1. No category

HSC Mathematics Ext. 2 (4 Unit)
SAMPLE LECTURE SLIDES
HSC Exam Preparation Programs
22-26 September, 2014
c 2014 Sci SchoolTM . All rights reserved.
Overview
1. Graphs & Curve
Sketching
2. Conics
3. Polynomials
4. Complex Numbers
1. Graphs & Curve Sketching
2. Conics
3. Polynomials
5. Complex Locus
Problems
4. Complex Numbers
6. Integration
5. Complex Locus Problems
7. Volumes
8. Mechanics
6. Integration
9. Harder 4 Unit
Practice
7. Volumes
8. Mechanics
9. Harder 4 Unit Practice
© 2014 Sci School
1. Graphs & Curve
Sketching
2. Conics
3. Polynomials
4. Complex Numbers
5. Complex Locus
Problems
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6. Integration
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
© 2014 Sci School
6.1 Integration by Substitution
1. Graphs & Curve
Sketching
2. Conics
3. Polynomials
4. Complex Numbers
5. Complex Locus
Problems
This technique is derived from the Fundamental Theorem of Calculus, which
states that, for continuous functions f (x) = F (x) and u(x),
b
(F (u)) dx = F (u(b)) − F (u(a))
a
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
u(b)
f (u) du.
=
u(a)
The Chain Rule tell us that
(F (u)) = f (u)u .
Hence, we arrive at the rule for Integration by Substitution:
b
u(b)
f (u)u dx =
a
f (u) du
u(a)
© 2014 Sci School
6.1 Integration by Substitution
1. Graphs & Curve
Sketching
The original integral (in x) is transformed to an equivalent integral in u.
2. Conics
3. Polynomials
x→u
4. Complex Numbers
5. Complex Locus
Problems
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
For example, solve
•
a
b
1
√
−1
x
u(a)
u(b)
u
x 2x + 1 dx using the substitution u = 2x + 3.
Step 1: Find an expression for u .
u = 2x + 3 =⇒ u = 2
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
•
Step 2: Substitute every x with u in the integrand.
√
√
1
(u − 3) u
x 2x + 3 −→
2
© 2014 Sci School
6.1 Integration by Substitution
1. Graphs & Curve
Sketching
•
2. Conics
u(−1) = 2(−1) + 3 = 1 and u(1) = 2(1) + 3 = 5
3. Polynomials
4. Complex Numbers
5. Complex Locus
Problems
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
Step 3: Evaluate the transformed limits.
•
Step 4: Transform the integral and solve.
f (u)
u
√
√
1
1
(u − 3) u (2) dx
x 2x + 3 dx =
2
2
−1
−1
1
1 5
=
(u − 3)u 2 du
4 1
1
1 5 3
u 2 − 3u 2 du
=
4 1
5
3
1 2 52
u − 2u 2
=
4 5
1
2
=
5
1
1
© 2014 Sci School
6.2 Integration By Parts
1. Graphs & Curve
Sketching
2. Conics
3. Polynomials
4. Complex Numbers
5. Complex Locus
Problems
This method is best applied to integrals of products of functions. The
theorem is derived from the Product Rule,
b
b
b
(uv) = uv + vu =⇒ [uv]a =
uv dx +
vu dx
a
a
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
Rearranging, we arrive at the formula for Integration by Parts.
a
b
uv dx =
[uv]ba
b
−
vu dx
a
To guide our choice for u and v , we use the DETAIL Rule:
v Dashed should be tried in order of Exponentials, Trig functions,
Algebraic functions, Inverse trig functions, then Logarithms.
© 2014 Sci School
6.2 Integration by Parts
1. Graphs & Curve
Sketching
2. Conics
3. Polynomials
For example, by integration by parts to evaluate
•
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
v = cos 3x
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
•
(1 − x) cos 3x dx.
=⇒
1
v = sin 3x
3
Step 2: Identify u and differentiate to find u .
u=1−x
•
0
Step 1: Identify v and integrate to find v.
4. Complex Numbers
5. Complex Locus
Problems
π
=⇒
u = −1
Step 3: Use the formula for integration by parts to rewrite the integral.
π π
π
1
1
sin 3x
sin 3x(−1) dx
(1 − x) cos 3x dx = (1 − x)
−
3
3
0
0
0
π
1
=0+
sin 3x dx
3 0
π
1
2
1
− cos 3x =
=
3
3
9
0
© 2014 Sci School
6.3 Recurrence Relationships
1. Graphs & Curve
Sketching
2. Conics
3. Polynomials
Sometimes several applications of the Integration by Parts formula is needed
to evaluate an integral. This often yields recurring expressions, which can be
used to quickly get to the solution.
4. Complex Numbers
5. Complex Locus
Problems
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
7. Volumes
8. Mechanics
For example, evaluate
e2x sin x dx.
v = sin x =⇒ v = − cos x
u = e2x =⇒ u = 2e2x
Using one application of integration by parts,
2x 2x
2x
dx
e sin x dx = e (− cos x) − (− cos x) 2e
= −e2x cos x + 2 e2x cos x dx
9. Harder 4 Unit
Practice
© 2014 Sci School
6.3 Recurrence Relationships
1. Graphs & Curve
Sketching
Applying integration by parts a second time,
2. Conics
3. Polynomials
4. Complex Numbers
5. Complex Locus
Problems
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
∴
v = cos x =⇒ v = sin x
e2x sin x dx = −e2x cos x + 2 e2x sin x − sin x 2e2x dx
= −e2x cos x + 2e2x sin x − 4 e2x sin x dx
Noticing that e2x sin x dx is a recurrence of the original integral, we can
move it to the left-hand side.
5 e2x sin x dx = −e2x cos x + 2e2x sin x
e2x
2x
(2 sin x − cos x) + C
∴ e sin x dx =
5
© 2014 Sci School
6.3 Recurrence Relationships
1. Graphs & Curve
Sketching
2. Conics
3. Polynomials
4. Complex Numbers
5. Complex Locus
Problems
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
Another powerful example occurs when u takes the form of an algebraic (or
power) function, e.g. xn .
n x
Consider applying integration by parts to In = x e dx.
v = ex =⇒ v = ex and u = xn =⇒ u = nxn−1
∴ xn ex dx = xn ex − n xn−1 ex dx
Written with different notation, we have
In = xn ex − nIn−1
For example,
x4 ex dx can be evaluated using only algebra, since
I4 = x4 ex − 4I3
3 x
2 x
4 x
x
x
= x e − 4 x e − 3 x e − 2 (xe − e ) + C
4
x
3
2
= e x − 4x + 12x − 24x + 24 + C
© 2014 Sci School
6.4 Rational Functions
1. Graphs & Curve
Sketching
2. Conics
3. Polynomials
4. Complex Numbers
When our integral involves the quotient of two functions, we can use the
following approaches:
•
Technique 1: Polynomial division, then integration term-by-term.
5. Complex Locus
Problems
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
•
Technique 2: Algebraic rearrangement to the form
f (x)
dx = ln |f (x)| + C
f (x)
•
Technique 3: Algebraic rearrangement to the form
f (x)
dx, then
f (x)
1
dx, then
2
2
x +a
x
1
1
−1
+C
dx
=
tan
x2 + a 2
a
a
© 2014 Sci School
6.4 Rational Functions
1. Graphs & Curve
Sketching
For example, use Techniques 2 and 3 to evaluate
2x−3
x2 −4x+5
dx.
2. Conics
3. Polynomials
4. Complex Numbers
5. Complex Locus
Problems
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
f (x)
2x − 4 +1
2x − 3
dx
=
dx
2
2
x − 4x + 5
x − 4x + 5
1
2x − 4
dx
+
dx
=
2
2
x − 4x + 5
x − 4x + 5
1
dx
= log |x2 − 4x + 5| +
2
x − 4x + 4 + 1
1
dx
= log |x2 − 4x + 5| +
2
(x − 2) + 1
= log |x2 − 4x + 5| + tan−1 (x − 1) + C
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
© 2014 Sci School
6.5 Partial Fractions
1. Graphs & Curve
Sketching
2. Conics
Partial fractions is anextension of the previous techniques, applied to
P (x)
dx.
integrals of the form Q(x)R(x)
3. Polynomials
4. Complex Numbers
5. Complex Locus
Problems
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
The fraction is expanded into simpler (partial) fractions, which are then
integrated term-by-term.
P (x)
dx =
Q(x)R(x)
A(x)
dx +
Q(x)
B(x)
dx
R(x)
For example, use the method of partial fractions to evaluate
2x−1
x2 −5x+6
dx.
• Step 1: Factorise the denominator.
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
2x − 1
2x − 1
=
x2 − 5x + 6
(x − 2)(x − 3)
© 2014 Sci School
6.4 Partial Fractions
1. Graphs & Curve
Sketching
•
2. Conics
2x − 1
A
B
=
+
(x − 2)(x − 3)
(x − 2) (x − 3)
3. Polynomials
4. Complex Numbers
5. Complex Locus
Problems
=⇒ 2x − 1 = A(x − 3) + B(x − 2)
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
Step 2: Split into partial fractions.
•
Step 3: Substitute well-chosen values of x to find A and B.
x = 2 : 2(2) − 1 = A(2 − 3) + 0
∴ A = −3
•
x = 3 : 2(3) − 1 = 0 + B(3 − 2)
∴ B=5
Step 4: Integrate term-by-term.
3
5
2x − 1
dx =
−
+
dx
(x − 2)(x − 3)
x−2 x−3
= −3 ln |x − 2| + 5 ln |x − 3| + C
© 2014 Sci School
6.6 t Formulae Substitution
1. Graphs & Curve
Sketching
2. Conics
A special category of Integration by Substitution is to use t = tan
solve integrals of the form,
3. Polynomials
4. Complex Numbers
5. Complex Locus
Problems
x
2
to
1
dx
a cos x + b sin x + c
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
Defining t = tan
x
2
, our essential substitution formulae are:
2 dt
1 + t2
sin x =
2t
1 + t2
1 − t2
cos x =
1 + t2
tan x =
2t
1 − t2
dx =
© 2014 Sci School
6.6 t Formulae Substitution
1. Graphs & Curve
Sketching
The first formula is derived using calculus as follows.
2. Conics
3. Polynomials
4. Complex Numbers
5. Complex Locus
Problems
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
x dt
d =
tan
dx
dx
2
1
2 x
= sec
2
2
x
1
tan2
+1
=
2
2
12
t +1
=
2
Separating the variables, we arrive at
12
dt =
t + 1 dx
2
or
dx =
2 dt
1 + t2
© 2014 Sci School
6.6 t Formulae Substitution
1. Graphs & Curve
Sketching
The remaining formulae are derived using geometry & double-angle identities.
2. Conics
3. Polynomials
4. Complex Numbers
5. Complex Locus
Problems
Consider a right-angled triangle consistent with t = tan θ2 . With respect to
the angle θ2 , the ratio of opposite & adjacent sides is t : 1. Hence, by
√
Pythagorus’ Theorem, the hypotenuse is 1 + t2 .
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
√
t
1 + t2
θ
2
1
Using double-angle identities, we can express
sin θ in terms of t,
θ
θ
sin θ = 2 sin
cos
2
2
t
1
=2· √
·√
1 + t2
1 + t2
2t
=
1 + t2
© 2014 Sci School
6.6 t Formulae Substitution
1. Graphs & Curve
Sketching
2. Conics
3. Polynomials
4. Complex Numbers
5. Complex Locus
Problems
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
Likewise for cos θ and tan θ,
θ
θ
2
2
− sin
cos θ = cos
2
2
1
t2
=
−
1 + t2
1 + t2
1 − t2
=
1 + t2
sin θ
cos θ
2t
1 + t2
=
×
2
1+t
1 − t2
2t
=
1 − t2
tan θ =
© 2014 Sci School
6.6 t Formulae Substitution
1. Graphs & Curve
Sketching
2. Conics
The t-formulae are used by substituting all expressions in x, e.g. dx, sin x,
cos x, and tan x, for corresponding expressions in t.
3. Polynomials
4. Complex Numbers
5. Complex Locus
Problems
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
For example,
1
dx =
3 + 2 cos x
1
3+2
·
2
1−t
1+t2
2 dt
t2 + 1
1
dt
=2
3(1 + t2 ) + 2(1 − t2 )
1
=2
dt
2
5+t
t
1
=2 √
tan−1 √
+C
5
5
x 1
2
+C
= √ tan−1 √ tan
2
5
5
© 2014 Sci School
6.7 HSC-Adapted Questions
1. Graphs & Curve
Sketching
Question 1 (3 Marks)
2. Conics
Evaluate
3. Polynomials
e2
2
dx.
x ln x
4. Complex Numbers
5. Complex Locus
Problems
e
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
Solution
Rewrite the integral in the form of
e
e2
2
dx = 2
x ln x
=
f (x)
f (x)
e
e2
dx.
1
x
ln x
dx
e2
2 [ln |ln x|]e
2
= 2 ln ln e − ln |ln e|
= 2 (ln 2 − ln 1)
= 2 ln 2
© 2014 Sci School
6.7 HSC-Adapted Questions
1. Graphs & Curve
Sketching
Question 2 (4 Marks)
2. Conics
Solve
3. Polynomials
2
4. Complex Numbers
4
x2
dx.
2
12 + 9x
5. Complex Locus
Problems
Solution
6. Integration
1
Rewrite the integral in the form of x2 +a
2 dx.
4 2
4
2
x
9x + 12 − 12
1
dx
=
dx
2
2
9 2
12 + 9x
2 12 + 9x
4
4
1
1−
dx
=
2
9 2
4 + 3x
√ 4
1
3x
2
=
x − √ tan−1
9
2
3
2
√
√
√
2
1 − 3 3 tan−1 2 3 + 3π
=
81
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
© 2014 Sci School
6.7 HSC-Adapted Questions
1. Graphs & Curve
Sketching
Question 3 (4 Marks)
2. Conics
Evaluate
3. Polynomials
4. Complex Numbers
5. Complex Locus
Problems
√
3
3
1
x2
√
1
dx.
2
x +1
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
Solution
Use integration by substitution, with x = tan u.
dx
= sec2 u
du
∴ dx = sec2 u du
Substitute every x with u in the integrand.
1
1
1
√
√ 2
−→
=
2
2
2
tan2 u sec u
x x +1
tan u tan u + 1
© 2014 Sci School
6.7 HSC-Adapted Questions
1. Graphs & Curve
Sketching
−1
When x = 1, u = tan
(1) =
π
4.
When x =
√
3
3 ,
2. Conics
3. Polynomials
4. Complex Numbers
5. Complex Locus
Problems
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
π
6.
Transforming the integral, we have
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
u = tan−1 ( √13 ) =
1
√
3
3
1
√
dx =
2
2
x x +1
π
6
π
4
1
2
·
sec
u du
tan2 u sec u
π
6
cos u
du
2
π
sin u
4
π
−1 6
= (− sin u)
π
=
4
=−
sin
1
π +
6
√
= 2−2
sin
1
π
4
© 2014 Sci School
6.7 HSC-Adapted Questions
1. Graphs & Curve
Sketching
Question 4 (5 Marks)
2. Conics
Using the substitution t = tan x2 , or otherwise, show that
√
2π
3
1
3
dx = loge
.
π
sin x + tan x
e
2
3. Polynomials
4. Complex Numbers
5. Complex Locus
Problems
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
Solution
We need the following three t formulae:
dx =
2 dt
,
2
1+t
sin x =
2t
2t
,
and
tan
x
=
.
2
2
1+t
1−t
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
Also, when x =
π
2,
t = tan
π
4
= 1. When x =
2π
3 ,
t = tan
π
3
√
= 3.
© 2014 Sci School
6.7 HSC-Adapted Questions
1. Graphs & Curve
Sketching
2. Conics
3. Polynomials
4. Complex Numbers
5. Complex Locus
Problems
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
2π
3
π
2
1
dx =
sin x + tan x
1
√
3
√
3
2t
1+t2
1
+
2t
1−t2
2 dt
·
1 + t2
(1 + t2 )(1 − t2 )
2 dt
=
·
2t − 2t3 + +2t + 2t3 1 + t2
1
√3 2
t
1
−
dt
=2
4t 4t
1
√3
1
1
ln |t| − t2
=
2
2 1
1 √
ln 3 − 1
=
2
1 √
ln 3 − ln e
=
2
√ 12
3
, as required.
= ln
e
© 2014 Sci School
6.7 HSC-Adapted Questions
1. Graphs & Curve
Sketching
Question 5 (7 Marks)
2. Conics
Let
3. Polynomials
cos2k+1 x dx,
Ik =
4. Complex Numbers
5. Complex Locus
Problems
π
0
where k is an integer, k ≥ 0.
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
(i) Show that
Ik = −π + (2k + 1)
9. Harder 4 Unit
Practice
x sin x cos2k x dx
0
for k ≥ 1. (3 Marks)
(ii) Show also that
Ik =
7. Volumes
8. Mechanics
π
2k
Ik−1
2k − 1
for k ≥ 1. (3 Marks)
(ii) Explain why Ik = 0 for k ≥ 1. (1 Mark)
© 2014 Sci School
6.7 HSC-Adapted Questions
1. Graphs & Curve
Sketching
Solution
2. Conics
(i) Using integration by parts, we have
3. Polynomials
4. Complex Numbers
5. Complex Locus
Problems
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
v = 1 =⇒ v = x
u = cos2k+1 x =⇒ u = (2k + 1) cos2k x(− sin x)
Substituting these values into the integration by parts formula,
π
π
Ik = x cos2k+1 x 0 −
x(2k + 1) cos2k x(− sin x) dx
0
π
= π · (−1)2k+1 + (2k + 1)
x sin x cos2k x dx
0
π
= −π + (2k + 1)
x sin x cos2k x dx,
0
since 2k + 1 is odd for all integers k ≥ 0.
© 2014 Sci School
6.7 HSC-Adapted Questions
1. Graphs & Curve
Sketching
2. Conics
3. Polynomials
4. Complex Numbers
5. Complex Locus
Problems
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
(ii) Rewrite the integral as
π
0
cos2k x cos x dx and use integration by parts.
v = cos x =⇒ v = − sin x
u = cos2k x =⇒ u = 2k cos2k−1 x(− sin x)
π
=⇒ Ik =
cos2k x cos x dx
0
π
π
= − cos2k x sin x 0 −
(− sin x)2k cos2k−1 x(− sin x) dx
0
π
= −2k
cos2k−1 x(1 − cos2 x) dx
0
= −2kIk−1 + 2kIk
∴ (1 − 2k)Ik = −2kIk−1
2k
Ik−1 , as required.
Ik =
2k − 1
© 2014 Sci School
6.7 HSC-Adapted Questions
1. Graphs & Curve
Sketching
(iii) Using Part (ii), we have
2. Conics
2k
Ik−1
Ik =
2k − 1
3. Polynomials
4. Complex Numbers
5. Complex Locus
Problems
for k ≥ 1.
6. Integration
6.1 Integration by
Substitution
6.2 Integration By
Parts
6.3 Recurrence
Relationships
6.4 Rational
Functions
6.5 Partial Fractions
6.6 t Formulae
Substitution
6.7 HSC-Adapted
Questions
Substituting k = 1 gives us,
2
I0
I1 =
2−1
π
=
cos x dx
0
= 0.
7. Volumes
8. Mechanics
9. Harder 4 Unit
Practice
Since Ik is a multiple of Ik−1 for all integers k ≥ 1, Ik is a multiple of zero.
Hence, Ik = 0 for all integers k ≥ 1.
© 2014 Sci School