AQA Further Paper 2 2024 June — Question 20 9 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2024
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReduction Formulae
TypeDefinite integral with special limits
DifficultyChallenging +1.3 This is a standard reduction formula question requiring integration by parts to establish the recurrence relation (6 marks suggests multiple steps but follows a well-known technique), followed by routine application to find a specific integral. While it requires careful algebraic manipulation and is from Further Maths, the method is textbook-standard with no novel insight needed. The evaluation at π/4 adds minor complexity but remains procedural.
Spec4.08a Maclaurin series: find series for function
The integral \(I_n\) is defined by $$I_n = \int_0^{\frac{\pi}{4}} \cos^n x \, dx \quad\quad (n \geq 0)$$
  1. Show that $$I_n = \left(\frac{n-1}{n}\right)I_{n-2} + \frac{1}{n\left(2^{\frac{n}{2}}\right)} \quad\quad (n \geq 2)$$ [6 marks]
  2. Use the result from part (a) to show that $$\int_0^{\frac{\pi}{4}} \cos^6 x \, dx = \frac{a\pi + b}{192}$$ where \(a\) and \(b\) are integers to be found. [3 marks]
Question 20:
AnswerMarks
20(a)Uses
integration by
AnswerMarks Guidance
parts.3.1a M1
4 4
I = ∫cosnxdx= ∫cosn−1xcosxdx
n
0 0
u =cosn–1x v'=cosx
u'= – ( n–1 ) cosn–2 xsinx v=sinx
π
π 4
I =  sinxcosn–1x  4 + ( n–1 )∫cosn–2 xsin2 xdx
n
0
0
π
I = 1   1   n–1 + ( n–1 )∫ 4 cosn–2 x ( 1–cos2 x ) dx
n  
2 2
0
π π
I =   1   n + ( n–1 )∫ 4 cosn–2 x dx– ( n–1 )∫ 4 cosn x dx
n  
2
0 0
1
( ) ( )
I = + n–1 I – n–1 I
n n n–2 n
22
1
( )
nI = + n–1 I
n n n–2
22
n–1 1
I = I +
n  n  n–2  n 
n22 
 
Obtains a
correct result
of integration
AnswerMarks Guidance
by parts.1.1b A1
Substitutes
limits into first
expression
on RHS and
AnswerMarks Guidance
simplifies.1.1a M1
Uses a trig
identity to
obtain an
equation
involving I n
AnswerMarks Guidance
and I n–23.1a M1
Rearranges
to make I n
the subject of
a three term
equation
AnswerMarks Guidance
.1.1a M1
Completes a
rigorous
argument to
reach the
required
result.
AnswerMarks Guidance
AG2.1 R1
Subtotal6
QMarking instructions AO
AnswerMarks
20(b)π
Obtains and uses I =
0
AnswerMarks Guidance
41.1b B1
I =
0
4
1π 1
I =  + (2–1)
2 24 2
π 1
= +
8 4
3π 1 1
I =  + + (2–2)
4 48 4 4
3π 3 1
= + +
32 16 16
3π 1
= +
32 4
53π 1 1
I =  + + (2–3)
6 632 4 6
15π 5 1
= + +
192 24 48
15π+44
=
192
AnswerMarks Guidance
Uses the formula three times.1.1a M1
Completes fully correct working
15π+44
to obtain
AnswerMarks Guidance
1922.1 R1
Subtotal3
Question total9
Question paper total100 
Question 20:
--- 20(a) ---
20(a) | Uses
integration by
parts. | 3.1a | M1 | π π
4 4
I = ∫cosnxdx= ∫cosn−1xcosxdx
n
0 0
u =cosn–1x v'=cosx
u'= – ( n–1 ) cosn–2 xsinx v=sinx
π
π 4
I =  sinxcosn–1x  4 + ( n–1 )∫cosn–2 xsin2 xdx

n
0
0
π
I = 1   1   n–1 + ( n–1 )∫ 4 cosn–2 x ( 1–cos2 x ) dx
n  
2 2
0
π π
I =   1   n + ( n–1 )∫ 4 cosn–2 x dx– ( n–1 )∫ 4 cosn x dx
n  
2
0 0
1
( ) ( )
I = + n–1 I – n–1 I
n n n–2 n
22
1
( )
nI = + n–1 I
n n n–2
22
n–1 1
I = I +
n  n  n–2  n 
n22 
 
Obtains a
correct result
of integration
by parts. | 1.1b | A1
Substitutes
limits into first
expression
on RHS and
simplifies. | 1.1a | M1
Uses a trig
identity to
obtain an
equation
involving I n
and I n–2 | 3.1a | M1
Rearranges
to make I n
the subject of
a three term
equation
. | 1.1a | M1
Completes a
rigorous
argument to
reach the
required
result.
AG | 2.1 | R1
Subtotal | 6
Q | Marking instructions | AO | Marks | Typical solution
--- 20(b) ---
20(b) | π
Obtains and uses I =
0
4 | 1.1b | B1 | π
I =
0
4
1π 1
I =  + (2–1)
2 24 2
π 1
= +
8 4
3π 1 1
I =  + + (2–2)
4 48 4 4
3π 3 1
= + +
32 16 16
3π 1
= +
32 4
53π 1 1
I =  + + (2–3)
6 632 4 6
15π 5 1
= + +
192 24 48
15π+44
=
192
Uses the formula three times. | 1.1a | M1
Completes fully correct working
15π+44
to obtain
192 | 2.1 | R1
Subtotal | 3
Question total | 9
Question paper total | 100
The integral $I_n$ is defined by
$$I_n = \int_0^{\frac{\pi}{4}} \cos^n x \, dx \quad\quad (n \geq 0)$$

\begin{enumerate}[label=(\alph*)]
\item Show that
$$I_n = \left(\frac{n-1}{n}\right)I_{n-2} + \frac{1}{n\left(2^{\frac{n}{2}}\right)} \quad\quad (n \geq 2)$$
[6 marks]

\item Use the result from part (a) to show that
$$\int_0^{\frac{\pi}{4}} \cos^6 x \, dx = \frac{a\pi + b}{192}$$

where $a$ and $b$ are integers to be found.
[3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 2 2024 Q20 [9]}}
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