Edexcel F3 2021 June — Question 5 10 marks

Exam BoardEdexcel
ModuleF3 (Further Pure Mathematics 3)
Year2021
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Parts
TypeReduction formula or recurrence
DifficultyChallenging +1.8 This is a challenging Further Maths reduction formula question requiring integration by parts with careful algebraic manipulation to derive the recurrence relation, followed by recursive application. The proof demands strategic choice of u and dv, use of trigonometric identities, and non-trivial rearrangement. Part (b) requires working backwards through multiple iterations (n=6,4,2,0) with exact arithmetic at π/4, making it substantially harder than standard A-level integration but typical of F3 material.
Spec1.08i Integration by parts4.08a Maclaurin series: find series for function
$$I_n = \int \sec^n x \, dx \quad n \geq 0$$
  1. Prove that for \(n \geq 2\) $$(n-1)I_n = \tan x \sec^{n-2} x + (n-2)I_{n-2}$$ [6]
  2. Hence, showing each step of your working, find the exact value of $$\int_0^{\pi/4} \sec^6 x \, dx$$ [4]
Question 5:
AnswerMarks Guidance
5(a)∫secn xdx=∫secn−2 xsec2 xdx Splits secn xinto secn−2 xsec2 x
∫secn xdx=secn−2 xtanx−∫(n−2)secn−2 xtan2 xdx
Depends on previous M mark
dM1: Uses integration by parts to obtain secn−2 xtanx−k∫secn−2 xtan2 xdx
AnswerMarks
A1: Correct integrationdM1A1
∫secn xdx=secn−2 xtanx−∫(n−2)secn−2 x(sec2 x−1)dx
AnswerMarks
Uses tan2 x=sec2 x−1B1
(M1 on
EPEN)
∫secn xdx=secn−2 xtanx−(n−2)∫secn xdx+(n−2)∫secn−2 xdx
=secn−2xtanx−(n−2)I +(n−2)I ⇒( n−1 ) I =...
n n−2 n
Depends on all previous M and B marks
Introduces I and I and makes progress to the given result.
AnswerMarks
n n-2ddM1
( n−1 ) I =tanxsecn−2x+( n−2 ) I *
AnswerMarks Guidance
n n−2Fully correct proof. A1cso
(6)
AnswerMarks Guidance
ALT∫secn xdx=∫secn−2 xsec2 xdx Splits secn xinto secn−2 xsec2 x
∫secn−2 xsec2 xdx=∫secn−2 x ( 1+tan2 x ) dx
AnswerMarks
=∫secn−2 xdx+∫tan2 xsecn−2 xdxUses sec2 x=1+tan2 xand splits into 2
integrals.B1
(4th mark
M1 on
EPEN)
1 1
∫tan2xsecn−2xdx= tanxsecn−2x− ∫secnxdx
(n−2) (n−2)
Uses integration by parts on ∫tan2 xsecn−2 xdx to obtain Atanxsecn−2 x−B∫secn xdx
AnswerMarks
Note this is the 2nd M on EPEN.dM1
1 1
∫secnxdx=∫secn−2xdx+ tanxsecn−2x− ∫secnxdx
(n−2) (n−2)
AnswerMarks
Fully correct integrationA1
1 1
∫secn xdx=I + tanxsecn−2x− I ⇒(n−1)I =...
n−2 (n−2) (n−2) n n
Depends on previous M and B marks
Introduces I and I and makes progress to the given result.
AnswerMarks
n n-2ddM1
( n−1 ) I =tanxsecn−2 x+( n−2 ) I *
AnswerMarks Guidance
n n−2Fully correct proof. A1cso
AnswerMarks
5(b)I =1
2Correct value for I seen or implied.
2B1
1 4
I = tanxsec4 x+ I
6 5 5 4
AnswerMarks Guidance
or e.g.Applies the given reduction formula once. M1
1 π π 4
I = tan sec4 + I
6 5 4 4 5 4
or e.g.
1 (1)( )4 4
I = 2 + I
6 5 5 4
1 41 2  1 ( )4 4 ( )2 8
= tanxsec4 x+  tanxsec2 x+ I  = ( 1 ) 2 + ( 1 ) 2 + ( 1 )
5 53 3 2  5 15 15
Applies the given reduction formula again and uses the limits
to reach a numerical expression for I
AnswerMarks
6M1
28
=
AnswerMarks Guidance
15Correct value A1
(4)
AnswerMarks
ALTI =1
2Correct value for I seen or implied.
2B1
1 2
I = tanxsec2 x+ I
4 3 3 2
or e.g.
1 π π 2
I = tan sec2 + I
4 3 4 4 3 2
or e.g.
1 (1)( )2 2
I = 2 + I
AnswerMarks Guidance
4 3 3 2Applies the given reduction formula once. M1
1 4 1 2  1 (1)( )4 4 (1)( )2 8
I = tanxsec4 x+  tanxsec2 x+ I = 2 + 2 +
6 5 53 3 2 5 15 15
Applies the given reduction formula again and uses the limits
to reach a numerical expression for I
AnswerMarks
6M1
28
=
AnswerMarks Guidance
15Correct value A1

Total 10

i j k
Normal to plane given by 1 0 3 =...
AnswerMarks
1 −2 1Attempt cross product of direction vectors. If
the method is unclear, look for at least 2
AnswerMarks Guidance
correct components.M1
=6i+2j−2kOr any multiple of this vector. A1
Substitute appropriate point into
6x+2y−2z =d
AnswerMarks
e.g. (1, 1, 1) or (2, 1, 4) to find “d”Use a valid point and use scalar product with
normal or substitute into Cartesian equation.M1
6x+2y−2z =6
AnswerMarks Guidance
3x+ y−z =3*Given answer. No errors seen A1* cso
(4)
AnswerMarks
6(a) ALTr =i+j+k+λ(i+3k)+µ(i−2j+k)
⇒ x=1+λ+µ, y =1−2µ, z =1+3λ+µ
M1: Forms equation of plane using (1, 1, 1) and direction vectors and extracts 3 equations
for x, y and z in terms of λ and µ
AnswerMarks
A1: Correct equationsM1A1
1 1 1 1 1
x=1+ − y+ z− + y
AnswerMarks Guidance
2 2 3 2 6Eliminates λ and µ and achieves an
equation in x, y and z only.M1
3x+ y−z =3*Given answer. No errors seen. A1 
Question 5:
--- 5(a) ---
5(a) | ∫secn xdx=∫secn−2 xsec2 xdx | Splits secn xinto secn−2 xsec2 x | M1
∫secn xdx=secn−2 xtanx−∫(n−2)secn−2 xtan2 xdx
Depends on previous M mark
dM1: Uses integration by parts to obtain secn−2 xtanx−k∫secn−2 xtan2 xdx
A1: Correct integration | dM1A1
∫secn xdx=secn−2 xtanx−∫(n−2)secn−2 x(sec2 x−1)dx
Uses tan2 x=sec2 x−1 | B1
(M1 on
EPEN)
∫secn xdx=secn−2 xtanx−(n−2)∫secn xdx+(n−2)∫secn−2 xdx
=secn−2xtanx−(n−2)I +(n−2)I ⇒( n−1 ) I =...
n n−2 n
Depends on all previous M and B marks
Introduces I and I and makes progress to the given result.
n n-2 | ddM1
( n−1 ) I =tanxsecn−2x+( n−2 ) I *
n n−2 | Fully correct proof. | A1cso
(6)
ALT | ∫secn xdx=∫secn−2 xsec2 xdx | Splits secn xinto secn−2 xsec2 x | M1
∫secn−2 xsec2 xdx=∫secn−2 x ( 1+tan2 x ) dx
=∫secn−2 xdx+∫tan2 xsecn−2 xdx | Uses sec2 x=1+tan2 xand splits into 2
integrals. | B1
(4th mark
M1 on
EPEN)
1 1
∫tan2xsecn−2xdx= tanxsecn−2x− ∫secnxdx
(n−2) (n−2)
Uses integration by parts on ∫tan2 xsecn−2 xdx to obtain Atanxsecn−2 x−B∫secn xdx
Note this is the 2nd M on EPEN. | dM1
1 1
∫secnxdx=∫secn−2xdx+ tanxsecn−2x− ∫secnxdx
(n−2) (n−2)
Fully correct integration | A1
1 1
∫secn xdx=I + tanxsecn−2x− I ⇒(n−1)I =...
n−2 (n−2) (n−2) n n
Depends on previous M and B marks
Introduces I and I and makes progress to the given result.
n n-2 | ddM1
( n−1 ) I =tanxsecn−2 x+( n−2 ) I *
n n−2 | Fully correct proof. | A1cso
--- 5(b) ---
5(b) | I =1
2 | Correct value for I seen or implied.
2 | B1
1 4
I = tanxsec4 x+ I
6 5 5 4
or e.g. | Applies the given reduction formula once. | M1
1 π π 4
I = tan sec4 + I
6 5 4 4 5 4
or e.g.
1 (1)( )4 4
I = 2 + I
6 5 5 4
1 41 2  1 ( )4 4 ( )2 8
= tanxsec4 x+  tanxsec2 x+ I  = ( 1 ) 2 + ( 1 ) 2 + ( 1 )
5 53 3 2  5 15 15
Applies the given reduction formula again and uses the limits
to reach a numerical expression for I
6 | M1
28
=
15 | Correct value | A1
(4)
ALT | I =1
2 | Correct value for I seen or implied.
2 | B1
1 2
I = tanxsec2 x+ I
4 3 3 2
or e.g.
1 π π 2
I = tan sec2 + I
4 3 4 4 3 2
or e.g.
1 (1)( )2 2
I = 2 + I
4 3 3 2 | Applies the given reduction formula once. | M1
1 4 1 2  1 (1)( )4 4 (1)( )2 8
I = tanxsec4 x+  tanxsec2 x+ I = 2 + 2 +
6 5 53 3 2 5 15 15
Applies the given reduction formula again and uses the limits
to reach a numerical expression for I
6 | M1
28
=
15 | Correct value | A1
Total 10
i j k
Normal to plane given by 1 0 3 =...
1 −2 1 | Attempt cross product of direction vectors. If
the method is unclear, look for at least 2
correct components. | M1
=6i+2j−2k | Or any multiple of this vector. | A1
Substitute appropriate point into
6x+2y−2z =d
e.g. (1, 1, 1) or (2, 1, 4) to find “d” | Use a valid point and use scalar product with
normal or substitute into Cartesian equation. | M1
6x+2y−2z =6
3x+ y−z =3* | Given answer. No errors seen | A1* cso
(4)
6(a) ALT | r =i+j+k+λ(i+3k)+µ(i−2j+k)
⇒ x=1+λ+µ, y =1−2µ, z =1+3λ+µ
M1: Forms equation of plane using (1, 1, 1) and direction vectors and extracts 3 equations
for x, y and z in terms of λ and µ
A1: Correct equations | M1A1
1 1 1 1 1
x=1+ − y+ z− + y
2 2 3 2 6 | Eliminates λ and µ and achieves an
equation in x, y and z only. | M1
3x+ y−z =3* | Given answer. No errors seen. | A1
$$I_n = \int \sec^n x \, dx \quad n \geq 0$$

\begin{enumerate}[label=(\alph*)]
\item Prove that for $n \geq 2$
$$(n-1)I_n = \tan x \sec^{n-2} x + (n-2)I_{n-2}$$
[6]

\item Hence, showing each step of your working, find the exact value of
$$\int_0^{\pi/4} \sec^6 x \, dx$$
[4]
\end{enumerate}

\hfill \mbox{\textit{Edexcel F3 2021 Q5 [10]}}
This paper (8 questions)
View full paper
Q1 6 Q2 7 Q3 9 Q4 8 Q5 10 Q6 13 Q7 8 Q8 14