 dx d 3

cx2 + dax + b

dx


a 3x2

ax + b dx b 3

x2

(ax + b)2 dx c 3

x2

(ax + b)3 dx d 3

x2

!ax + b dx

Master



UNCORRECTED PAGE P

ROOFS

Other substitutionsNon-linear substitutionsThe basic idea of a non-linear substitution is to reduce the integrand to one of the standard u forms shown in the table below. Remember that after making a substitution, x or the original variable should be eliminated. The integral must be entirely in terms of the new variable u.

8.4

f(u) 3f(u)du

un n ≠ −1un+1

n + 1

1u

loge a∣u∣b

eu eu

cos(u) sin(u)

sin(u) −cos(u)

sec2(u) tan(u)

Find 33x

(x2 + 9)2 dx.

THINK WRITE

1 Write the integrand as a power using index laws.

33x(x2 + 9)−2dx

2 Use a non-linear substitution. Let u = x2 + 9.

33x(x2 + 9)−2dx = 33xu−2dx


dudx

= 2x

dxdu

= 12x

dx = 12x

du

4 Substitute for dx, noting that the terms involving x will cancel.

33xu−2dx = 33xu−2 × 12x

du

5 Transfer the constant factors outside the front of the integral sign.

= 323u−2du

6 The integral can now be done. Antidifferentiate

using 3undu = un+1

n + 1 with n = −2,

so that n + 1 = −1.

= −32

u−1 + c


AOS 3

Topic 2

Concept 3

Integration by substitutionConcept summaryPractice questions



UNCORRECTED PAGE P

ROOFS

Integrals involving the logarithm function

The result 3undu = un+1

n + 1 is true provided that n ≠ −1. When n = −1 we have the

special case 31u

du = loge a∣u∣b + c.

7 Write the expression with positive indices. = − 32u

+ c

8 Substitute back for x, and state the � nal result. 33x

(x2 + 9)2 dx = − 3

2(x2 + 9)+ c

Find 3x − 3

x2 − 6x + 13 dx.

THINK WRITE/DRAW

Method 1

1 Use a non-linear substitution. Let u = x2 − 6x + 13.

dudx

= 2x − 6

= 2(x − 3)

dxdu

= 12(x − 3)

dx = 12(x − 3)

du

2 Substitute for dx and u, noting that the terms involving x cancel.

3x − 3

x2 − 6x + 13 dx = 3

x − 3u × 1

2(x − 3) du

3 Transfer the constant outside the front of the integral sign.

= 123

1u du

4 The integration can now be done.

Antidifferentiate using 31u

du = loge a∣u∣b .

= 12 loge a∣u∣b + c

5 In this case, since x2 − 6x + 13 = (x − 3)2 + 4 > 0, for all values of x, the modulus is not needed. Substitute back for x, and state the � nal result.

Note that since ddx

[loge (

f(x))] =f ′(x)

f(x),

it follows that 3f ′(x)

f(x) dx = loge a∣ f(x)∣b + c.

3x − 3

x2 − 6x + 13 dx = 1

2 loge (x2 − 6x + 13) + c




UNCORRECTED PAGE P

ROOFS

Examples involving trigonometric functions

For trigonometric functions we use the results 3cos(u)du = sin(u) + c and

3sin(u)du = −cos(u) + c.

Method 2

1 To make the numerator the derivative of the denominator, multiply both the numerator and the denominator by 2, and take the constant factor outside the front of the integral sign.

3x − 3

x2 − 6x + 13 dx

= 32(x − 3)

2(x2 − 6x + 13) dx

= 123

2x − 6x2 − 6x + 13

dx

2 Use the result 3f ′ (x)

f(x) dx = loge a∣ f(x)∣b + c,

with f(x) = x2 − 6x + 13.

3x − 3

x2 − 6x + 13 dx = 1

2 loge (x2 − 6x + 13) + c

Find 31x2

cosa1xbdx.

THINK WRITE

1 Use a non-linear substitution. Let u = 1x.

We choose this as the derivative of 1x is − 1

x2,

which is present in the integrand.

u = 1x

= x−1

dudx

= −x−2

= − 1x2

dxdu

= −x2

dx = −x2dx

2 Substitute for u and dx, noting that the x2 terms cancel.

31x2

cosa1xbdx

= 31x2

cos(u) × −x2du

3 Transfer the negative sign outside the integral sign. = −3cos(u)du

4 Antidifferentiate, using 3cos(u)du = sin(u) + c. = −sin(u) + c

5 Substitute back for x and state the answer. 31x2

cosa1xbdx = −sina1

xb + c




UNCORRECTED PAGE P

ROOFS

examples involving exponential functions

For exponential functions we use the result 3eudu = eu + c.

Find:

a 3sin(2x) ecos(2x)dx b 3sin(!x)

!x dx.

tHINK WrIte

a 1 Use a non-linear substitution. Let u = cos(2x). Choose this as the derivative of cos(2x) is −2 sin(2x), which is present in the integrand. Note that the substitution u = sin(2x) will not work.

a u = cos(2x)

dudx

= −2 sin(2x)

dxdu

= −12 sin(2x)

dx = −12 sin(2x)

du

2 Substitute for u and dx. 3sin(2x)ecos(2x)dx

= 3sin(2x)eu −12 sin(2x)

du

3 Transfer the constant factor outside the integral sign. = −123eudu

4 Antidifferentiate using 3eudu = eu + c. = −12eu + c

5 Substitute u = cos(2x) and state the answer. 3sin(2x)ecos(2x)dx = −12ecos(2x) + c

b 1 Use a non-linear substitution. Let u = !x, but only replace u in the numerator.

Express dx in terms of du by inverting both sides.

b u = !x = x 12

dudx

= 12

x−1

2 = 12!x

dxdu

= 2!x

dx = 2!x du

2 Substitute for u and dx, noting that the !x terms cancel.

3sin(!x)

(!x) dx = 3sin(!x) × 1

!x dx

= 3sin(u) × 1!x

dx

3 Transfer the constant factor outside the integral sign. = 23sin(u)du

4 Antidifferentiate using 3sin(u)du = −cos(u) + c. = −2 cos(u) + c

5 Substitute u = !x and state the answer. 3sin(!x)

!x dx = −2 cos(!x) + c

WOrKeD eXaMPle 181818



UNCORRECTED PAGE P

ROOFS

Definite integrals involving non-linear substitutionsWhen evaluating a de� nite integral, recall that the result is a number independent of the original or dummy variable. In these cases, instead of substituting the original variable back into the integral, change the terminals and work with the new de� nite integral obtained.

Evaluate 3

!5

0

t

"t2 + 4 dt.

THINK WRITE

1 Write the integrand as a power using index laws.

3

!5

0

t

"t2 + 4 dt

3

!5

0

t(t2 + 4)−1

2dt

2 Use a non-linear substitution. Let u = t2 + 4.

3

!5

0

t(t2 + 4)−1

2dt = 3

t=!5

t=0

tu−1

2dt

3 The integral cannot be done in this form, so differentiate. Express dt in terms of du by inverting both sides.

dudt

= 2t

dtdu

= 12t

dt = 12t

du

4 Change the terminals to the new variable. When t = 0, u = 4, and when t = !5, u = 9.

5 Substitute for dt and the new terminals, noting that the dummy variable t cancels.

3

t=!5

t=0

tu−1

2dt = 3

u=9

u=4

tu−1

2 12t

du

6 Transfer the multiplying constant outside the front of the integral sign.

= 123

9

4

u−1

2du

7 Perform the integration using 3undu = un+1

n + 1

with n = −12, so that n + 1 = 1

2.

= 12c2u

12 d

9

4

= cu 12 d

9

4




UNCORRECTED PAGE P

ROOFS

Defi nite integrals involving inverse trigonometric functions

Recall that ddx

asin−1axabb = 1

"a2 − x2, d

dxacos−1ax

abb = −1

"a2 − x2 for a > 0 and

∣x∣ < a and ddx

atan−1axabb = a

a2 + x2 for x ∈ R.

8 Evaluate the de� nite integral. = [ !9 − !4]= 3 − 2= 1

9 State the answer. 3

!5

0

t

"t2 + 4 dt = 1

Evaluate 3

12

0

cos−1(2x)!1 − 4x2

dx.

THINK WRITE

1 Use a non-linear substitution. Let u = cos−1(2x).

3

12

0

cos−1 (2x)

"1 − 4x2 dx = 3

x =

12

x=0

u

"1 − 4x2 dx


dudx

= −2

"1 − 4x2

dxdu

= −12"1 − 4x2

dx = −12"1 − 4x2

du

3 Change the terminals to the new variable. When x = 12, u = cos−1(1) = 0, and

when x = 0, u = cos−1(0) = π2

.

4 Substitute for dx and the new terminals, noting that the x terms cancel. Transfer the constant multiple outside the front of the integral sign.

3

x =

12

x = 0

u

"1 − 4x2 dx

= −12 3

u=0

u = π2

1

"1 − 4x2 u"1 − 4x2

du

= −12 3

0

π2

u du




UNCORRECTED PAGE P

ROOFS

Other substitutions

1 WE15 Find 38x

(x2 + 16)3 dx.

2 Find 35x

"2x2 + 3 dx.

3 WE16 Find 3x + 2

x2 + 4x + 29 dx.

4 Find 3x2

x3 + 9 dx.

5 WE17 Find 31x2

sina1xbdx.

6 Find 3x cos(x2)dx.

7 WE18 Find:

a 3cos(3x)esin(3x)dx b 3cos(!x)

!x dx.

8 a Find 3sec2(2x)etan(2x)dx. b Find 3 loge (3x)

4x dx.

9 WE19 Evaluate 3

2!2

0

s

"2s2 + 9 ds.

ExErcisE 8.4

PractisE

5 Using the properties of the definite integral, swap the terminals to change the sign.

3

b

a

f 1 t 2dt = − 3

a

b

f 1 t 2dt

= 123

π2

0

u du

6 Perform the integration. = 12c12u2 d

π2

0

= 14cu2 d

π2

0

7 Evaluate the definite integral. = 14c aπ

2b

2

− 02 d

8 State the answer.3

12

0

cos−1 (2x)

"1 − 4x2 dx = π2

16



UNCORRECTED PAGE P

ROOFS

10 Evaluate 3

1

0

p

(3p2 + 5)2 dp.

11 WE20 Evaluate 3

1

3

0

sin−1(3x)

"1 − 9x2 dx.

12 Evaluate 3

4

0

tan−1ax4b

16 + x2 dx.

For questions 13–17, find each of the indefinite integrals shown.

13 a 3x(x2 + 4)5dx b 3x

x2 + 4 dx

c 3x

(x2 + 9)2 dx d 3

x

"x2 + 9 dx

14 a 3x2

(x3 + 27)3 dx b 3

x2

"x3 + 27 dx

c 3x2

x3 + 8 dx d 3x2(x3 + 8)3dx

15 a 3(x − 2)(x2 − 4x + 13)3dx b 3x − 2

(x2 − 4x + 13)2 dx

c 34 − x

"x2 − 8x + 25 dx d 3

4 − xx2 − 8x + 25

dx

16 a 3e2x

4e2x + 5 dx b 3

e−3x

(2e−3x − 5)2 dx

c 3e−2x

(3e−2x + 4)3 dx d 3

2e2x + 1

(e2x + x)2 dx

17 a 31x

sin(loge (4x))dx b 31x

cos(loge (3x))dx

c 3

sin−1ax2b

"4 − x2 dx d 3

tan−1(2x)

1 + 4x2 dx

18 a A certain curve has a gradient given by x sin(x2). Find the equation of the particular curve that passes through the origin.

b If dy

dx= 1

x2 sec2a1

xb and when x = 4

π, y = 0, find y when x = 3π.

c Given that f ′ (x) = 5 − x

x2 − 10x + 29 and f(0) = 0, find f(1).

d If dy

dx= sin(2x)ecos(2x) and when x = π

4, y = 0, find y when x = 0.

ConsolidatE



UNCORRECTED PAGE P

ROOFS

For questions 19–21, evaluate each of the definite integrals shown.

19 a 3

5

3

3t

"t2 − 9 dt b 3

2

0

x4x2 + 9

dx

c 3

2

1

x(2x2 + 1)2

dx d 3

1

−1

s

"2s2 + 3 ds

20 a 3

5

4

3 − x

x2 − 6x + 34 dx b 3

3

2

2 − x(x2 − 4x + 5)2

dx

c 3

4

1

e!p

!p dp d 3

π8

0

sec2(2θ)e

tan (2θ)

dθ

21 a 3

π4

0

sin(2θ )ecos(2θ)dθ b 3

e2

12

loge (2t)

3t dt

c 3

3π

6π

1x2

cosa1xbdx d 3

4

1

sin!x

!x dx

22 a The graph of y = xx2 − c

has vertical asymptotes at x = ±2. Find the value

of c and determine the area bounded by the curve, the x-axis and the lines x = 3 and x = 5.

b Find the area bounded by the curve y = x cos(x2), the x-axis and the

lines x = 0 and x = !π2

.

c Find the area bounded by the curve y = xe−x2, the x-axis and the

lines x = 0 and x = 2.

d Find the area bounded by the curve y = x

"x2 + 4, the x-axis and the

lines x = 0 and x = 2!3.

23 If a, b ∈ R\{0}, then find each of the following.

a 3x

"ax2 + b dx b 3

x(ax2 + b)2

dx

c 3x(ax2 + b)ndx n ≠ −1 d 3x

ax2 + b dx

24 Deduce the following indefinite integrals, where f(x) is any function of x.

a 3f ′(x)

f(x) dx b 3

f ′(x)

(f(x))2 dx c 3

f ′ (x)

!f(x) dx d 3f ′(x)e

f (x)dx

Master



UNCORRECTED PAGE P

ROOFS

Integrals of powers of trigonometric functionsIntroductionIn this section we examine inde� nite and de� nite integrals involving powers of

trigonometric functions of the form 3sinm(kx)cosn(kx)dx where m, n ∈ J.

Integrals involving sin2(kx) and cos2(kx)The trigonometric double-angle formulas

(1) sin(2A) = 2 sin(A)cos(A)

and

(2) cos(2A) = cos2(A) − sin2(A)

= 2 cos2(A) − 1

= 1 − 2 sin2(A)

are useful in integrating certain powers of trigonometric functions. Rearranging (2) gives

(3) sin2(A) = 12(1 − cos(2A)) and

(4) cos2(A) = 12(1 + cos(2A))

To integrate sin2(kx) use (3); to integrate cos2(kx) use (4).

8.5AOS 3

Topic 2

Concept 6

Antiderivatives with trigonometric identitiesConcept summaryPractice questions

Find:

a 3cos2(3x)dx b 3sin2(3x)cos2(3x)dx.

THINK WRITE

a 1 Use the double-angle formula cos2(A) = 1

2(1 + cos(2A)) with A = 3x.

Transfer the constant factor outside the front of the integral sign.

a 3cos2(3x)dx

= 123(1 + cos(6x))dx

2 Integrate term by term,

using 3cos(kx) = 1k

sin(kx) + c with k = 6.

= 12c x + 1

6 sin(6x) d + c

3 Expand and state the � nal result. 3cos2(3x)dx = x2

+ 112

sin(6x) + c

b 1 Use the double-angle formula 2 sin(A)cos(A) = sin(2A) with A = 3x, so that sin2(A)cos2(A) = 1

4 sin2(2A).

b 3sin2(3x)cos2(3x)dx

= 143(2 sin(3x)cos(3x))2dx

= 143sin2(6x)dx




UNCORRECTED PAGE P

ROOFS

Note that as well as the double-angle formulas, there are many other relationships between trigonometric functions, for example sin2(A) + cos2(A) = 1.

Often answers to integrals involving trigonometric functions can be expressed in several different ways, for example, as powers or multiple angles. Answers derived from CAS calculators may appear different, but often they are actually identical and differ in the constant term only.

Integrals involving sin(kx)cosm(kx) and cos(kx)sinm(kx) where m > 1Integrals of the forms sin(kx)cosm(kx) and cos(kx)sinm(kx) where m > 1 can be performed using non-linear substitution, as described in the previous section.

2 Use the double-angle formula sin2(A) = 1

2(1 − cos(2A)) with A = 6x.

= 143

12(1 − cos(12x))dx

= 183(1 − cos(12x))dx

3 Integrate term by term, using

3cos(kx) = 1k

sin(kx) + c with k = 12.

= 18c x − 1

12 sin(12x) d + c

4 Expand and state the � nal result. 3sin2(3x)cos2(3x)dx = x8

− 196

sin(12x) + c

Find 3sin(3x)cos4(3x)dx.

THINK WRITE

1 Use a non-linear substitution. Let u = cos(3x). We choose this as the derivative of cos(3x) is −3 sin(3x), which is present in the integrand.

u = cos(3x)

dudx

= −3 sin(3x)

dxdu

= −13 sin(3x)

dx = −13 sin(3x)

du

2 Substitute for u and dx, noting that the x terms cancel.

3sin(3x)cos4(3x)dx

= 3sin(3x)u4 × −13 sin(3x)

du

3 Transfer the constant factor outside the integral sign.

= −133u4du

4 Antidifferentiate. = −13

× u5

5+ c

= − 115

u5 + c

5 Substitute back for x and state the � nal result. 3sin(3x)cos4(3x)dx = − 115

cos5(3x) + c




UNCORRECTED PAGE P

ROOFS

Integrals involving even powersAntidifferentiation of sinm(kx)cosn(kx) when both m and n are even powers must be performed using the double-angle formulas.

Integrals involving odd powersAntidifferentiation of sinm(kx)cosn(kx) when at least one of m or n is an odd power can be performed using a non-linear substitution and the formula sin2(A) + cos2(A) = 1.

Find an antiderivative of sin3(2x)cos4(2x).

THINK WRITE

1 Write the required antiderivative. 3sin3(2x)cos4(2x)dx

2 Break the odd power. sin3(2x) = sin(2x)sin2(2x), and sin2(A) = 1 − cos2(A).

3sin(2x)sin2(2x)cos4(2x)dx

= 3sin(2x)(1 − cos2(2x))cos4(2x)dx

3 Use a non-linear substitution. Let u = cos(2x). We choose this as the derivative of cos(2x) is −2 sin(2x), which is present in the integrand and will cancel.

u = cos(2x)

dudx

= −2 sin(2x)

dxdu

= −12 sin(2x)

dx = −12 sin(2x)

du

4 Substitute for u and dx, noting that the sin(2x) terms cancel.

= 3sin(2x)(1 − u2)u4 × −12 sin(2x)

du

5 Transfer the constant factor outside the integral sign and expand.

= −123(1 − u2)u4du

= −123(u4 − u6)du

6 Antidifferentiate. = −12

× a15u5 − 1

7u7b + c

= 114

u7 − 110

u5 + c

7 Substitute back for x and state the � nal result. 3sin3(2x)cos4(2x)dx

= 114

cos7(2x) − 110

cos5(2x) + c


Find an antiderivative of cos4(2x).

THINK WRITE

1 Write the required antiderivative. 3cos4(2x)dx




UNCORRECTED PAGE P

ROOFS

2 Since there is no odd power, we must use the double-angle formula cos2(A) = 1

2(1 + cos(2A)) with A = 2x.

3(cos2(2x))2dx

= 3a12(1 + cos(4x))b

2

dx

3 Expand the integrand = 143(1 + 2 cos(4x) + cos2(4x))dx

4 Replace cos2(A) = 12(1 + cos(2A)) with

A = 4x and expand. The integrand is now in a form that we can integrate term by term.

= 143a1 + 2 cos(4x) + 1

2(1 + cos(8x))bdx

= 143a3

2+ 2 cos(4x) + 1

2 cos(8x)bdx

5 Antidifferentiate term by term. = 14c3x

2+ 2

4 sin(4x) + 1

16 sin(8x) d + c

6 State the � nal result. 3cos4(2x)dx = 3x8

+ 18

sin(4x) + 164

sin(8x) + c

Integrals involving powers of tan (kx)In this section we examine inde� nite and de� nite integrals involving powers of the

tangent function, that is, integrals of the form 3tann(kx)dx where n ∈ J.

The result tan(A) =sin(A)

cos(A) is used to integrate tan(A).

Find 3tan(2x)dx.

THINK WRITE

1 Use tan(A) =sin(A)

cos(A) with A = 2x. 3tan(2x)dx = 3

sin(2x)cos(2x)

dx

2 Use a non-linear substitution. Let u = cos(2x). We choose this as the derivative of cos(2x) is −2 sin(2x), which is present in the numerator integrand and will cancel.

u = cos(2x)

dudx

= −2 sin(2x)

dxdu

= −12 sin(2x)

dx = −12 sin(2x)

du

3 Substitute for u and dx, noting that the x terms cancel.

3tan(2x)dx = 3sin(2x)

u × −12 sin(2x)

du

4 Transfer the constant factor outside the integral sign.

= −123

1u du




UNCORRECTED PAGE P

ROOFS

Integrals involving tan2 (kx)

To � nd 3tan2(kx)dx, we use the results 1 + tan2(A) = sec2(A) and

ddx

(tan(kx)) = k sec2(kx), so that 3sec2(kx) = 1k tan(kx) + c.

5 Antidifferentiate. = −12 loge a∣u∣b + c

6 Substitute back for x and state the � nal result. Again note that there are different answers using log laws and trigonometric identities.

3tan(2x)dx = −12 loge a∣cos(2x)∣b + c

= 12 loge a∣cos(2x)∣b

−1

+ c

= 12 loge a∣sec(2x)∣b + c

Find 3tan2(2x)dx.

THINK WRITE

1 Use 1 + tan2(A) = sec2(A) with A = 2x. 3tan2(2x)dx = 3(sec2(2x) − 1)dx

2 Use 3sec2(kx) = 1k

tan(kx) + c and integrate

term by term. State the � nal result.

3tan2(2x)dx = 12

tan(2x) − x + c


Integrals of powers of trigonometric functions1 WE21 Find:

a 3sin2ax4bdx b 3sinax

4bcosax

4bdx.

2 Evaluate:

a 3

π6

0

4 sin2(2x)dx b 3

π3

0

sin2(2x)cos2(2x)dx.

3 WE22 Find 3cos(4x)sin5(4x)dx.

4 Evaluate 3

π4

0

sin(2x)cos3(2x)dx.

5 WE23 Find an antiderivative of cos5(4x)sin2(4x).

EXERCISE 8.5

PRACTISE



UNCORRECTED PAGE P

ROOFS

6 Evaluate 3

π12

0

sin3(3x)dx.

7 WE24 Find an antiderivative of sin4(2x).

8 Evaluate 3

π12

0

cos4(3x)dx.

9 WE25 Find 3 tanax2bdx.

10 Evaluate 3

π12

0

tan(4x)dx.

11 WE26 Find 3tan2ax3bdx.

12 Evaluate 3

π16

0

tan2(4x)dx.


a 3cos(2x)sin(2x)dx b 3(cos2(2x) + sin2(2x))dx

c 3cos3(2x)sin(2x)dx d 3cos(2x)sin3(2x)dx


a 3

π4

0

cos(2x)sin4(2x)dx b 3

π4

0

cos2(2x)sin3(2x)dx

c 3

π4

0

cos2(2x)sin2(2x)dx d 3

π4

0

cos3(2x)sin2(2x)dx


a 3cos2(4x)sin2(4x)dx b 3cos2(4x)sin3(4x)dx

c 3cos3(4x)sin2(4x)dx d 3cos3(4x)sin4(4x)dx

16 Find an antiderivative of each of the following.

a cosec2(2x)cos(2x) b sec2(2x)sin(2x)

c sin(2x)

cos3(2x)d

cos(2x)

sin3(2x)

ConsolidatE



UNCORRECTED PAGE P

ROOFS


a 3

π6

0

sin2(3x)dx b 3

π6

0

cos3(3x)dx

c 3

π6

0

sin4(3x)dx d 3

π6

0

cos5(3x)dx

18 a Find 3(cos(2x) + sin(2x))2dx.

b Find 3cos3(2x) + sin3(2x)dx.

c Consider 3sin3(2x)cos3(2x)dx. Show that this integration can be done using:

i a double-angle formulaii the substitution u = cos(2x)iii the substitution u = sin(2x).


a 3tan(3x)dx b 3cot(3x)dx

c 3tan(3x)sec2(3x)dx d 3tan2(3x)sec2(3x)dx


a 3

π20

0

tan(5x)dx b 3

π20

0

tan2(5x)dx

c 3

π20

0

tan3(5x)sec2(5x)dx d 3

π20

0

tan2(5x)sec4(5x)dx

21 Given that n ≠ −1 and a ∈ R, find each of the following.

a 3sin(ax)cosn(ax)dx b 3cos(ax)sinn(ax)dx

c 3tann(ax)sec2(ax)dx

22 Integrate each of the following where a ∈ R\{0}.

a tan(ax) b tan2(ax) c tan3(ax) d tan4(ax)


a sin2(ax) b sin3(ax) c sin4(ax) d sin5(ax)


a cos2(ax) b cos3(ax) c cos4(ax) d cos5(ax)

MastEr

topic 8 IntegraL CaLCuLus 417


UNCORRECTED PAGE P

ROOFS

Integrals involving the inverse cosine function

Since ddx

acos−1axabb = −1

"a2 − x2, it follows that 3

−1

"a2 − x2 dx = cos−1ax

ab + c

where a > 0 and ∣x∣ < a.

Find:

a 312

"36 − x2 dx b 3

4

"49 − 36x2 dx.

tHinK WritE

a Use 31

"a2 − x2 dx = sin−1ax

ab + c with a = 6.

a3

12

"36 − x2 dx

= 1231

"36 − x2 dx

= 12 sin−1ax6b + c

b 1 Use a linear substitution with u = 6x and express dx in terms of du by inverting both sides.

b u = 6x u2 = 36x2

dudx

= 6

dxdu

= 16

dx = 16

du

2 Substitute for dx and u. 34

"49 − 36x2 dx

= 34

"49 − u2× 1

6 du

3 Use the properties of indefi nite integrals to transfer the constant factor outside the front of the integral sign.

= 233

1

"49 − u2 du

4 Use 31

"a2 − u2 du = sin−1au

ab + c with a = 7. = 2

3 sin−1au

7b + c

5 Substitute back for u and express the fi nal answer in terms of x only and an arbitrary constant +c.

34

"49 − 36x2 dx = 2

3 sin−1a6x

7b + c

WOrKeD eXaMPLe 272727

Integrals involving inverse trigonometric functionsIntegrals involving the inverse sine function

Since ddx

asin−1axabb = 1

"a2 − x2, it follows that 3

1

"a2 − x2 dx = sin−1ax

ab + c for

a > 0 and ∣x∣ < a.

8.6



UNCORRECTED PAGE P

ROOFS

On a certain curve the gradient is given by −4

"81 − 25x2. Find the equation of

the curve that passes through the origin.

THINK WRITE

1 Recognise that the gradient of a curve is given by dy

dx.

dy

dx= −4

"81 − 25x2

2 Integrating both sides gives an expression for y. y = 3−4

"81 − 25x2 dx

3 Use a linear substitution with u = 5x and express dx in terms of du by inverting both sides.

u = 5x u2 = 25x2

dudx

= 5

dxdu

= 15

dx = 15

du

4 Substitute for dx and u. y = 3−4

"81 − u2× 1

5 du


y = 453

−1

"81 − u2 du

6 Use 3−1

"a2 − u2 du = cos−1au

ab + c with a = 9. y = 4

5 cos−1au

9b + c

7 Substitute back for u and express the � nal answer in terms of x only and an arbitrary constant +c.

y = 45

cos−1a5x9b + c

8 Since the curve passes through the origin, we can let y = 0 when x = 0 to � nd c.

0 = 45

cos−1(0) + c

0 = 45

× π2

+ c

c = −2π5

9 Substitute back for c. y = 45

cos−1a5x9b − 2π

5

10 State the equation of the particular curve in a factored form.

y = 45acos−1a5x

9b − π

2b

11 Note that an alternative answer is

possible, since sin−1(x) + cos−1(x) = π2

.

y = −45

sin−1a5x9b


Integrals involving the inverse tangent function

Since ddx

atan−1axabb = a

a2 + x2, it follows that 3

1a2 + x2

= 1a

tan−1axab + c for x ∈ R.



UNCORRECTED PAGE P

ROOFS

Antidifferentiate each of the following with respect to x.

a 12

36 + x2 b 4

49 + 36x2

THINK WRITE

a 1 Write the required antiderivative. a 312

36 + x2 dx

2 Use 31

a2 + x2= 1

a tan−1ax

ab + c with a = 6. 123

136 + x2

dx

= 12 × 16

tan−1ax6b + c

= 2 tan−1ax6b + c

b 1 Write the required antiderivative. b 34

49 + 36x2 dx

2 Use a linear substitution with u = 6x and express dx in terms of du by inverting both sides.

u = 6x u2 = 36x2

dudx

= 6

dxdu

= 16

dx = 16

du


49 + 36x2 dx = 3

449 + u2

× 16

du


= 233

149 + u2

du

5 Use 31

a2 + x2= 1

a tan−1ax

ab + c with a = 7. = 2

3× 1

7 tan−1au

7b + c


34

49 + 36x2 dx = 2

21 tan−1a6x

7b + c


Defi nite integrals involving inverse trigonometric functions

Evaluate:

a 3

2

9

0

−2

"16 − 81x2 dx b 3

4

9

0

216 + 81x2

dx.




UNCORRECTED PAGE P

ROOFS

THINK WRITE

a 1 Use a linear substitution with u = 9x and express dx in terms of du by inverting both sides.

a u = 9x u2 = 81x2

dudx

= 9

dxdu

= 19

dx = 19 du


3 Substitute for dx, u and the new terminals.3

2

9

0

−2

"16 − 81x2 dx

= 3 2

0

−2

"16 − u2 × 1

9 du


= 293

2

0

−1

"16 − u2 du

5 Perform the integration using

3−1

"a2 − u2 du = cos−1au

ab + c with a = 4.

= 29ccos−1au

4b d

2

0

6 Evaluate the definite integral. = 29ccos−1a1

2b − cos−1(0) d

= 29cπ3

− π2d

= 29aπ(2 − 3)

6b

7 State the final result. Note that since this is just evaluating a definite integral and not finding an area, the answer stays as a negative number.

3

2

9

0

−2

"81 − 16x2 dx = − π

27

b 1 Use a linear substitution with u = 9x and express dx in terms of du by inverting both sides.

b u = 9x u2 = 81x2

dudx

= 9

dxdu

= 19

dx = 19

du




UNCORRECTED PAGE P

ROOFS

3 Substitute for dx, u and the new terminals. 3

4

9

0

216 + 81x2

dx

= 3

4

0

216 + u2

× 19

du


= 293

4

0

116 + u2

du

5 Perform the integration using

31

a2 + x2= 1

a tan−1ax

ab + c with a = 4.

= 29c14

tan−1au4b d

4

0

= 118

c tan−1au4b d

4

0

6 Evaluate this de� nite integral. = 118

3 tan−1(1) − tan−1(0) 4

= 118

cπ4

− 0 d


4

9

0

216 + 81x2

dx = π72

Integrals involving completing the squareA quadratic expression in the form ax2 + bx + c can be expressed in the form a(x + h)2 + k (the completing the square form). This can be used to integrate

expressions of the form 1ax2 + bx + c

with a > 0 and Δ = b2 − 4ac < 0, which

will then involve the inverse tangent function.

Integrating expressions of the form 1

"ax2 + bx + c with a < 0 and Δ = b2 − 4ac > 0

will involve the inverse sine or cosine functions.

Find each of the following with respect to x.

a 31

9x2 + 12x + 29 dx b 3

1

"21 − 12x − 9x2 dx

THINK WRITE

a 1 Express the quadratic factor in the denominator in the completing the square form by making it into a perfect square.

a 9x2 + 12x + 29 = (9x2 + 12x + 4) + 25 = (3x + 2)2 + 25




UNCORRECTED PAGE P

ROOFS

2 Write the denominator as the sum of two squares.

31

9x2 + 12x + 29 dx

= 31

(3x + 2)2 + 25 dx

3 Use a linear substitution with u = 3x + 2 and express dx in terms of du by inverting both sides.

u = 3x + 2

dudx

= 3

dxdu

= 13

dx = 13

du


(3x + 2)2 + 25 dx = 3

1u2 + 25

× 13

du


= 133

1u2 + 25

du

6 Use 31

a2 + u2= 1

a tan−1au

ab + c with a = 5. = 1

3× 1

5 tan−1au

5b + c

= 115

tan−1au5b + c

7 Substitute back for u and express the final answer in terms of x only and an arbitrary constant +c.

31

9x2 + 12x + 29 dx = 1

15 tan−1a3x + 2

5b + c

b 1 Express the quadratic factor in the denominator in the completing the square form by making it into the difference of two squares.

b 21 − 12x − 9x2

= 21 − (9x2 + 12x) = 21 − (9x2 + 12x + 4) + 4 = 25 − (3x + 2)2

2 Write the denominator as the difference of two squares under the square root.

31

"21 − 12x − 9x2 dx

= 31

"25 − (3x + 2)2 dx

3 Use a linear substitution with u = 3x + 2 and express dx in terms of du by inverting both sides.

u = 3x + 2

dudx

= 3

dxdu

= 13

dx = 13

du


"25 − (3x + 2)2 dx

= 31

"25 − u2 × 1

3 du



UNCORRECTED PAGE P

ROOFS

Integrals involving substitutions and inverse trigonometric functionsWe can break up complicated integrals into two manageable integrals using the

following property of inde� nite integrals: 3(

f(x) ± g(x))dx = 3f(x)dx ± 3g(x)dx.


= 133

1

"25 − u2 du

6 Use 31

"a2 − u2= sin−1au

ab + c with a = 5. = 1

3 sin−1au

5b + c


31

"21 − 12x − 9x2 dx = 1

3 sin−1a3x + 2

5b + c

Find 34x + 5

"25 − 16x2 dx.

THINK WRITE

1 If the x was not present in the numerator, the integral would involve an inverse sine function. If the 5 was not present in the numerator, the integral would involve a non-linear substitution. Break the integral into two distinct problems: one involving a non-linear substitution and one involving the inverse trigonometric function.

34x + 5

"25 − 16x2 dx

= 34x

"25 − 16x2 dx + 3

5

"25 − 16x2 dx

2 Write the � rst integral as a power using index laws. 34x

"25 − 16x2 dx

= 34x(25 − 16x2)−1

2dx

3 Use a non-linear substitution with u = 25 − 16x2 and express dx in terms of du by inverting both sides.

u = 25 − 16x2

dudx

= −32x

dxdu

= − 132x

dx = − 132x

du

4 Substitute for dx and u, noting that the x terms cancel.

34x(25 − 4x2)−1

2dx

= 34xu−1

2 × − 132x

du




UNCORRECTED PAGE P

ROOFS

5 Transfer the constant factor outside the integral sign. = −183u

−12du

6 Perform the integration using 3undu = 1n + 1

un+1 with

n = −12, so that n + 1 = 1

2.

= −18

× 112

u12

= −14!u

7 Substitute back for u. 3x

"25 − 4x2 dx = −1

4"25 − 16x2

8 Consider the second integral. Use a linear substitution with v = 4x and express dx in terms of dv by inverting both sides.

35

"25 − 16x2 dx

v = 4x

dvdx

= 4

dxdv

= 14

dx = 14

dv

9 Substitute for dx and v. 35

"25 − 15x2 dx

= 35

"25 − v2 × 1

4 dv

10 Use the properties of indefinite integrals to transfer the constant factor outside the front of the integral sign.

= 543

1

"25 − v2 dv

11 Use 31

"a2 − v2 dv = sin−1av

ab + c with a = 5. = 5

4 sin−1av

5b

12 Substitute back for v. 35

"25 − 4x2 dx = 5

4 sin−1a4x

5b

13 Express the original integral as the sum as the two integrals, adding in only one arbitrary constant + c.

34x + 5

"25 − 16x2 dx

= 54

sin−1a4x5b − 1

4"25 − 16x2 + c

Integrals involving inverse trigonometric functions1 WE27 Find:

a 31

"100 − x2 dx b 3

12

"64 − 9x2 dx.


a 1

"36 − 25x2b x

"36 − 25x2

ExErcisE 8.6

PractisE



UNCORRECTED PAGE P

ROOFS

3 WE28 On a certain curve the gradient is given by −2

"25 − 16x2. Find the equation

of the curve that passes through the origin.

4 On a certain curve the gradient is given by −2

"4 − x2. Find the equation of the

curve that passes through the point (2, 3).

5 WE29 Antidifferentiate each of the following with respect to x.

a 1100 + x2

b 1264 + 9x2


a 136 + 25x2

b 5x

64 + 25x2

7 WE30 Evaluate:

a 3

98

0

1

"81 − 16x2 dx b 3

94

0

181 + 16x2

dx.

8 Evaluate:

a 3

1

"2

0

13 + 2x2

dx b 3

1

#2

0

1

"2 − 3x2 dx.

9 WE31 Find each of the following with respect to x.

a 31

4x2 − 12x + 25 dx b 3

1

"7 + 12x − 4x2 dx

10 Find each of the following with respect to x.

a 31

25x2 − 20x + 13 dx b 3

1

"12 + 20x − 25x2 dx

11 WE32 Find 35 − 3x

"25 − 9x2 dx.

12 Find 33x + 5

9x2 + 25 dx.


a 1

"16 − x2b 1

"1 − 16x2c 10

"49 − 25x2d 10x

"49 − 25x2


a 3−1

"4 − x2 dx b 3

−2

"1 − 4x2 dx

c 3−3x

"36 − 49x2 dx d 3

−3

"36 − 49x2 dx

AOS 3

Topic 2

Concept 1

Antiderivatives involving inverse circular functionsConcept summaryPractice questions

ConsolidatE



UNCORRECTED PAGE P

ROOFS

15 Find a set of antiderivatives for each of the following.

a 116 + x2

b x49 + 36x2

c 2149 + 36x2

d 81 + 16x2


a 3

3

0

x

"9 − x2 dx b 3

3

0

1

"9 − x2 dx c 3

3

0

19 + x2

dx d 3

3

0

x9 + x2

dx

17 a On a certain curve the gradient is given by 1

"4 − x2. Find the equation of the

curve that passes through the point (!3, π).

b On a certain curve the gradient is given by 11 + 4x2

. Find the equation of the

curve that passes through the point a12, πb.

c If dy

dx+ 1

3 + x2= 0, and when x = 1, y = 0, find y when x = 0.

d If dy

dx+ 1

"6 − x2= 0, and when x = !3, y = 0, find y when x = 0.


a 3

5

6

0

1

"25 − 9x2 dx b 3

5

3

0

x25 + 9x2

dx

c 3

5

3

0

x

"25 − 9x2 dx d 3

5

3

0

125 + 9x2

dx


a 3

1

0

1

"4 − 3x2 dx b 3

1

0

11 + 3x2

dx c 3

1

0

x1 + 3x2

dx d 3

1

0

x

"4 − 3x2 dx


a 32

"5 − 4x − x2 dx b 3

2x2 + 4x + 13

dx

c 36

"24 − 30x − 9x2 dx d 3

674 + 30x + 9x2

dx


a 33x − 4

"9 − 16x2 dx b 3

3 + 4x

9 + 16x2 dx

c 35 − 2x

"5 − 2x2 dx d 3

5 − 2x

25 + 2x2 dx



UNCORRECTED PAGE P

ROOFS

22 If a, b, p and q are positive real constants, find each of the following.

a 31

"p2 − q2x2 dx b 3

1p2 + q2x2

dx

c 3ax + b

"p2 − q2x2 dx d 3

ax + b

p2 + q2x2 dx

23 a i Use the substitution x = 4 sin(θ ) to find 3"16 − x2 dx.

ii Evaluate 3

4

0

"16 − x2 dx. What area does this represent?

b i Use the substitution x = 65 cos(θ ) to find 3"36 − 25x2

dx.

ii Evaluate 3

6

5

0

"36 − 25x2 dx. What area does this represent?

c Prove that the total area inside the circle x2 + y2 = r2 is given by πr2.

d Prove that the total area inside the ellipse x2

a2+

y2

b2= 1 is given by πab.


a 3

5

3

x − 4x2 − 6x + 13

dx b 3

7

2

32

2x + 1

4x2 − 12x + 25 dx.

c 3

4

1

3x + 2

"8 + 2x − x2 dx d 3

1

−2

2x − 3

"12 − 8x − 4x2 dx

Integrals involving partial fractionsIntegration of rational functionsA rational function is a ratio of two functions, both of which are polynomials. For

example, mx + k

ax2 + bx + c is a rational function; it has a linear function in the numerator

and a quadratic function in the denominator.

In the preceding section, we integrated certain expressions of this form when a > 0 and Δ = b2 − 4ac < 0, meaning that the quadratic function was expressed as the sum of two squares. In this section, we examine cases when a ≠ 0 and Δ = b2 − 4ac > 0. This means that the quadratic function in the denominator can now be factorised into linear factors. Integrating expressions of this kind does not involve a new integration technique, just an algebraic method of expressing the integrand into its partial fractions decomposition.

Converting expressions into equivalent forms is useful for integration. We have seen this when expressing a quadratic in the completing the square form or converting trigonometric powers into multiple angles.

Master

8.7AOS 3

Topic 2

Concept 7

Antiderivatives with partial fractionsConcept summaryPractice questions



UNCORRECTED PAGE P

ROOFS

However, if the derivative of the denominator is equal to the numerator or

a constant multiple of it, that is, if ddx

(ax2 + bx + c) = 2ax + b, then

32ax + b

ax2 + bx + c dx = loge a∣ax2 + bx + c∣b.

The method of partial fractions involves factorising the denominator, for example, mx + k

ax2 + bx + c= mc + k

(ax + α)(x + β), and then expressing as A

(ax + α)+ B

(x + β), where

A and B are constants to be found. We can � nd A and B by equating coef� cients and then solving simultaneous equations, or by substituting in speci� c values of x. See the following worked examples.

Equating coef� cients means, for example, that if Ax + B = 3x − 4, then A = 3 from equating the coef� cient of x, since Ax = 3x, and B = −4 from the term independent of x (the constant term).

Find:

a 312

36 − x2 dx b 3

449 − 36x2

dx.

THINK WRITE

a 1 Factorise the denominator into linear factors using the difference of two squares.

a12

36 − x2= 12

(6 + x)(6 − x)

2 Write the integrand in its partial fractions decomposition, where A and B are constants to be found.

= A6 + x

+ B6 − x

3 Add the fractions by forming a common denominator.

= A(6 − x) + B(6 + x)(6 + x)(6 − x)

4 Expand the numerator, factor in x and expand the denominator.

= 6A − 6Ax + 6B + 6Bx(6 + x)(6 − x)

= 6x(B − A) + 6(A + B)

36 − x2

5 Because the denominators are equal, the numerators are also equal. Equate the coef� cients of x and the term independent of x. This gives two simultaneous equations for the two unknowns, A and B.

6x(B − A) + 6(A + B) = 12(1) B − A = 0(2) 6(A + B) = 12

6 Solve the simultaneous equations (1) ⇒ A = B; substitute into (2) 12A = 12 ⇒ A = B = 1

7 An alternative method can be used to � nd the unknowns A and B. Equate the numerators from the working above.

12 = A(6 − x) + B(x + 6)

8 Substitute an appropriate value of x. Substitute x = 6:12 = 12B B = 1


AOS 2

Topic 1

Concept 1

Partial fractions with a quadratic denominator that has linear factorsConcept summaryPractice questions



UNCORRECTED PAGE P

ROOFS

9 Substitute an appropriate value of x. Substitute x = −6:12 = 12A A = 1

10 Express the integrand as its partial fractions decomposition.

1236 − x2

= 16 + x

+ 16 − x

11 Instead of integrating the original expression, we integrate the partial fractions expression, since these expressions are equal.

312

36 − x2 dx = 3a 1

6 + x+ 1

6 − xb dx


the result 31

ax + b dx = 1

a loge a∣ax + b∣b.

= loge a∣6 + x∣b − loge a∣6 − x∣b + c

13 Use the log laws to express the final answer as a single log term.

312

36 − x2 dx = loge a ∣ 6 + x

6 − x ∣ b + c

b 1 Factorise the denominator into linear factors using the difference of two squares.

b4

49 − 36x2= 4

(7 − 6x)(7 + 6x)


= A7 − 6x

+ B7 + 6x


= A(7 + 6x) + B(7 − 6x)(7 − 6x)(7 + 6x)


= 7A + 6Ax + 7B − 6Bx(7 − 6x)(7 + 6x)

= 6x(A − B) + 7(A + B)

49 − 36x2

5 Because the denominators are equal, the numerators are also equal. Equate coefficients of x and the term independent of x. This gives two simultaneous equations for the two unknowns, A and B.

6x(A − B) + 7(A − B) = 4(1) A − B = 0(2) 7(A + B) = 4

6 Solve the simultaneous equations (1) ⇒ A = B: substitute into (2) 14A = 4 ⇒ A = B = 2

7


449 − 36x2

= 27(7 − 6x)

+ 27(7 + 6x)

8 Instead of integrating the original expression, we integrate the partial fractions expression, since these expressions are equal. Transfer the constant factors outside the integral sign.

34

49 − 36x2 dx

= 3a 27(7 − 6x)

+ 27(7 + 6x)

bdx

= 273a

17 − 6x

+ 17 + 6x

bdx



UNCORRECTED PAGE P

ROOFS


the result 31

ax + b dx = 1

a loge a∣ax + b∣b.

= 27c−1

6 loge a∣7 − 6x∣b + 1

6 loge a∣7 + 6x∣b d + c

10 Use the log laws to express the � nal answer as a single log term.

34

49 − 36x2 dx = 1

21 loge a `7 + 6x

7 − 6x` b + c

Compare the previous worked example to Worked examples 27 and 29.

Find 3x + 10

x2 − x − 12 dx.

THINK WRITE

1 Factorise the denominator into linear factors.x + 10

x2 − x − 12= x + 10

(x − 4)(x + 3)


= Ax − 4

+ Bx + 3


=A(x + 3) + B(x − 4)

(x − 4)(x + 3)


= Ax + 3A + Bx − 4B(x − 4)(x + 3)

= x(A + B) + 3A − 4B

x2 − x − 12

5 Because the denominators are equal, the numerators are also equal. Equate the coef� cients of x and the term independent of x to give two simultaneous equations for the two unknowns, A and B.

x(A + B) + 3A − 4B = x + 10(1) A + B = 1(2) 3A − 4B = 10

6 Solve the simultaneous equations by elimination. Add the two equations to eliminate B and solve for A.

4 × (1) 4A + 4B = 4 (2) 3A − 4B = 10

4 × (1) + (2) 7A = 14A = 2

7 Substitute back into (1) to � nd B. 2 + B = 1 B = −1

8 An alternative method can be used to � nd the unknowns A and B. Equate the numerators from the working above.

A(x + 3) + B(x − 4) = x + 10

9 Let x = 4. 7A = 14A = 2

10 Let x = −3. −7B = 7 B = −1




UNCORRECTED PAGE P

ROOFS

Perfect squaresIf an expression is of the form

mx + k

ax2 + bx + c when a ≠ 0 and Δ = b2 − 4ac = 0, the

quadratic function in the denominator is now a perfect square.

Integrating by the method of partial fractions in this case involves writing mx + k

ax2 + bx + c= mc + k

( px + α)2 and then expressing it as A

( px + α)+ B

( px + α)2, where

A and B are constants to be found. We can � nd A and B by equating coef� cients and then solving the simultaneous equations, as before.

AOS 2

Topic 1

Concept 2

Partial fractions with perfect square denominatorsConcept summaryPractice questions


3x + 10

x2 − x − 12 dx = 3a 2

x − 4− 1

x + 3bdx

12 Integrate term by term using the result

31

ax + b dx = 1

a loge a∣ax + b∣b, and state

the � nal simpli� ed answer using log laws.

3x + 10

x2 − x − 12 dx

= 2 loge a∣x − 4∣b − loge a∣x + 3∣b + c

= loge a (x − 4)2

∣x + 3∣ b

Find 36x − 5

4x2 − 12x + 9 dx.

THINK WRITE

1 Factorise the denominator as a perfect square.

6x − 54x2 − 12x + 9

= 6x − 5(2x − 3)2


= A2x − 3

+ B(2x − 3)2

3 Add the fractions by forming the lowest common denominator and expanding the numerator.

=A(2x − 3) + B

(2x − 3)2

= 2Ax + B − 3A(2x − 3)2

4 Because the denominators are equal, the numerators are also equal. Equate coef� cients of x and the term independent of x to give two simultaneous equations for the two unknowns, A and B.

2Ax + B − 3A = 6x − 5(1) 2A = 6(2) B − 3A = −5

5 Solve the simultaneous equations. (1) ⇒ A = 3: substitute into (2):B − 9 = −5 ⇒ B = 4


36x − 5

4x2 − 12x + 9 dx

= 3a 32x − 3

+ 4(2x − 3)2

bdx




UNCORRECTED PAGE P

ROOFS

Note that this example can also be done using a linear substitution with a back substitution. See Worked example 12b.

Rational functions involving ratios of two quadratic functionsWhen the degree of the polynomial in the numerator is greater than or equal to the degree of the polynomial in the denominator, the rational function is said to be a non-proper rational function. In this case we have to divide the denominator into the numerator to obtain a proper rational function.

For example, when we have quadratic functions in both the numerator and

denominator, that is, the form rx2 + sx + t

ax2 + bx + c where r ≠ 0 and a ≠ 0, we can use long

division to divide the denominator into the numerator to express the function as rx2 + sx + t

ax2 + bx + c= q + mx + k

ax2 + bx + c where q = r

a.

7 Integrate term by term, using the result

31

ax + b dx = 1

a loge a∣ax + b∣b in the � rst term and

31

(ax + b)2 dx = − 1

a(ax + b) for the second term.

= 32

loge a∣2x − 3∣b − 22x − 3

+ c

Find 32x2 + 5x + 3x2 + 3x − 4

dx.

THINK WRITE

1 Use long division to express the rational function as a proper rational function.

2x2 + 5x + 3x2 + 3x − 4

=2(x2 + 3x − 4) + 11 − x

x2 + 3x − 4

= 2 + 11 − xx2 + 3x − 4

2 Factorise the denominator into linear factors. = 2 + 11 − x(x − 1)(x + 4)


= 2 + Ax − 1

+ Bx + 4

4 Add the fractions by forming a common denominator and expanding the numerator.

= 2 +A(x + 4) + B(x − 1)

(x − 1)(x + 4)

= 2 +x(A + B) + 4A − B

x2 + 3x − 4

5 Because the denominators are equal, the numerators are also equal. Equate coef� cients of x and the term independent of x to give two simultaneous equations for the two unknowns, A and B.

x(A + B) + 4A − B = 11 − x(1) A + B = −1(2) 4A − B = 11




UNCORRECTED PAGE P

ROOFS

Rational functions involving non-linear factorsIf the denominator does not factorise into linear factors, then we proceed with

px2 + qx + r

(x + α)(x2 + a2)= A

x + α+ Bx + C

x2 + a2. Note that we will need three simultaneous

equations to solve for the three unknowns: A, B and C.

AOS 2

Topic 1

Concept 3

Partial fractions with quadratic factors in the denominatorConcept summaryPractice questions

Find 3x2 + 7x + 2

x3 + 2x2 + 4x + 8 dx.

THINK WRITE

1 First try to factorise the cubic in the denominator using the factor theorem.

f(x) = x3 + 2x2 + 4x + 8 f(1) = 1 + 2 + 4 + 8 ≠ 0 f(2) = 8 + 8 + 8 + 8 ≠ 0 f(−2) = −8 + 8 − 8 + 8 = 0

(x + 2) is a factor.

2 Factorise the denominator into factors. x3 + 2x2 + 4x + 8 = (x + 2)(x2 + 4)

3 Write the integrand in its partial fractions decomposition, where A, B and C are constants to be found.

x2 + 7x + 2x3 + 2x2 + 4x + 8

= Ax + 2

+ Bx + Cx2 + 4

4 Add the fractions by forming a common denominator and expanding the numerator.

=A(x2 + 4) + (x + 2)(Bx + C)

(x + 2)(x2 + 4)

= Ax2 + 4A + Bx2 + 2Bx + Cx + 2C(x + 2)(x2 + 4)

=x2(A + B) + x(2B + C) + 4A + 2C

x3 + 2x2 + 4x + 8


6 Solve the simultaneous equations by elimination. Add the two equations to eliminate B.

5A = 10 A = 2 ⇒ B = −3


32x2 + 5x + 3x2 + 3x − 4

dx

= 3a2 + 2x − 1

− 3x + 4

b dx

8 Integrate term by term using the result

31

ax + b dx = 1

a loge a∣ax + b∣b and state

the � nal answer.

32x2 + 5x + 3x2 + 3x − 4

dx

= 2x + 2 loge a∣x − 1∣b − 3 loge a∣x + 4∣b + c

= 2x + loge a (x − 1)2

(x + 4)3b + c



UNCORRECTED PAGE P

ROOFS

5 Because the denominators are equal, the numerators are also equal. Equate the coefficients of x2 and x and the term independent of x to give three simultaneous equations for the three unknowns: A, B and C.

(1) A + B = 1(2) 2B + C = 7(3) 4A + 2C = 2

6 Solve the simultaneous equations by elimination.

2 × (1) 2A + 2B = 2 (2) 2B + C = 7Subtracting gives (4) C − 2A = 5

2 × (4) 2C − 4A = 10

(3) 4A + 2C = 2 Adding gives 4C = 12

C = 2

7 Back substitute to find the remaining unknowns.

Substitute into (2):2B + 3 = 7 ⇒ B = 2

Substitute into (1): ⇒ A = −1


x2 + 7x + 2x3 + 2x2 + 4x + 8

= −1x + 2

+ 2x + 3x2 + 4

9 Separate the last term into two expressions. 3x2 + 7x + 2

x3 + 2x2 + 4x + 8 dx

= 3a −1x + 2

+ 2x + 3x2 + 4

bdx

= 3a 2xx2 + 4

− 1x + 2

+ 3x2 + 4

bdx

10 Integrate term by term, using the

results 3f ′ (x)

f(x) dx = loge a∣ f(x)∣b,

31

ax + b dx = 1

a loge a∣ax + b∣b and

31

x2 + a2 dx = 1

a tan−1ax

ab, and state the

final answer.

3x2 + 7x + 2

x3 + 2x2 + 4x + 8 dx

= loge (x2 + 4) − loge a∣x + 2∣b

+ 32

tan−1ax2b + c

Integrals involving partial fractions1 WE33 Find:

a 31

100 − x2 dx b 3

1264 − 9x2

dx.

2 Find:

a 31

36 − 25x2 dx b 3

x36 − 25x2

dx.

ExErcisE 8.7

PractisE



UNCORRECTED PAGE P

ROOFS

3 WE34 Find 3x + 13

x2 + 2x − 15 dx.

4 Find 3x − 11

x2 + 3x − 4 dx.

5 WE35 Find 32x + 1

x2 + 6x + 9 dx.

6 Find 32x − 1

9x2 − 24x + 16 dx.

7 WE36 Find 33x2 + 10x − 4

x2 + 3x − 10 dx.

8 Find 3−2x2 − x + 20

x2 + x − 6 dx.

9 WE37 Find 319 − 3x

x3 − 2x2 + 9x − 18 dx.

10 Find 325

x3 + 3x2 + 16x + 48 dx.


a x + 11

x2 + x − 12b 5x − 9

x2 − 2x − 15c 2x − 19

x2 + x − 6d 11

x2 − 3x − 28


a 1x2 − 4

b 216 + x2

c xx2 − 25

d 2x − 3x2 − 36


a 2x + 3

x2 − 6x + 9b 2x − 5

x2 + 4x + 4

c 4x4x2 + 12x + 9

d 6x − 199x2 − 30x + 25


a x2 − 4x − 11x2 + x − 12

b −3x2 − 4x − 5x2 + 2x − 3

c 4x2 − 17x − 26x2 − 4x − 12

d −2x3 + 12x2 − 17x

x2 − 6x + 8

15 a Determine 31

x2 + kx + 25 dx for the cases when:

i k = 0 ii k = 26 iii k = −10.

b Determine 31

4x2 − 12x + k dx for the cases when:

i k = 8 ii k = 9 iii k = 25.

consolidatE

436 MaThs QuesT 12 sPecIalIsT MaTheMaTIcs Vce units 3 and 4


UNCORRECTED PAGE P

ROOFS


a 3

2

1

1x2 + 4x

dx b 3

6

5

1x2 − 16

dx

c 3

1

−1

3x + 8

x2 + 6x + 8 dx d 3

4

3

x − 6x2 − 4x + 4

dx

17 a Find the area bounded by the curve y = 5x2 + x − 6

, the coordinate axes and x = −1.

b Find the area bounded by the curve y = 2x − 34 + 3x − x2

, the x-axis and the lines x = 2 and x = 3.

c Find the area bounded by the curve y = 2140 − 11x − 2x2

, the coordinate axes and x = −5.

d Find the area bounded by the curve y = x3 − 9x + 9

x2 − 9, the x-axis and the

lines x = 4 and x = 6.

18 a Find the value of a if 3

2

1

24x − x2

dx = loge (a).

b Find the value of b if 3

2

1

2

"4x − x2 dx = πb.

c Find the value of c if 3

4

3

3x

x2 − x − 2 dx = loge (c).

d Find the value of d if 3

2

1

2

3x2 − x + 1

dx = πd.

19 a If y = 16x2 + 25

16x2 − 25, find:

i 3y dx ii 31y

dx.

b If y = 36x2 − 4936x2 + 49

, find:

i 3y dx ii 31y

dx.

20 Show that 3

1

0

x4(1 − x)4

1 + x2 dx = 22

7− π.



UNCORRECTED PAGE P

ROOFS


a 3

1

0

27x2

16 − 9x2 dx b 3

2

0

20x2

4x2 + 4x + 1 dx

c 3

3

!3

x2 − 2x + 9

x3 + 9x dx d 3

3

2

4x2 − 16x + 19

(2x − 3)3 dx

22 If a, b, p and q are all non-zero real constants, find each of the following.

a 31

b2x2 − a2 dx b 3

xb2x2 − a2

dx

c 3x

(ax − b)2 dx d 3

1( px + a)(qx + b)

dx

MastEr

438 MaThs QuesT 12 sPecIalIsT MaTheMaTIcs Vce units 3 and 4


UNCORRECTED PAGE P

ROOFS

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Units 3 & 4 Integral calculus

Sit topic test


UNCORRECTED PAGE P

ROOFS

8 AnswersExErcisE 8.21 26

2 163

3 923

4 16

5 4

6 12

7 2(e6 − 1)

8 6(e2 − e−2)

9 16

10 137227

11 8

12 128

13 12112

14 36

15 a 9 b 37 12

16 a i 32 ii 22

b i 16623

ii 132

17 a 36π

b 16π

c 2an

18 a i loge (4) ii 1 iii loge (a)

b i 23

ii 1 − 1a

iii 85 13

19 a 13912

b 8113

c i 648 ii 465 34

20 a 4a3

3b 1

2a4

21 a 16

b 12

c 310

d 23

22 a i 288 ii 288

b i 12112

ii 12112

23 a 154

− 2 loge (4) ≈ 0.9774

b 4027

− 13 loge (9) ≈ 0.7491

24 Check with your teacher.

25 a i (3.4443, 3.9543) ii 4.6662

b i (3.8343, 1.9172) ii 8.6558

26 a i (2.6420, 86.1855) ii 64.8779

b i (7.5882, –1.7003), (24.2955, –4.6781)

ii 384.3732

ExErcisE 8.31 1

35(5x − 9)7 + c

2 124

(3x + 4)8 + c

3 y = 18(!16x + 25 − 5)

4 114

5 13 loge a∣3x − 5∣b + c

6 −12 loge a∣7 − 2x∣b + c

7 2

8 8

9 a 1175

(5x − 9)7 + 350

(5x − 9)6 + c

= 1350

(10x + 3)(5x − 9)6

b 29

loge a∣3x − 4∣b − 59(3x − 4)

+ c

10 a −(4x + 7)

8(2x + 7)2+ c b 1

27(3x − 5)!6x + 5 + c

11 4

12 53 loge a

52b − 1

13 2x − 83 loge a∣3x + 4∣b + c

14 2 + 5 loge a35b

15 a 121

(3x + 5)7 + c b −1

3(3x + 5)+ c

c −1

6(3x + 5)2+ c d 1

2"3 (3x + 5)2 + c

16 a 154

(6x + 7)9 + c b 13!6x + 7 + c

c 16 loge a∣6x + 7∣b + c d

−16(6x + 7)

+ c

17 a 172

(3x + 5)8 − 563

(3x + 5)7 + c

= 1504

(21x − 5)(3x + 5)7 + c

b 19

loge a∣3x + 5∣b + 59(3x + 5)

+ c

c −(6x + 5)

18(3x + 5)2+ c

d 110

(2x − 5)(3x + 5)

2

3 + c

18 a 1360

(6x + 7)10 − 7324

(6x + 7)9 + c

= 13240

(54x − 7)(6x + 7)9 + c

b 127

(3x − 7)!6x + 7 + c

c x6

− 736

loge a∣6x + 7∣b + c

d 7

36(6x + 7)+ 1

36 loge a∣6x + 7∣b + c

19 a t

2(2 − 5t)b y = −!3 − 2x

c 32 loge (3) d y = 1

3(x − 9)!2x + 9 + 9



UNCORRECTED PAGE P

ROOFS

20 a 72

b 477

c 6 d 33310

21 a 263

b 13 loge a

145b c 1

35d 6

22 a i 163

ii 323

b 323

c 25615

d i 43"a3

ii 815"a5

23 a 23a

(ax + b)

3

2 + c

b 2

15a2(3ax − 3b) (ax + b)

3

2 + c

c 1a

loge a∣ax + b∣b + c

d xa

− b

a2 loge a∣ax + b∣b + c

24 a 2a!ax + b + c

b 2

3a2(ax − 2b)!ax + b + c

c −1

a(ax + b)+ c

d b

a2(ax + b)+ 1


25 a ad − bc

a2 loge a∣ax + b∣b + cx

a+ k

b c

a2 loge a∣ax + b∣b − ad − bc

a2(ax + b)+ k

c cx

a2− 2bc

a3 loge a∣ax + b∣b − a2d + cb2

a3(ax + b)+ k

d a2d + cb2

a3 loge a∣ax + b∣b + cx2

2a− bcx

a2+ k

26 a 1

2a3(ax − 3b) (ax + b) + b2


b x(xa + 2b)

a2(ax + b)− 2b


c 1

a3 loge a∣ax + b∣b +

b(4ax + 3b)

2a3(ax + b)2+ c

d 2

15a3(3a2x2 − 4abx + 8b2)!ax + b + c

ExErcisE 8.4

1 −2

(x2 + 16)2+ c

2 52"2x2 + 3 + c

3 12 loge (x2 + 4x + 29) + c

4 13 loge a∣ x3 + 9∣b + c

5 cosa1xb + c

6 12 sin(x2) + c

7 a 13esin(3x) + c b 2 sin(!x) + c

8 a 12etan(2x) + c b 1

8( loge (3x))2 + c

9 1

10 180

11 π2

24

12 π2

128

13 a 112

(x2 + 4)6 + c b 12 loge (x2 + 4) + c

c −1

2(x2 + 9)+ c d "x2 + 9 + c

14 a −1

6(x3 + 27)2+ c b 2

3"x3 + 27 + c

c 13 loge a∣ x3 + 8∣b + c d 1

12(x3 + 8)4 + c

15 a 18(x2 − 4x + 13)4 + c b

−1

2(x2 − 4x + 13)+ c

c −"x2 − 8x + 25 + c

d −12 loge (x2 − 8x + 25) + c

16 a 18

loge (4e2x + 5) + c b 1

6(2e−3x − 5)+ c

c 1

12(3e−2x + 4)2+ c d

−1

e2x + x+ c

17 a −cos(loge (4x)) + c b sin(loge (3x)) + c

c 12asin−1ax

2bb

2

+ c d 14(tan−1(2x))2 + c

18 a y = 12(1 − cos(x2)) b 1 − !3

c 12 loge a

2920

b d 12(1 − e)

19 a 12 b 14 loge a

53b c 1

18 d 0

20 a 12 loge a

2629

b b −14

c 2e(e − 1) d 12(e − 1)

21 a 12(e − 1) b 1

6

c 12(1 − !3) d 2(cos(1) − cos(2))

22 a c = 4, 12 loge a

215b b !2

4

c 12(1 − e−4) d 2

23 a 1a"ax2 + b + c b

−1

2a(ax2 + b)+ c

c 1

2a(n + 1)(ax2 + b)n+1 + c d

12a

loge a∣ax2 + b∣b + c

24 a loge a∣ f(x)∣b + c b − 1f(x)

+ c

c 2!f(x) + c d e f(x) + c

Topic 8 InTegral calculus 441Topic 8 InTegral calculus 441


UNCORRECTED PAGE P

ROOFS

ExErcisE 8.5

1 a x2

− sinax2b + c b −cosax

2b + c

2 a 4π − 3!3

12b

π24

− !3

128

3 124

sin6 (4x) + c

4 18

5 112

sin3(4x) − 110

sin5(4x) + 128

sin7(4x) + c

6 8 − 5!2

36

7 3x8

− 18

sin(4x) + 164

sin(8x) + c

8 π32

+ 112

9 2 loge a `secax2b ` b + c

10 14 loge (2)

11 3 tanax2b − x + c

12 4 − π

16

13 a −18 cos(4x) + c b x + c

c −18

cos4(2x) + c d 18

sin4(2x) + c

14 a 110

b 115

c π32

d 115

15 a x8

− 1128

sin(16x) + c

b 120

cos5(4x) − 112

cos3(4x) + c

c 112

sin3(4x) − 120

sin5(4x) + c

d 120

sin5(4x) − 128

sin7(4x) + c

16 a −12

cosec(2x) + c b 12

sec(2x) + c

c 14

sec2(2x) + c d −14

cosec2(2x) + c

17 a π12

b 29

c π16

d 845

18 a x − 14

cos(4x) + c

b 12(sin(2x) − cos(2x)) + 1

6(cos3(2x) − sin3(2x)) + c

c i 196

cos3(4x) − 132

cos(4x) + c1

ii 112

cos6(2x) − 18

cos4(2x) + c3

iii 18 sin4(2x) − 1

12 sin6(2x) + c3

19 a 13

loge a∣sec(3x)∣b + c b 13

loge a∣sin(3x)∣b + c

c 16

tan2(3x) + c d 19

tan3(3x) + c

20 a 110

loge (2) b 4 − π

20

c 120

d 875

21 a −1

a(n + 1) cosn+1(ax) + c

b 1

a(n + 1) sinn+1(ax) + c

c 1

a(n + 1) tann+1(ax) + c

22 a 1a

loge a∣sec(ax)∣b + c

b 1a

tan(ax) − x + c

c 12a

tan2(ax) + 1a

loge a∣cos(ax)∣b + c

d 13a

tan3(ax) − 1a

tan(ax) + x + c

23 a x2

− 14a

sin(2ax) + c

b 13a

cos3(ax) − 1a

cos(ax) + c

c 3x8

− 14a

sin(2ax) + 132a

sin(4ax) + c

d −1a

cos(ax) + 23a

cos3(ax) − 15a

cos5(ax) + c

24 a x2

+ 14a

sin(2ax) + c

b 1a

sin(ax) − 13a

sin3(ax) + c

c 3x8

+ 14a

sin(2ax) + 132a

sin(4ax) + c

d 1a

sin(ax) − 23a

sin3(ax) + 15a

sin5(ax) + c

ExErcisE 8.6

1 a sin−1a x10

b + c b 4 sin−1a3x8b + c

2 a 15

sin−1a5x6b + c b − 1

25"36 − 25x2 + c

3 −12

sin−1a4x5b = 1

4a2 cos−1a4x

5b − πb

4 2 cos−1ax2b + 3

5 a 110

tan−1a x10

b + c b 12

tan−1a3x8b + c

6 a 130

tan−1a5x6b + c b 1

10 loge (64 + 25x2) + c

7 a π24

b π

144

8 a π!6

36b

π!3

9

9 a 18

tan−1a2x − 34

b + c b 12

sin−1a2x − 34

b + c



UNCORRECTED PAGE P

ROOFS

10 a 115

tan−1a5x − 23

b + c b 15

sin−1a5x − 23

b + c

11 53

sin−1a3x5b + 1

3"25 − 9x2 + c

12 16

loge (9x2 + 25) + 13

tan−1a3x5b + c

13 a sin−1ax4b + c b 1

4 sin−1(4x) + c

c 2 sin−1a5x7b + c d −2

5"49 − 25x2 + c

14 a cos−1ax2b + c b cos−1(2x) + c

c 349"36 − 49x2 + c d

37

cos−1a7x6b + c

15 a 14

tan−1ax4b + c b 1

72 loge (36x2 + 49) + c

c 12

tan−1a6x7b + c d 2 tan−1(4x) + c

16 a 3 b π2

c π12

d 12 loge (2)

17 a y = sin−1ax2b + 2π

3b y = 1

2 tan−1(2x) + 7π

8

c π!3

18d

π4

18 a π18

b 118

loge (2)

c 59

d π60

19 a !3 π

9b

!3 π9

c 13

loge (2) d 13

20 a 2 sin−1ax + 2

3b + c b

23

tan−1ax + 2

3b + c

c 2 sin−1a3x + 5

7b + c d

27

tan−1a3x + 5

7b + c

21 a cos−1a4x3b − 3

16"9 − 16x2 + c

b 14

tan−1a4x3b + 1

8 loge (16x2 + 9) + c

c "5 − 2x2 + 5!2

2 sin−1a!10 x

5b + c

d !2

2 tan−1a!2 x

5b − 1

2 loge (2x2 + 25) + c

22 a 1q

sin−1aqx

pb + c b

1pq

tan−1aqx

pb + c

c bq

sin−1aqx

pb − a

q2"p2 − q2x2 + c

d bpq

tan−1aqx

pb + a

2q2 loge (q2x2 + p2) + c

23 a i 8 sin−1ax4b + x

2"16 − x2 + c

ii 4π; one-quarter of the area of a circle of radius 4

b i x2"36 − 25x2 − 18

5 cos−1a5x

6b + c

ii 9π5

; one-quarter of the area of an ellipse with

semi-minor and major axes of 65 and 6

24 a 12

loge (2) − π8

b 14

loge (2) + π8

c 9 + 5π2

d !3 − 5π3

ExErcisE 8.7

1 a 120

loge a `x + 10

x − 10` b + c

b 14

loge a `3x + 8

3x − 8` b + c

2 a 160

loge a `5x + 6

5x − 6` b + c

b − 150

loge a∣25x2 − 36∣b + c

3 loge a (x − 3)2

∣x + 5∣b + c

4 loge a ∣x + 4∣3

(x − 1)2b + c

5 5

x + 3+ 2 loge a∣x + 3∣b + c

6 29

loge a∣3x − 4∣b − 59(3x − 4)

+ c

7 3x + loge a (x − 2)4

∣x + 5∣3b + c

8 loge a (x − 2)2

∣x + 3∣b − 2x + c

9 loge a ∣x − 2∣"x2 + 9

b − 53

tan−1ax3b + c

10 loge a ∣x + 3∣"x2 + 16

b + 34

tan−1ax4b + c

11 a loge a (x − 3)2

∣x + 4∣b + c

b loge a (x − 5)2 ∣x + 3∣3b + c

c loge a ∣x + 3∣5

∣x − 2∣3b + c

d loge a ∣ x − 7x + 4 ∣ b + c

Topic 8 InTegral calculus 443Topic 8 InTegral calculus 443


UNCORRECTED PAGE P

ROOFS

12 a 14

loge a ∣ x − 2x + 2 ∣ b + c

b 12

tan−1ax4b + c

c 12 loge a∣x2 − 25∣b + c

d 14 loge a∣(x − 6)3(x + 6)5∣b + c

13 a 2 loge a∣x − 3∣b − 9x − 3

+ c

b 2 loge a∣x + 2∣b + 9x + 2

+ c

c loge a∣2x + 3∣b + 32x + 3

+ c

d 23

loge a∣3x − 5∣b + 33x − 5

+ c

14 a x − loge a∣(x − 3)2(x + 4)3∣b + c

b −3x + loge a `(x + 3)5

(x − 1)3` b + c

c 4x + loge a (x − 6)2

∣x + 2∣3b + c

d −x2 + loge a ∣x − 2∣(x − 4)2

b + c

15 a i 15

tan−1ax5b + c

ii 124

loge a ` x + 1

x + 25` b + c

iii − 1

x − 5+ c

b i 14

loge a `x − 2x − 1

` b + c

ii 1

2(3 − 2x)+ c

iii 18

tan−1a2x − 34

b + c

16 a 14 loge a

53b b 1

8 loge a

95b

c loge a253b d loge (2) − 2

17 a loge a94b b loge a

32b

c loge (8) d 10 + 32 loge a

73b

18 a !3 b 13 c 5 d 2!3

3

19 a i x + 54

loge a `4x − 54x + 5

` b + c

ii x − 52

tan−1a4x5b + c

b i x − 73

tan−1a6x7b + c

ii x + 76

loge a `6x − 76x + 7

` b + c

20 Check with your teacher.

21 a 2 loge (7) − 3 b 12 − 5 loge (5)

c 12

loge (3) − π18

d 29

+ 12 loge (3)

22 a 1

2ab loge a ∣bx − a∣

∣bx + a∣b + c

b 1

2b2 loge a∣b2x2 − a2∣b + c

c 1

a2 loge a∣ax − b∣b − b

a2(ax − b)+ c

d 1

aq − bp loge a ∣qx + b∣

∣ px + a∣b + c



UNCORRECTED PAGE P

ROOFS


UNCORRECTED PAGE P

ROOFS

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