Edexcel F3 2021 June — Question 7 8 marks

Exam BoardEdexcel
ModuleF3 (Further Pure Mathematics 3)
Year2021
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration using inverse trig and hyperbolic functions
TypeCompleting square then standard inverse trig
DifficultyChallenging +1.2 This is a Further Maths F3 question testing standard inverse trig and hyperbolic integration techniques. Part (i) requires completing the square to get arctan form (routine for FM students). Part (ii) requires a trigonometric or hyperbolic substitution with some algebraic manipulation. While these are non-trivial techniques, they are core syllabus methods for F3 with straightforward execution once the appropriate form/substitution is identified. The 8 total marks reflect moderate length rather than exceptional difficulty.
Spec1.08h Integration by substitution4.08h Integration: inverse trig/hyperbolic substitutions
Using calculus, find the exact values of
  1. \(\int_1^2 \frac{1}{x^2 - 4x + 5} \, dx\) [3]
  2. \(\int_{\sqrt{3}}^3 \frac{\sqrt{x^2 - 3}}{x^2} \, dx\) [5]
Question 7:
AnswerMarks Guidance
7(i)x2 −4x+5=(x−2)2 +1 Attempts to complete the square.
Allow for (x – 2)2 + c, c > 0M1
1
∫ dx=arctan(x−2)
AnswerMarks Guidance
(x−2)2+1Allow for karctan f (x). M1
 π π
[arctan(x−2)]2 =0− − =
AnswerMarks
1  4 4π
cao
AnswerMarks
4A1
(3)
AnswerMarks
7(ii)x2 −3 x2 −3 1
∫ dx=− +∫ dx
x2 x x2 −3
x2 −3 1
Uses integration by parts and obtains A +B∫ dx
AnswerMarks
x x2 −3M1
x2 −3 x
=− +arcosh
AnswerMarks
x 31
B∫ dx=karcosh f(x)
AnswerMarks
x2−3M1
All correctA1
3
3 x2 −3  x2 −3 x   6 
∫ dx=− +arcosh  =− +arcosh 3−(0+arcosh1)
3 x2  x 3   3  
3
Applies the limits 3 and Ö3
AnswerMarks
Depends on both previous M marksdM1
1 1
arcosh 3− 6 =ln( 2+ 3)− 6
AnswerMarks Guidance
3 3Accept either of these forms. A1
(5)
7(ii)
AnswerMarks
ALT 1x2−3 3cosh2u−3
∫ dx=∫ 3sinhudu
AnswerMarks Guidance
x2 3cosh2uA complete substitution using x = Ö3 cosh u M1
=∫tanh2uduObtains k∫tanh2udu M1
=∫(1−sech2u)du =u−tanhuCorrect integration A1
x2−3
∫ 3 dx =[ u−tanhu ]arcosh 3 =arcosh 3−tanh ( arcosh 3 ) −0
3 x2 0
Applies the limits 0 and arcoshÖ3
AnswerMarks
Depends on both previous M marksdM1
1 1
arcosh 3− 6 =ln( 2+ 3)− 6
AnswerMarks Guidance
3 3Accept either of these forms. A1
7(ii)
AnswerMarks
ALT 2x2−3 3sec2u−3
∫ dx=∫ 3secutanudu
AnswerMarks Guidance
x2 3sec2uA complete substitution using x = Ö3 sec u M1
tan2u
= ∫ du
AnswerMarks
secutan2u
Obtains k∫ du
AnswerMarks Guidance
secuM1
=ln(secu+tanu)−sinuCorrect integration A1
x2−3
∫ 3 dx =[ ln(secu+tanu)−sinu ]arcsec 3
3 x2 0
=ln ( sec ( arcsec 3 ) +tan ( arcsec 3 )) −ln ( sec ( 0 )+tan ( 0 )) −sin ( arcsec 3 )
Applies the limits 0 and arcsecÖ3
AnswerMarks
Depends on both previous M marksdM1
3 x2−3 1
∫ dx =ln( 2+ 3)− 6
AnswerMarks Guidance
3 x2 3Correct answer. A1

Total 8

Question 7:
--- 7(i) ---
7(i) | x2 −4x+5=(x−2)2 +1 | Attempts to complete the square.
Allow for (x – 2)2 + c, c > 0 | M1
1
∫ dx=arctan(x−2)
(x−2)2+1 | Allow for karctan f (x). | M1
 π π
[arctan(x−2)]2 =0− − =
1  4 4 | π
cao
4 | A1
(3)
--- 7(ii) ---
7(ii) | x2 −3 x2 −3 1
∫ dx=− +∫ dx
x2 x x2 −3
x2 −3 1
Uses integration by parts and obtains A +B∫ dx
x x2 −3 | M1
x2 −3 x
=− +arcosh
x 3 | 1
B∫ dx=karcosh f(x)
x2−3 | M1
All correct | A1
3
3 x2 −3  x2 −3 x   6 
∫ dx=− +arcosh  =− +arcosh 3−(0+arcosh1)
3 x2  x 3   3  
3
Applies the limits 3 and Ö3
Depends on both previous M marks | dM1
1 1
arcosh 3− 6 =ln( 2+ 3)− 6
3 3 | Accept either of these forms. | A1
(5)
7(ii)
ALT 1 | x2−3 3cosh2u−3
∫ dx=∫ 3sinhudu
x2 3cosh2u | A complete substitution using x = Ö3 cosh u | M1
=∫tanh2udu | Obtains k∫tanh2udu | M1
=∫(1−sech2u)du =u−tanhu | Correct integration | A1
x2−3
∫ 3 dx =[ u−tanhu ]arcosh 3 =arcosh 3−tanh ( arcosh 3 ) −0
3 x2 0
Applies the limits 0 and arcoshÖ3
Depends on both previous M marks | dM1
1 1
arcosh 3− 6 =ln( 2+ 3)− 6
3 3 | Accept either of these forms. | A1
7(ii)
ALT 2 | x2−3 3sec2u−3
∫ dx=∫ 3secutanudu
x2 3sec2u | A complete substitution using x = Ö3 sec u | M1
tan2u
= ∫ du
secu | tan2u
Obtains k∫ du
secu | M1
=ln(secu+tanu)−sinu | Correct integration | A1
x2−3
∫ 3 dx =[ ln(secu+tanu)−sinu ]arcsec 3
3 x2 0
=ln ( sec ( arcsec 3 ) +tan ( arcsec 3 )) −ln ( sec ( 0 )+tan ( 0 )) −sin ( arcsec 3 )
Applies the limits 0 and arcsecÖ3
Depends on both previous M marks | dM1
3 x2−3 1
∫ dx =ln( 2+ 3)− 6
3 x2 3 | Correct answer. | A1
Total 8
Using calculus, find the exact values of

\begin{enumerate}[label=(\roman*)]
\item $\int_1^2 \frac{1}{x^2 - 4x + 5} \, dx$
[3]

\item $\int_{\sqrt{3}}^3 \frac{\sqrt{x^2 - 3}}{x^2} \, dx$
[5]
\end{enumerate}

\hfill \mbox{\textit{Edexcel F3 2021 Q7 [8]}}
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Q1 6 Q2 7 Q3 9 Q4 8 Q5 10 Q6 13 Q7 8 Q8 14