OCR Further Additional Pure 2021 November — Question 7 8 marks

Exam BoardOCR
ModuleFurther Additional Pure (Further Additional Pure)
Year2021
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReduction Formulae
TypeTrigonometric power reduction
DifficultyChallenging +1.2 This is a standard reduction formula question from Further Maths requiring integration by parts to derive the formula, then recursive application to find I₈, and finally a substitution using sin²x = 1-cos²x. While it involves multiple steps and is from an advanced module, the techniques are well-practiced and follow predictable patterns without requiring novel insight.
Spec1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)8.06a Reduction formulae: establish, use, and evaluate recursively
7 Let \(\mathrm { I } _ { \mathrm { n } } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \cos ^ { \mathrm { n } } \mathrm { xdx }\) for integers \(n \geqslant 0\).
  1. Show that, for \(n \geqslant 2 , \quad \mathrm { nl } _ { \mathrm { n } } = ( \mathrm { n } - 1 ) \mathrm { I } _ { \mathrm { n } - 2 }\).
  2. Use this reduction formula to deduce the exact value of \(I _ { 8 }\).
  3. Use the results of parts (a) and (b) to determine the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \cos ^ { 6 } x \sin ^ { 2 } x d x\).
Question 7:
AnswerMarks Guidance
7(a)  12  12
I = n  c o s n x dx =  − c o s n 1 x c o s x dx
0 0
1
c  2
= o s n − 1 x s in x − (n−1)cosn − 2 x.(−sin x) sin xdx
0
1
2 ( )
= 0 + (n−1)cosn − 2 x1−cos2 x dx = (n – 1){I n – 2 – I n }
0
 n I = (n – 1) I
AnswerMarks
n n – 2M1
A1
M1
A1
AnswerMarks
[4]1.1
1.1
3.1a
AnswerMarks
1.1Correct splitting and attempted use of ‘parts’
Correct unsimplified
Setting up into I terms only
k
AG legitimately obtained
AnswerMarks
(b)1  n − 1 
I = π evaluated and I = I used repeatedly
0 n n – 2
2 n
1 3 5 35
gives I = π, I = π, I = π, I = π
2 4 6 8
AnswerMarks
4 16 32 256M1
A1
AnswerMarks
[2]1.1
1.1Note: candidates may choose to evaluate I = 14 π
2
directly as a starting-point
AnswerMarks
(c) 12
 c o s 6 x s in 2 x dx = {I – I } 6 8
0
5 35 5
= π− π = π
AnswerMarks
32 256 256M1
A1
AnswerMarks
[2]3.1a
2.2aConverting into I terms and using the previous results
k
from (b)
NB Missing factor of 2 and/or incorrect sign penalised
only here at the end
FT I – I = 1 I = 1 (answer fom (b))
6 8 8
7 7
Question 7:
7 | (a) |  12  12
I = n  c o s n x dx =  − c o s n 1 x c o s x dx
0 0
1
c  2
= o s n − 1 x s in x − (n−1)cosn − 2 x.(−sin x) sin xdx
0
1
2 ( )
= 0 + (n−1)cosn − 2 x1−cos2 x dx = (n – 1){I n – 2 – I n }
0
 n I = (n – 1) I
n n – 2 | M1
A1
M1
A1
[4] | 1.1
1.1
3.1a
1.1 | Correct splitting and attempted use of ‘parts’
Correct unsimplified
Setting up into I terms only
k
AG legitimately obtained
(b) | 1  n − 1 
I = π evaluated and I = I used repeatedly
0 n n – 2
2 n
1 3 5 35
gives I = π, I = π, I = π, I = π
2 4 6 8
4 16 32 256 | M1
A1
[2] | 1.1
1.1 | Note: candidates may choose to evaluate I = 14 π
2
directly as a starting-point
(c) |  12
 c o s 6 x s in 2 x dx = {I – I } 6 8
0
5 35 5
= π− π = π
32 256 256 | M1
A1
[2] | 3.1a
2.2a | Converting into I terms and using the previous results
k
from (b)
NB Missing factor of 2 and/or incorrect sign penalised
only here at the end
FT I – I = 1 I = 1 (answer fom (b))
6 8 8
7 7
7 Let $\mathrm { I } _ { \mathrm { n } } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \cos ^ { \mathrm { n } } \mathrm { xdx }$ for integers $n \geqslant 0$.
\begin{enumerate}[label=(\alph*)]
\item Show that, for $n \geqslant 2 , \quad \mathrm { nl } _ { \mathrm { n } } = ( \mathrm { n } - 1 ) \mathrm { I } _ { \mathrm { n } - 2 }$.
\item Use this reduction formula to deduce the exact value of $I _ { 8 }$.
\item Use the results of parts (a) and (b) to determine the exact value of $\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \cos ^ { 6 } x \sin ^ { 2 } x d x$.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Additional Pure 2021 Q7 [8]}}
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