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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
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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?
2 Use CAS to � nd each of the following.
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?
3 Use CAS to � nd each of the following.
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?
4 Use CAS to � nd each of the following.
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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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.
2 Draw a diagram to identify the required area and shade this area.
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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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
1 Draw a diagram to identify the required area and shade this area.
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
4 Perform the integration using square bracket notation.
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.
382 MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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3 Perform the integration using square bracket notation.
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
1 Draw a diagram to identify the required area and shade this area.
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
3 Perform the integration using square bracket notation.
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
Topic 8 INTEGRAL CALCULUS 383
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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
1 Factorise the quadratic to � nd the x-intercepts.
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
384 MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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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
4 Perform the integration using square bracket notation.
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
Topic 8 INTEGRAL CALCULUS 385
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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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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
1 Factorise the quadratic to � nd the x-intercepts.
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
Topic 8 INTEGRAL CALCULUS 387
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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
5 Perform the integration using square bracket notation.
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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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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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
.
i Find the coordinates of the point of intersection between the graphs.
ii Determine the area between the curves, the y-axis and this point of intersection, giving your answer correct to 4 decimal places.
Master
390 Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
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26 a Consider the graphs of y = 23e
x
2 and y = 45 sina2x
3b + 42 for x ≥ 0.
i Find the coordinates of the point of intersection between the graphs.
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
Topic 8 INTEGRAL CALCULUS 391
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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
5 Express dx in terms of du by inverting both sides. dxdu
= 14
dx = 14
du
6 Substitute for dx. y = 3 u−
12 1
4du
7 Use the properties of inde� nite integrals to transfer the constant factor outside the front of the integral sign.
y = 143u
−12du
WORKED EXAMPLE 999
392 MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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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
4 Express dx in terms of du by inverting both sides. dxdu
= 15
dx = 15
du
5 Substitute for dx. 31
5x + 4 dx = 3
1u
× 15
du
6 Use the properties of inde� nite integrals to transfer the constant factor outside the front of the integral sign.
= 153
1u du
WORKED EXAMPLE 101010
Topic 8 INTEGRAL CALCULUS 393
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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
WORKED EXAMPLE 111111
394 MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 394 14/08/15 6:40 PM
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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
WORKED EXAMPLE 121212
Topic 8 INTEGRAL CALCULUS 395
c08IntegralCalculus.indd 395 14/08/15 6:15 PM
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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
396 Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
c08IntegralCalculus.indd 396 14/08/15 4:35 PM
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4 Substitute for 6x − 5, u and dx. 36x − 5
(2x − 3)2 dx = 3
3u + 4u2
× 12
du
5 Use the properties of inde� nite integrals to transfer the constant factor outside the front of the integral sign.
= 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
3 The integral cannot be done in this form, so differentiate. Express dx in terms of du by inverting both sides.
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.
WORKED EXAMPLE 131313
Defi nite integrals using a back substitution
Topic 8 INTEGRAL CALCULUS 397
c08IntegralCalculus.indd 397 14/08/15 6:15 PM
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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
11 State the � nal result. 3
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
WORKED EXAMPLE 141414
398 MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 398 14/08/15 4:36 PM
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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
Topic 8 InTegral calculus 399
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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
17 Integrate each of the following with respect to x.
a x(3x + 5)6 b x(3x + 5)2
c x(3x + 5)3
d x
"3 3x + 5
18 Find each of the following.
a 3x(6x + 7)8dx b 3x
!6x + 7 dx
c 3x
6x + 7 dx d 3
x(6x + 7)2
dx
ConsolidatE
400 Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
c08IntegralCalculus.indd 400 14/08/15 4:36 PM
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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
24 Given that a and b are non-zero real constants, find each of the following.
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
26 Given that a and b are non-zero real constants, find each of the following.
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
Topic 8 InTegral calculus 401
c08IntegralCalculus.indd 401 14/08/15 6:23 PM
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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
3 The integral cannot be done in this form, so differentiate. Express dx in terms of du by inverting both sides.
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
WORKED EXAMPLE 151515
AOS 3
Topic 2
Concept 3
Integration by substitutionConcept summaryPractice questions
402 MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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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
WORKED EXAMPLE 161616
Topic 8 INTEGRAL CALCULUS 403
c08IntegralCalculus.indd 403 14/08/15 6:17 PM
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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
WORKED EXAMPLE 171717
404 MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 404 14/08/15 6:17 PM
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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
Topic 8 InTegral calculus 405
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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
WORKED EXAMPLE 191919
406 MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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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
2 The integral cannot be done in this form, so differentiate. Express dx in terms of du by inverting both sides.
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
WORKED EXAMPLE 202020
Topic 8 INTEGRAL CALCULUS 407
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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
408 Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
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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
Topic 8 InTegral calculus 409
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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
410 Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
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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
WORKED EXAMPLE 212121
Topic 8 INTEGRAL CALCULUS 411
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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
WORKED EXAMPLE 222222
412 MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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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
WORKED EXAMPLE 232323
Find an antiderivative of cos4(2x).
THINK WRITE
1 Write the required antiderivative. 3cos4(2x)dx
WORKED EXAMPLE 242424
Topic 8 INTEGRAL CALCULUS 413
c08IntegralCalculus.indd 413 14/08/15 4:38 PM
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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
WORKED EXAMPLE 252525
414 MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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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
WORKED EXAMPLE 262626
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
Topic 8 INTEGRAL CALCULUS 415
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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.
13 Find each of the following.
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
14 Evaluate each of the following.
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
15 Find each of the following.
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
416 Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
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17 Evaluate each of the following.
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).
19 Find each of the following.
a 3tan(3x)dx b 3cot(3x)dx
c 3tan(3x)sec2(3x)dx d 3tan2(3x)sec2(3x)dx
20 Evaluate each of the following.
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)
23 Integrate each of the following where a ∈ R\{0}.
a sin2(ax) b sin3(ax) c sin4(ax) d sin5(ax)
24 Integrate each of the following where a ∈ R\{0}.
a cos2(ax) b cos3(ax) c cos4(ax) d cos5(ax)
MastEr
topic 8 IntegraL CaLCuLus 417
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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
418 Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
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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
5 Use the properties of inde� nite integrals to transfer the constant factor outside the front of the integral sign.
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
WORKED EXAMPLE 282828
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.
Topic 8 INTEGRAL CALCULUS 419
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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
3 Substitute for dx and u. 34
49 + 36x2 dx = 3
449 + u2
× 16
du
4 Use the properties of inde� nite integrals to transfer the constant factor outside the front of the integral sign.
= 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
6 Substitute back for u and express the � nal answer in terms of x only and an arbitrary constant +c.
34
49 + 36x2 dx = 2
21 tan−1a6x
7b + c
WORKED EXAMPLE 292929
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.
WORKED EXAMPLE 303030
420 MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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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
2 Change the terminals to the new variable. When x = 29, u = 2, and when x = 0, u = 0.
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
4 Transfer the constant factor outside the front of the integral sign.
= 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
2 Change the terminals to the new variable. When x = 49, u = 4, and when x = 0, u = 0.
Topic 8 InTegral calculus 421
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ROOFS
3 Substitute for dx, u and the new terminals. 3
4
9
0
216 + 81x2
dx
= 3
4
0
216 + u2
× 19
du
4 Transfer the constant factor outside the front of the integral sign.
= 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
7 State the � nal result. 3
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
WORKED EXAMPLE 313131
422 MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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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
4 Substitute for dx and u. 31
(3x + 2)2 + 25 dx = 3
1u2 + 25
× 13
du
5 Transfer the constant factor outside the front of the integral sign.
= 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
4 Substitute for dx and u. 31
"25 − (3x + 2)2 dx
= 31
"25 − u2 × 1
3 du
Topic 8 InTegral calculus 423
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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.
5 Transfer the constant factor outside the front of the integral sign.
= 133
1
"25 − u2 du
6 Use 31
"a2 − u2= sin−1au
ab + c with a = 5. = 1
3 sin−1au
5b + c
7 Substitute back for u and express the � nal answer in terms of x only and an arbitrary constant +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
WORKED EXAMPLE 323232
424 MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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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.
2 Integrate each of the following with respect to x.
a 1
"36 − 25x2b x
"36 − 25x2
ExErcisE 8.6
PractisE
Topic 8 InTegral calculus 425
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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
6 Integrate each of the following with respect to x.
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.
13 Integrate each of the following with respect to x.
a 1
"16 − x2b 1
"1 − 16x2c 10
"49 − 25x2d 10x
"49 − 25x2
14 Find each of the following.
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
426 Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
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15 Find a set of antiderivatives for each of the following.
a 116 + x2
b x49 + 36x2
c 2149 + 36x2
d 81 + 16x2
16 Evaluate each of the following.
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.
18 Evaluate each of the following.
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
19 Evaluate each of the following.
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
20 Find each of the following.
a 32
"5 − 4x − x2 dx b 3
2x2 + 4x + 13
dx
c 36
"24 − 30x − 9x2 dx d 3
674 + 30x + 9x2
dx
21 Find each of the following.
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
Topic 8 InTegral calculus 427
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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.
24 Evaluate each of the following.
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
428 Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
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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
WORKED EXAMPLE 333333
AOS 2
Topic 1
Concept 1
Partial fractions with a quadratic denominator that has linear factorsConcept summaryPractice questions
Topic 8 INTEGRAL CALCULUS 429
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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
12 Integrate term by term, using
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)
2 Write the integrand in its partial fractions decomposition, where A and B are constants to be found.
= A7 − 6x
+ B7 + 6x
3 Add the fractions by forming a common denominator.
= A(7 + 6x) + B(7 − 6x)(7 − 6x)(7 + 6x)
4 Expand the numerator, factor in x and expand the denominator.
= 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
7 Express the integrand as its partial fractions decomposition.
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
430 Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
c08IntegralCalculus.indd 430 14/08/15 4:43 PM
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ROOFS
9 Integrate term by term, using
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)
2 Write the integrand in its partial fractions decomposition, where A and B are constants to be found.
= Ax − 4
+ Bx + 3
3 Add the fractions by forming a common denominator.
=A(x + 3) + B(x − 4)
(x − 4)(x + 3)
4 Expand the numerator, factor in x and expand the denominator.
= 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
WORKED EXAMPLE 343434
Topic 8 INTEGRAL CALCULUS 431
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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
11 Express the integrand as its partial fractions decomposition.
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
2 Write the integrand in its partial fractions decomposition, where A and B are constants to be found.
= 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
6 Express the integrand as its partial fractions decomposition.
36x − 5
4x2 − 12x + 9 dx
= 3a 32x − 3
+ 4(2x − 3)2
bdx
WORKED EXAMPLE 353535
432 MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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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)
3 Write the integrand in its partial fractions decomposition, where A and B are constants to be found.
= 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
WORKED EXAMPLE 363636
Topic 8 INTEGRAL CALCULUS 433
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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
WORKED EXAMPLE 373737
6 Solve the simultaneous equations by elimination. Add the two equations to eliminate B.
5A = 10 A = 2 ⇒ B = −3
7 Express the integrand as its partial fractions decomposition.
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
434 MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 434 14/08/15 6:21 PM
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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
8 Express the integrand as its partial fractions decomposition.
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
Topic 8 InTegral calculus 435
c08IntegralCalculus.indd 435 14/08/15 6:22 PM
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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.
11 Integrate each of the following with respect to x.
a x + 11
x2 + x − 12b 5x − 9
x2 − 2x − 15c 2x − 19
x2 + x − 6d 11
x2 − 3x − 28
12 Find an antiderivative of each of the following.
a 1x2 − 4
b 216 + x2
c xx2 − 25
d 2x − 3x2 − 36
13 Integrate each of the following with respect to x.
a 2x + 3
x2 − 6x + 9b 2x − 5
x2 + 4x + 4
c 4x4x2 + 12x + 9
d 6x − 199x2 − 30x + 25
14 Find an antiderivative of each of the following.
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
c08IntegralCalculus.indd 436 14/08/15 4:44 PM
UNCORRECTED PAGE P
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16 Evaluate each of the following.
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.
Topic 8 InTegral calculus 437
c08IntegralCalculus.indd 437 14/08/15 4:44 PM
UNCORRECTED PAGE P
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21 Evaluate each of the following.
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
c08IntegralCalculus.indd 438 14/08/15 4:44 PM
UNCORRECTED PAGE P
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studyON is an interactive and highly visual online tool that helps you to clearly identify strengths and weaknesses prior to your exams. You can then con� dently target areas of greatest need, enabling you to achieve your best results.
ONLINE ONLY 8.8 Review www.jacplus.com.au
The Maths Quest Review is available in a customisable format for you to demonstrate your knowledge of this topic.
The Review contains:• Short-answer questions — providing you with the
opportunity to demonstrate the skills you have developed to ef� ciently answer questions using the most appropriate methods
• Multiple-choice questions — providing you with the opportunity to practise answering questions using CAS technology
• Extended-response questions — providing you with the opportunity to practise exam-style questions.
A summary of the key points covered in this topic is also available as a digital document.
REVIEW QUESTIONSDownload the Review questions document from the links found in the Resources section of your eBookPLUS.
Topic 8 INTEGRAL CALCULUS 439
Units 3 & 4 Integral calculus
Sit topic test
c08IntegralCalculus.indd 439 14/08/15 4:45 PM
UNCORRECTED PAGE P
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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
440 Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
c08IntegralCalculus.indd 440 14/08/15 4:46 PM
UNCORRECTED PAGE P
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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
a2 loge a∣ax + b∣b + c
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
a3 loge a∣ax + b∣b + c
b x(xa + 2b)
a2(ax + b)− 2b
a3 loge a∣ax + b∣b + c
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
c08IntegralCalculus.indd 441 14/08/15 5:20 PM
UNCORRECTED PAGE P
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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
442 Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
c08IntegralCalculus.indd 442 14/08/15 6:22 PM
UNCORRECTED PAGE P
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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
c08IntegralCalculus.indd 443 14/08/15 5:21 PM
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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
444 Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
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