Edexcel CP1 2021 June — Question 9 11 marks

Exam BoardEdexcel
ModuleCP1 (Core Pure 1)
Year2021
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration using inverse trig and hyperbolic functions
TypeHyperbolic substitution to evaluate integral
DifficultyChallenging +1.2 This is a structured Further Maths question requiring hyperbolic substitution (x = cosh u) and integration by parts, but follows standard techniques taught in CP1. Part (a) is a guided proof of a standard integral result, and part (b) applies it with integration by parts in a straightforward manner. The multi-step nature and hyperbolic functions place it above average difficulty, but the clear guidance and standard methods prevent it from being truly challenging.
Spec4.07f Inverse hyperbolic: logarithmic forms4.08h Integration: inverse trig/hyperbolic substitutions
  1. Use a hyperbolic substitution and calculus to show that $$\int \frac{x^2}{\sqrt{x^2 - 1}} dx = \frac{1}{2}\left[x\sqrt{x^2 - 1} + \text{arcosh } x\right] + k$$ where \(k\) is an arbitrary constant. [6]
\includegraphics{figure_1} Figure 1 shows a sketch of part of the curve \(C\) with equation $$y = \frac{4}{15}x \text{ arcosh } x \quad x \geq 1$$ The finite region \(R\), shown shaded in Figure 1, is bounded by the curve \(C\), the \(x\)-axis and the line with equation \(x = 3\)
  1. Using algebraic integration and the result from part (a), show that the area of \(R\) is given by $$\frac{1}{15}\left[17\ln\left(3 + 2\sqrt{2}\right) - 6\sqrt{2}\right]$$ [5]
Question 9:
AnswerMarks
9(a) x 2
d x →  f ( u ) d u
x 2 − 1
Uses the substitution x=coshu fully to achieve an integral in terms
AnswerMarks Guidance
of u only, including replacing the dxM1 3.1a
 c o s h 2 u
( )
s i n h u d u
AnswerMarks Guidance
c o s h 2 u − 1A1 1.1b
Uses correct identities
c o s h 2 u − 1 = s i n h 2 u and c o s h 2 u = 2 c o s h 2 u − 1
to achieve an integral of the form
AnswerMarks Guidance
A  ( c o s h 2 u  1 ) d u A0M1 3.1a
 1 
Integrates to achieve A  sinh2uu (+c ) A  0
AnswerMarks Guidance
 2 M1 1.1b
Uses the identity sinh2u=2sinhucoshu and c o s h 2 u − 1 = s i n h 2 u
AnswerMarks Guidance
→ s i n h 2 u = 2 x x 2 − 1M1 2.1
1  x x 2 − 1 + a r c o s h x  + k * cso
AnswerMarks Guidance
2A1* 1.1b
(6)
AnswerMarks
(b)Uses integration by parts the correct way around to achieve
 4  x2
 xarcoshxdx= Px2arcoshx−Q dx
AnswerMarks Guidance
15  x2 −1M1 2.1
 
4 1 1  x 2
= x 2 a r c o s h x − d x
AnswerMarks Guidance
1 5 2 2 x 2 − 1A1 1.1b
 
4 1 11 
=  x2arcoshx−   x x2 −1+arcoshx  
   
15 2 22 
AnswerMarks Guidance
 B1ft 2.2a
Uses the limits x =1 and x = 3 the correct way around and subtracts
= 4  1 ( 3 ) 2 a r c o s h 3 − 1  1  3 ( 3 ) 2 − 1 + a r c o s h 3    − 4 ( 0 )
AnswerMarks Guidance
1 5 2 2 2 1 5dM1 1.1b
 
4 9 ( ) 3 8 1 ( )
= l n 3 + 8 − − l n 3 + 8
1 5 2 4 4
= 1  1 7 l n ( 3 + 2 2 ) − 6 2  *
AnswerMarks Guidance
1 5A1* 1.1b
(5)
(11 marks)
Notes:
(a)
M1: Uses the substitution x = c o s h u fully to achieve an integral in terms of u only. Must have
replaced the dx but allow if the du is missing.
A1: Correct integral in terms of u. (Allow if the du is missing.)
M1: Uses correct identities c o s h 2 u − 1 = s i n h 2 u and c o s h 2 u = 2 c o s h 2 u − 1 to achieve an
integrand of the required form
M1: Integrates to achieve the correct form, may be sign errors.
M1: Uses the identities s i n h 2 u = 2 s i n h u c o s h u and c o s h 2 u − 1 = s i n h 2 u to attempt to find
sinh2 u in terms of x. If using exponentials there must be a full and complete method to attempt the
correct form.
A1*: Achieves the printed answer with no errors seen, cso
NB attempts at integration by parts are not likely to make progress – to do so would need to split the
x
integrand as x . If you see any attempts that you feel merit credit, use review.
x 2 − 1
(b)
M1: Uses integration by parts the correct way around to achieve the required form.
A1: Correct integration by parts
B1ft: Deduces the integral by using the result from part (a). Follow through on their ‘uv’
dM1: Dependent on previous method mark. Uses the limits x = 1 and x = 3 the correct way around
and subtracts
A1*cso: Achieves the printed answer with at least one intermediate step showing the evaluation of
the arcosh 3, and no errors seen.
Question 9:
--- 9(a) ---
9(a) |  x 2
d x →  f ( u ) d u
x 2 − 1
Uses the substitution x=coshu fully to achieve an integral in terms
of u only, including replacing the dx | M1 | 3.1a
 c o s h 2 u
( )
s i n h u d u
c o s h 2 u − 1 | A1 | 1.1b
Uses correct identities
c o s h 2 u − 1 = s i n h 2 u and c o s h 2 u = 2 c o s h 2 u − 1
to achieve an integral of the form
A  ( c o s h 2 u  1 ) d u A0 | M1 | 3.1a
 1 
Integrates to achieve A  sinh2uu (+c ) A  0
 2  | M1 | 1.1b
Uses the identity sinh2u=2sinhucoshu and c o s h 2 u − 1 = s i n h 2 u
→ s i n h 2 u = 2 x x 2 − 1 | M1 | 2.1
1  x x 2 − 1 + a r c o s h x  + k * cso
2 | A1* | 1.1b
(6)
(b) | Uses integration by parts the correct way around to achieve
 4  x2
 xarcoshxdx= Px2arcoshx−Q dx
15  x2 −1 | M1 | 2.1
 
4 1 1  x 2
= x 2 a r c o s h x − d x
1 5 2 2 x 2 − 1 | A1 | 1.1b
 
4 1 11 
=  x2arcoshx−   x x2 −1+arcoshx  
   
15 2 22 
  | B1ft | 2.2a
Uses the limits x =1 and x = 3 the correct way around and subtracts
= 4  1 ( 3 ) 2 a r c o s h 3 − 1  1  3 ( 3 ) 2 − 1 + a r c o s h 3    − 4 ( 0 )
1 5 2 2 2 1 5 | dM1 | 1.1b
 
4 9 ( ) 3 8 1 ( )
= l n 3 + 8 − − l n 3 + 8
1 5 2 4 4
= 1  1 7 l n ( 3 + 2 2 ) − 6 2  *
1 5 | A1* | 1.1b
(5)
(11 marks)
Notes:
(a)
M1: Uses the substitution x = c o s h u fully to achieve an integral in terms of u only. Must have
replaced the dx but allow if the du is missing.
A1: Correct integral in terms of u. (Allow if the du is missing.)
M1: Uses correct identities c o s h 2 u − 1 = s i n h 2 u and c o s h 2 u = 2 c o s h 2 u − 1 to achieve an
integrand of the required form
M1: Integrates to achieve the correct form, may be sign errors.
M1: Uses the identities s i n h 2 u = 2 s i n h u c o s h u and c o s h 2 u − 1 = s i n h 2 u to attempt to find
sinh2 u in terms of x. If using exponentials there must be a full and complete method to attempt the
correct form.
A1*: Achieves the printed answer with no errors seen, cso
NB attempts at integration by parts are not likely to make progress – to do so would need to split the
x
integrand as x . If you see any attempts that you feel merit credit, use review.
x 2 − 1
(b)
M1: Uses integration by parts the correct way around to achieve the required form.
A1: Correct integration by parts
B1ft: Deduces the integral by using the result from part (a). Follow through on their ‘uv’
dM1: Dependent on previous method mark. Uses the limits x = 1 and x = 3 the correct way around
and subtracts
A1*cso: Achieves the printed answer with at least one intermediate step showing the evaluation of
the arcosh 3, and no errors seen.
\begin{enumerate}[label=(\alph*)]
\item Use a hyperbolic substitution and calculus to show that

$$\int \frac{x^2}{\sqrt{x^2 - 1}} dx = \frac{1}{2}\left[x\sqrt{x^2 - 1} + \text{arcosh } x\right] + k$$

where $k$ is an arbitrary constant.
[6]
\end{enumerate}

\includegraphics{figure_1}

Figure 1 shows a sketch of part of the curve $C$ with equation

$$y = \frac{4}{15}x \text{ arcosh } x \quad x \geq 1$$

The finite region $R$, shown shaded in Figure 1, is bounded by the curve $C$, the $x$-axis and the line with equation $x = 3$

\begin{enumerate}[label=(\alph*)]
\setcounter{enumii}{1}
\item Using algebraic integration and the result from part (a), show that the area of $R$ is given by

$$\frac{1}{15}\left[17\ln\left(3 + 2\sqrt{2}\right) - 6\sqrt{2}\right]$$
[5]
\end{enumerate}

\hfill \mbox{\textit{Edexcel CP1 2021 Q9 [11]}}
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Q1 6 Q2 7 Q3 6 Q4 9 Q5 7 Q6 12 Q7 8 Q8 9 Q9 11