AQA Further Paper 2 Specimen — Question 8 5 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
SessionSpecimen
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReduction Formulae
TypeTrigonometric power reduction
DifficultyStandard +0.8 This is a standard reduction formula derivation using integration by parts, a core Further Maths technique. While it requires careful algebraic manipulation and use of the identity cos²x + sin²x = 1, it follows a well-established method that students practice extensively. The 5 marks reflect moderate length rather than exceptional difficulty—it's above average but not requiring novel insight.
Spec1.08i Integration by parts4.08a Maclaurin series: find series for function
Given that \(I_n = \int_0^{\frac{\pi}{2}} \sin^n x \, dx\) \quad \(n \geq 0\) show that \(n I_n = (n-1)I_{n-2}\) \quad \(n \geq 2\) [5 marks]
Question 8:
AnswerMarks Guidance
8Selects suitable split in choosing u
and v using integration by partsAO3.1a M1
I  2sinnx dx
n
0
 2(sinn1x)(sin x) dx
0
du
usinn1x n1sinn2xcosx
dx
dv
sinx vcosx
dx
I cosxsinn1x2
n  
0
n12sinn2xcos2x dx
0
I 0n12sinn2x(1sin2x) dx
n
0
n1 2sinn2xsinnx dx
0
n1I I 
n2 n
I n1I nI I
n n2 n n
nI n1I
n n2
Uses integration by parts (allow one
AnswerMarks Guidance
sign error)AO1.1a M1
Integrates fully correctly (no need to
AnswerMarks Guidance
be simplified)AO1.1b A1
Uses the identity cos2x=1sin2x
AnswerMarks Guidance
in  cos2xsinn2x dx (PI)AO1.1b B1
Completes rigorous argument to
show result with all steps in the
AnswerMarks Guidance
argument clearly set out AGAO2.1 R1
QMarking Instructions AO
ALT
π
I 2sinn2xsin2x dx
n
0
 π 2sinn2x  1cos2x  dx
0
π
 I 2sinn2xcos2x dx
n2
0
π
 I 2(sinn2xcosx)cosx dx
n2
0
du
ucos x sin x
dx
dv 1
sinn2xcosx v sinn1x
dx n1
π
 1 2
I  I  sinn1xcos x
n n2 n1
 
0
π 1
2 sin xsinn2x dx
n1
0
1 π
 I  0  2sinnx dx
n2 n1
0
1
 I  I
n2 n1 n
n1I n1I I
n n2 n
nI n1I
n n2
AnswerMarks Guidance
Total5
QMarking Instructions AO 
Question 8:
8 | Selects suitable split in choosing u
and v using integration by parts | AO3.1a | M1 | 
I  2sinnx dx
n
0

 2(sinn1x)(sin x) dx
0
du
usinn1x n1sinn2xcosx
dx
dv
sinx vcosx
dx

I cosxsinn1x2
n  
0

n12sinn2xcos2x dx
0

I 0n12sinn2x(1sin2x) dx
n
0

n1 2sinn2xsinnx dx
0
n1I I 
n2 n
I n1I nI I
n n2 n n
nI n1I
n n2
Uses integration by parts (allow one
sign error) | AO1.1a | M1
Integrates fully correctly (no need to
be simplified) | AO1.1b | A1
Uses the identity cos2x=1sin2x
in  cos2xsinn2x dx (PI) | AO1.1b | B1
Completes rigorous argument to
show result with all steps in the
argument clearly set out AG | AO2.1 | R1
Q | Marking Instructions | AO | Marks | Typical Solution
ALT
π
I 2sinn2xsin2x dx
n
0
 π 2sinn2x  1cos2x  dx
0
π
 I 2sinn2xcos2x dx
n2
0
π
 I 2(sinn2xcosx)cosx dx
n2
0
du
ucos x sin x
dx
dv 1
sinn2xcosx v sinn1x
dx n1
π
 1 2
I  I  sinn1xcos x
n n2 n1
 
0
π 1
2 sin xsinn2x dx
n1
0
1 π
 I  0  2sinnx dx
n2 n1
0
1
 I  I
n2 n1 n
n1I n1I I
n n2 n
nI n1I
n n2
Total | 5
Q | Marking Instructions | AO | Marks | Typical Solution
Given that $I_n = \int_0^{\frac{\pi}{2}} \sin^n x \, dx$ \quad $n \geq 0$

show that $n I_n = (n-1)I_{n-2}$ \quad $n \geq 2$
[5 marks]

\hfill \mbox{\textit{AQA Further Paper 2  Q8 [5]}}
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