2.5· 31 questions · 245 marks · 294 min · 2008–2025· Structured questions
Every Cambridge A Level Mathematics Paper 2 question on integration, laid out as 38 A4 pages with the mark scheme below. Nothing is left out. Free to read, no account.
![Question 1: 4 18 −5 Show that dx = ln 25. [6] x 2x + 1 1](https://img.pastlit.com/crops/f4198a1f-0cee-4541-bbf6-e9a383d01bad/q5.webp)
![Question 2: 3 Show that 2e dx 12e2 2. [5] ã = + −3 0 (ex + 1)2](https://img.pastlit.com/crops/9f23fc01-0acc-43bc-b604-48e2a8de678c/q3.webp)
![Question 3: 3 Show that 2e dx 12e2 2. [5] ã = + −3 0 (ex + 1)2](https://img.pastlit.com/crops/a55466cd-b8d6-4219-b467-70415da7e069/q3.webp)
![Question 4: 2 5 2 Show that dx ln 3. [5] 4x 1 = ä 2 +](https://img.pastlit.com/crops/4cc97a69-9a61-42c3-91ee-3b299f1ace0d/q2.webp)
![Question 5: (i) By first expanding show that cos(2x + x), cos 3x cos3x cos x. ≡4 −3 [5] (ii) Hence show that 16π cos3x dx 12.5 ã 0 (2 −cos x) = [5]](https://img.pastlit.com/crops/4cc97a69-9a61-42c3-91ee-3b299f1ace0d/q8.webp)
![Question 6: (i) Show that sin x cos can be written in the form 5 2 sin 2x cos 2x. [5] 2 2 (2 + x)2 + −3 14π (ii) Hence find the exact value of sin x cos…](https://img.pastlit.com/crops/57e0a1ae-48ed-409c-83a9-5615a6e7d331/q7.webp)
1 / 38![Question 8: (a) Find 4 cos2 1 [3] Ó 21 d1. 6 1 (b) Find the exact value of dx. [4] Ô 2x 3 −1 +](https://img.pastlit.com/crops/1f45d539-201c-4c61-b287-e5a3a3875ca3/q3.webp)
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38 / 38Answers below. Sit the paper first if you are practising.
Pastlit
Mathematics 9709 · Integration — Paper 2
A Level · topical answer key — answer key (teacher use)
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Answer
Marks
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8| Question | Answer | Marks | From |
|---|---|---|---|
| 1 | see sheet | 6 | 9709/21 Oct/Nov 2008 |
| 2 | see sheet | 5 | 9709/21 Oct/Nov 2010 |
| 3 | see sheet | 5 | 9709/22 Oct/Nov 2010 |
| 4 | see sheet | 5 | 9709/21 Oct/Nov 2011 |
| 5 | see sheet | 10 | 9709/21 Oct/Nov 2011 |
| 6 | see sheet | 9 | 9709/21 May/June 2012 |
| 7 | see sheet | 7 | 9709/22 Oct/Nov 2012 |
| 8 | see sheet | 7 | 9709/21 Oct/Nov 2014 |
| 9 | see sheet | 7 | 9709/21 Oct/Nov 2015 |
| 10 | see sheet | 5 | 9709/21 May/June 2017 |
| 11 | see sheet | 9 | 9709/22 May/June 2017 |
| 12 | see sheet | 9 | 9709/23 May/June 2017 |
| 13 | see sheet | 8 | 9709/21 Oct/Nov 2017 |
| 14 | see sheet | 10 | 9709/22 Oct/Nov 2017 |
| 15 | see sheet | 8 | 9709/23 Oct/Nov 2017 |
| 16 | see sheet | 11 | 9709/22 May/June 2018 |
| 17 | see sheet | 5 | 9709/21 Oct/Nov 2018 |
| 18 | see sheet | 7 | 9709/21 May/June 2019 |
| 19 | see sheet | 10 | 9709/22 May/June 2020 |
| 20 | see sheet | 10 | 9709/23 May/June 2020 |
| 21 | see sheet | 9 | 9709/22 Feb/March 2021 |
| 22 | see sheet | 4 | 9709/21 Oct/Nov 2021 |
| 23 | see sheet | 9 | 9709/23 Oct/Nov 2021 |
| 24 | see sheet | 10 | 9709/22 May/June 2023 |
| 25 | see sheet | 10 | 9709/23 May/June 2023 |
| 26 | see sheet | 8 | 9709/22 Oct/Nov 2023 |
| 27 | see sheet | 6 | 9709/23 Oct/Nov 2023 |
| 28 | see sheet | 10 | 9709/22 May/June 2024 |
| 29 | see sheet | 10 | 9709/23 May/June 2024 |
| 30 | see sheet | 8 | 9709/23 Oct/Nov 2024 |
| 31 | see sheet | 8 | 9709/22 Feb/March 2025 |
1 4 18 −5 Show that dx = ln 25. [6] x 2x + 1 1
6 marks
Mark scheme: 5 Integrate and state term ln x B1 Obtain term of the form kln (2x + 1) M1 State correct term –2ln (2x + 1) A1 Substitute limits correctly M1 Use law for the logarithm of a product, quotient or power M1 Obtain given answer correctly A1 [6] − 1 x 1 x
1 3 Show that 2e dx 12e2 2. [5] ã = + −3 0 (ex + 1)2
5 marks
Mark scheme: 1 2 x 3 Integrate and obtain e term B1 2 Obtain 2e x term B1 Obtain x B1 Use limits correctly, allow use of limits x = 1 and x = 0 into an incorrect form M1 Obtain given answer A1 [5] S. R. Feeding limits into original integrand, 0/5 dx 1 dy 2
1 3 Show that 2e dx 12e2 2. [5] ã = + −3 0 (ex + 1)2
5 marks
Mark scheme: 1 2 x 3 Integrate and obtain e term B1 2 Obtain 2e x term B1 Obtain x B1 Use limits correctly, allow use of limits x = 1 and x = 0 into an incorrect form M1 Obtain given answer A1 [5] S. R. Feeding limits into original integrand, 0/5 dx 1 dy 2
6 2 5 2 Show that dx ln 3. [5] 4x 1 = ä 2 +
5 marks
Mark scheme: 2 Integrate and obtain term of the form kln(4x + 1) M1 1 State correct term ln( 4 x + )1 A1 2 Substitute limits correctly M1 Use law for the logarithm of a quotient or a power M1 Obtain given answer correctly A1 [5] 1 2
8 (i) By first expanding show that cos(2x + x), cos 3x cos3x cos x. ≡4 −3 [5] (ii) Hence show that 16π cos3x dx 12.5 ã 0 (2 −cos x) = [5]
10 marks
Mark scheme: 8 (i) Make relevant use of the cos(A + B) formula M1* Make relevant use of the cos 2A and sin 2A formulae M1* Obtain a correct expression in terms of cos x and sin x A1 Use sin2 x = 1 – cos2 x to obtain an expression in terms of cos x M1(dep*) Obtain given answer correctly A1 [5] 1 1 (ii) Replace integrand by cos 3 x + cos x , or equivalent B1 2 2 1 1 Integrate, obtaining sin 3 x + sin x , or equivalent B1 + B1√ 6 2 Use limits correctly M1 Obtain given answer A1 [5]
7 (i) Show that sin x cos can be written in the form 5 2 sin 2x cos 2x. [5] 2 2 (2 + x)2 + −3 14π (ii) Hence find the exact value of sin x cos dx. [4] ã 0 (2 + x)2
9 marks
Mark scheme: 7 (i) Expand to obtain 4 sin2 x + 4 sin x cos x + cos2 x B1 Use 2 sin x cos x = sin 2x B1 Attempt to express sin2 x or cos2 x (or both) in terms of cos 2x M1 Obtain correct 12 k 1( − cos 2 x ) for their k sin2 x or equivalent A1√ Confirm given answer 52 + 2 sin 2 x − 32 cos 2 x A1 [5] (ii) Integrate to obtain form px + q cos 2x + r sin 2x M1 Obtain 52 x − cos 2 x − 34 sin 2 x A1 Substitute limits in integral of form px + q cos 2x + r sin 2x and attempt simplification DM1 Obtain 85 π + 14 or exact equivalent A1 [4]
2x6 (a) Find 4e−1 dx. [2] ã 3 6 (b) Show that dx ln 16. [5] 3x = ä1 −1
7 marks
Mark scheme: 6 (a) Obtain integral ke 2 with any non-zero k M1 Correct integral A1 [2] (b) State indefinite integral of the form k ln (3x – 1), where k = 2, 6 or 3 M1 State correct integral 2 ln (3x – 1) A1 Substitute limits correctly (must be a function involving a logarithm) M1 Use law for the logarithm of a power or a quotient M1 Obtain given answer correctly A1 [5] dy 2
3 (a) Find 4 cos2 1 [3] Ó 21 d1. 6 1 (b) Find the exact value of dx. [4] Ô 2x 3 −1 +
7 marks
Mark scheme: 3 (a) Express integrand in the form p cos θ + 2 M1 State correct 2 cos θ + 2 A1 Integrate to obtain 2 sin θ + 2θ (+ c) A1 [3] (b) Integrate to obtain form k ln (2 x + 3) M1 1 Obtain correct ln (2 x + 3) A1 2 Apply limits correctly DM1 1 Obtain ln 15 A1 [4] 2
5 (a) Find tan2x sin 2x dx. [3] Ó + 1 (b) Find the exact value of dx. [4] Ó 0 3e1−2x
7 marks
Mark scheme: 5 (a) Use tan 2 x = sec 2 x − 1 B1 Obtain integral of form p tan x + qx + r cos 2 x M1 1 Obtain tan x − x − cos 2 x + c A1 [3] 2 (b) Obtain integral of form k1e− 2 x M1* 3 − 2 x Obtain − 1e A1 2 Apply both limits the correct way round M1 dep 3 −1 3 Obtain − e + e or exact equivalent A1 [4] 2 2
a 3 Given that 4e Ó0 12x+3 dx = 835, find the value of the constant a correct to 3 significant figures. [5] … … … … … … … … … … … … … … … … … … … … … … … … …
5 marks
Mark scheme: 3 Integrate to obtain form 1 2 3 + x ke where k is constant not equal to 4 M1 Obtain correct 1 2 3 8 + x e A1 Allow unsimplified for A1 Obtain 1 2 3 3 8 8 835 + − = a e e or equivalent A1 Carry out correct process to find a from equation of form 1 2 3 + = a ke c M1 Obtain 3.65 A1 If 3.65 seen with no actual attempt at integration, award B1 if it is thought that trial and improvement with calculator has been used. Total: 5
7 (a) Find 2 cos cos 1 [4] Ó 1 −3 1 + d1. … … … … … … … … … … … … … … … … … … … … … … … … … @ A 4 1 (b) (i) Find dx. [2] 2x Ô 2x 1 + + … … … … … … … … … … … … … … 4 @ A 4 1 (ii) Hence find dx, giving your answer in the form ln k. [3] 2x 1 2x + Ô 1 + … … … … … … … …
9 marks
Mark scheme: 7(a) 2 2cos cos 3 d θ θ θ − − ∫ B1 Attempt use of identity to obtain integrand involving cos2θ and cosθ M1 Integrate to obtain form 1 2 3 sin2 sin k k k θ θ θ + + for non-zero constants M1 Obtain 1 2 sin2 sin 2 c θ θ θ − − + A1 Total: 4 Question Answer Marks Guidance 7(b)(i) Integrate to obtain form ( ) ( ) 1 2 ln 2 1 ln k x k x + + or ( ) ( ) 1 2 ln 2 1 ln 2 k x k x + + M1 Obtain ( ) 1 2 2ln 2 1 ln x x + + or ( ) ( ) 1 2 2ln 2 1 ln 2 x x + + A1 Total: 2 7(b)(ii) Use relevant logarithm power law for expression obtained from application of limits M1 Use relevant logarithm addition / subtraction laws M1 Obtain ln18 A1 Total: 3
7 (a) Find 2 cos cos 1 [4] Ó 1 −3 1 + d1. … … … … … … … … … … … … … … … … … … … … … … … … … @ A 4 1 (b) (i) Find dx. [2] 2x Ô 2x 1 + + … … … … … … … … … … … … … … 4 @ A 4 1 (ii) Hence find dx, giving your answer in the form ln k. [3] 2x 1 2x + Ô 1 + … … … … … … … …
9 marks
Mark scheme: 7(a) 2 2cos cos 3 d θ θ θ − − ∫ B1 Attempt use of identity to obtain integrand involving cos2θ and cosθ M1 Integrate to obtain form 1 2 3 sin2 sin k k k θ θ θ + + for non-zero constants M1 Obtain 1 2 sin2 sin 2 c θ θ θ − − + A1 Total: 4 Question Answer Marks Guidance 7(b)(i) Integrate to obtain form ( ) ( ) 1 2 ln 2 1 ln k x k x + + or ( ) ( ) 1 2 ln 2 1 ln 2 k x k x + + M1 Obtain ( ) 1 2 2ln 2 1 ln x x + + or ( ) ( ) 1 2 2ln 2 1 ln 2 x x + + A1 Total: 2 7(b)(ii) Use relevant logarithm power law for expression obtained from application of limits M1 Use relevant logarithm addition / subtraction laws M1 Obtain ln18 A1 Total: 3
4 4 (a) Find [4] + sin21 Ô 1 d1. −sin21 … … … … … … … … … … … a 2 (b) Given that dx ln 16, find the value of the positive constant a. [4] 3x 1 = Ô0 + … … … … … … … … … … …
8 marks
Mark scheme: 4(a) Obtain integrand of form a sec 2 θ+ b M1 Obtain correct 5sec 2 θ− 1 A1 Integrate to obtain form a tanθ+ bθ M1 Obtain 5tanθ− θ+ c A1 4 4(b) Obtain integral of form k ln(3 x + 1) *M1 Apply limits and obtain 23 ln(3a + 1) = ln16 A1 Obtain equation with no presence of ln DM1 Obtain 21 A1 4
1 406 (a) Find the exact value of sin x 4 sin x 6 cosx dx. [5] Ó 0 + … … … … … … … … … … … … … … … … … … … … … … … … … a 6 (b) Given that dx ln 49, find the value of the positive constant a. [5] 3x 2 = Ô0 + … … … … … … … … … … … … … … … … … … … … … … … …
10 marks
Mark scheme: 6(a) Obtain 2 − 2cos2x as part of integrand B1 Obtain 3sin 2x as part of integrand B1 Allow second B1 for writing Integrate to obtain form M1 1 2 , M1 6sin x cos x d x = 6 sin x k1 x + k 2 sin 2 x + k 3 cos2 x ∫ 2 may then be implied by subsequent work Obtain 2 x − sin2 x − 32 cos2 x or A1 2 x − sin 2 x + 3sin 2 x Apply limits to obtain 12 π+ 12 A1 5 6(b) B1 6 Integrate to obtain 2ln(3 x + 2) Allow ln ( 3 x + 2 ) for B1 3 Use at least one relevant logarithm property *M1 3a + 2 (3a + 2) 2 A1 Obtain = 7 or = 49 or 2 4 equivalent without ln Solve relevant equation to find a DM1 Dep on *M1, allow for 49 = ( 3a + 2 ) 2 OE or correct working involving ( 3a + 2 ) Obtain a = 4 only A1 5
4 4 (a) Find [4] + sin21 Ô 1 d1. −sin21 … … … … … … … … … … … a 2 (b) Given that dx ln 16, find the value of the positive constant a. [4] 3x 1 = Ô0 + … … … … … … … … … … …
8 marks
Mark scheme: 4(a) Obtain integrand of form a sec 2 θ+ b M1 Obtain correct 5sec 2 θ− 1 A1 Integrate to obtain form a tanθ+ bθ M1 Obtain 5tanθ− θ+ c A1 4 4(b) Obtain integral of form k ln(3 x + 1) *M1 Apply limits and obtain 23 ln(3a + 1) = ln16 A1 Obtain equation with no presence of ln DM1 Obtain 21 A1 4
a 1 2x 26 It is given that 1 e dx 10, where a is a positive constant. Ó 0 + = P Q 15 (i) Show that a 2 ln −a1 . [6] = 2a 4 e + … … … … … … … … … … … … … … … … … … … … … … … (ii) Use the equation in part (i) to show by calculation that 1.5 a 1.6. [2] < < … … … … … … … … … … … … (iii) Use an iterative formula based on the equation in part (i) to find the value of a correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] … … … … … … … … … … …
11 marks
Mark scheme: 6(i) Rewrite integrand as 1 2 1 2e e + + x B1 Integrate to obtain form 1 2 1 2 e e + + x x x k k M1 Obtain 1 2 4e e + + x x x A1 Use limits to obtain 1 2 4e e 5 10 + + − = a a a A1 Rearrange as far as 1 2e ... = a including use of 1 1 1 2 2 2 4e e e (4 e ) + = + a a a a M1 Confirm 1 2 15 2ln 4 e − = + a a a A1 AG; necessary detail needed 6 6(ii) Consider sign of 1 2 15 2ln 4 e − − + a a a for 1.5 and 1.6 or equivalent M1 Obtain 0.08... − and 0.06… or equivalents and justify conclusion A1 2 6(iii) Use iterative process correctly at least once M1 Obtain final answer 1.56 A1 Show sufficient iterations to 5 sf to justify answer or show sign change in interval (1.555,1.565) A1 3
7 6 2 Show dx ln 125. [5] 2x 1 = thatÔ1 + … … … … … … … … … … … … … … … … … … … … … … … …
5 marks
Mark scheme: 2 Integrate to obtain form ln(2 1) k M1 Obtain correct 3ln(2 1) x + A1 Use subtraction law of logarithms correctly M1 Dependent on first M1 Use power law of logarithms correctly M1 Dependent on first M1 Confirm ln125 A1 5
4 (a) Find tan2 3x dx. [3] Ó … … … … … … … … … 1 e3x 4 (b) Find the exact value dx. Show all necessary working. [4] + ex ofÔ0 … … … … … … … … … … … … …
7 marks
Mark scheme: 4(a) Use identity 2 2 tan 3 sec 3 1 x x = − B1 Integrate to obtain form 1 2 tan3 k x k x + M1 Obtain correct 1 3 tan3x x c − + A1 3 4(b) Express integrand as 2 e 4e x x − + B1 Integrate to obtain form 2 3 4 e e x x k k − + M1 Obtain correct 2 1 2 e 4e x x − − A1 Use limits to obtain 2 1 7 1 2 2 e 4e− − + or similarly simplified equivalent A1 4
8 (a) Show that 3 sin cot 6 [2] 21 1 cos21. … … … … … … … … … … … (b) Solve the equation 3 sin cot 5 for 0 [3] 21 1 = < 1 < π. … … … … … … … … … … … … … 1 2π 1(c) Find the exact value of 3 sin x cot 2x dx. [5] 1 Ó 4π … … … … … … … … … … … … … … … … … … … … … … … …
10 marks
Mark scheme: 8(a) Use at least one of sin 2 2sin cos θ θ θ = and cos cot sin θ θ θ = B1 Use both and conclude 2 6cos θ AG B1 2 8(b) Attempt solution of 2 5 cos 6 θ = to find at least one value M1 Obtain 0.421 A1 Obtain 2.72 A1 3 Question Answer Marks 8(c) Express integrand in form cos + a b x M1 Obtain correct integrand 3 3cos + x A1 Integrate to obtain sin + px q x *M1 Apply limits correctly DM1 Obtain 3 3 3 4 2 π + − or exact equivalent A1 5
8 (a) Show that 3 sin cot 6 [2] 21 1 cos21. … … … … … … … … … … … (b) Solve the equation 3 sin cot 5 for 0 [3] 21 1 = < 1 < π. … … … … … … … … … … … … … 1 2π 1(c) Find the exact value of 3 sin x cot 2x dx. [5] 1 Ó 4π … … … … … … … … … … … … … … … … … … … … … … … …
10 marks
Mark scheme: 8(a) Use at least one of sin 2 2sin cos θ θ θ = and cos cot sin θ θ θ = B1 Use both and conclude 2 6cos θ AG B1 2 8(b) Attempt solution of 2 5 cos 6 θ = to find at least one value M1 Obtain 0.421 A1 Obtain 2.72 A1 3 Question Answer Marks 8(c) Express integrand in form cos + a b x M1 Obtain correct integrand 3 3cos + x A1 Integrate to obtain sin + px q x *M1 Apply limits correctly DM1 Obtain 3 3 3 4 2 π + − or exact equivalent A1 5
7 (a) Express 5 3 cos x 5 sin x in the form R cos x , where R 0 and 0 1 [3] + −! > < ! < 2π. … … … … … … … … … … … … … … … (b) As x varies, find the least possible value of 4 5 3 cos x 5 sin x, + + and determine the corresponding value of x where x [3] −π < < π. … … … … … … … … … … … … … 1 (c) Find [3] 2 Ô 5 3 cos 5 sin d1. 31 + 31 … … … … … … … … … … … … … … … … …
9 marks
Mark scheme: 7(a) State 10 R = B1 Use appropriate trigonometry to find α M1 Obtain 1 6 α = π A1 3 7(b) State 6 − B1 FT Following their value of R. Attempt to find x from their cos( ) 1 x α − = − M1 Obtain 1 6 x − π = −π and hence 5 π 6 − A1 3 7(c) State integrand of form 2 1 1 sec 3 π 6 k θ − *M1 Integrate to obtain form 2 1 tan 3 π 6 k θ − DM1 Obtain 1 1 tan 3 π 300 6 c θ − + A1 3
2 1 Find the exact value of 4e2x dx. [4] Ó −2e−x −1 … … … … … … … … … … … … … … … … … … … … … … … … …
4 marks
Mark scheme: 1 Integrate to obtain 2 Integrate to obtain 2e−x B1 Apply limits correctly to integral of the form 2 1 2 e e− + x x k k M1 1 4 ≠ k . Condone one error. Obtain 4 2e 2e − A1 or exact equivalent. 4
7 (a) By first expanding cos 2 , show that cos 3 4 cos3 3 cos . [3] … … … … … … … … … … … … … (b) Find the exact value of 2 cos3 5 cos 5 . [2] 18π −32 18π … … … … … … … … … … (c) Find 12 cos3x cos3 3x dx. [4] −4 … … … … … … … … … … … … … … … … … … … … … … … … … Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … …
9 marks
Mark scheme: 7(a) θ θ θ θ − B1 Attempt correct relevant identities to express in terms of cosθ only M1 M0 if moving terms from side to side. Confirm 3 4cos 3cos θ θ − with sufficient detail A1 AG 3 7(b) Use identity with 5 18 π θ = M1 Obtain 5 1 2 6 cos π and hence 1 4 3 − A1 2 7(c) Express integrand in form 1 2 (cos3 3cos ) (cos9 3cos3 ) + + + k x x k x x M1 Obtain correct integrand 9cos cos9 − x x A1 OE (allow unsimplified). Integrate to obtain form 3 4 sin sin9 + k x k x M1 Obtain correct 1 9 9sin sin9 − x x A1 Now simplified; condone missing ... + c . 4
6 (a) Show that 4 sin 1 cos 3 2 sin [4] 1 + 3π 1 −13π + 21. … … … … … … … … … … … … … … … … (b) Find the exact value of 4 sin 17 cos 1 [2] 24π 24π. … … … … … … … 18π 1(c) Find the exact value of 4 sin 2x cos 2x dx. [4] Ó 0 + 3π −13π … … … … … … … … … … … … … … … … … … … … … … … … …
10 marks
Mark scheme: 6(a) Obtain at least either 1 1 2 2 ( sin 3cos ) or 1 1 2 2 ( cos 3sin ) Expand and simplify with correct use of 2 2 sin cos 1 M1 Use 1 2 sin cos sin 2 M1 Confirm given result 3 2sin2 A1 AG necessary detail required. 4 6(b) Identify value of is 3 8 π *B1 OE Obtain 3 4 3 2sin π and conclude 3 2 DB1 or exact equivalent. 2 6(c) Identify integrand as 3 2sin4 x B1 Integrate to obtain form 1 2 cos4 k x k x M1 where 1 2 0 k k . Obtain correct 1 2 3 cos4 x x A1 Obtain 1 1 8 2 π 3 A1 or exact equivalent. 4
6 (a) Show that 4 sin 1 cos 3 2 sin [4] 1 + 3π 1 −13π + 21. … … … … … … … … … … … … … … … … (b) Find the exact value of 4 sin 17 cos 1 [2] 24π 24π. … … … … … … … 18π 1(c) Find the exact value of 4 sin 2x cos 2x dx. [4] Ó 0 + 3π −13π … … … … … … … … … … … … … … … … … … … … … … … … …
10 marks
Mark scheme: 6(a) Obtain at least either 1 1 2 2 ( sin 3cos ) or 1 1 2 2 ( cos 3sin ) B1 Allow if implied by decimal values. Expand and simplify with correct use of 2 2 sin cos 1 M1 Use 1 2 sin cos sin 2 M1 Confirm given result 3 2 sin 2 A1 AG necessary detail required. 4 6(b) Identify value of is 3 8 π *B1 OE Obtain 3 4 3 2sin π and conclude 3 2 DB1 or exact equivalent. 2 6(c) Identify integrand as 3 2 sin 4 x B1 Integrate to obtain form 1 2 cos 4 k x k x M1 where 1 2 0 k k . Obtain correct 1 2 3 cos4 x x A1 Obtain 1 1 8 2 π 3 A1 or exact equivalent. 4
5 (a) Find the quotient when 6x3 is divided by 2x 1 , and show that the remainder is 6. −5x2 −24x −4 + [3] … … … … … … … … … … … … … … … … … … … … … … … … … (b) Hence find 7 6x3 dx, −5x2 −24x −4 2x 1 Ô2 + giving your answer in the form a ln b, where a and b are integers. [5] + … … … … … … … … … … … … … … … … … … … … … … …
8 marks
Mark scheme: 5(a) Carry out division at least as far as 3x 2 + k1 x M1 OE (e.g. by inspection). Obtain quotient 3 x 2 − 4 x − 10 A1 Confirm given result of remainder is 6 with sufficient detail A1 AG SC If remainder = 6 shown using remainder theorem allow B1. 3 5(b) Integrate to obtain at least 3x and term of form k 2 ln(2 x + 1) *M1 ln term must be added. Obtain x 3 − 2 x 2 − 10 x + 3ln(2 x + 1) A1 Apply limits correctly to expression with four terms DM1 Apply appropriate logarithm properties correctly to obtain the form k3 ln a DM1 Obtain 195 + ln27 A1 5
3 (a) Find dx, giving your answer in the form ln a, where a is an integer. [4] 2x 4 −5 … … … … … … … … … … … … … … 10 (b) Find the exact value of [2] e2x−5 dx. 4 … … … … … … … …
6 marks
Mark scheme: 3(a) Obtain 2ln(2 x − 5) B1 Apply limits correctly M1 For integral of form k ln(2 x − 5) . Use one relevant logarithm property correctly M1 For integral of form k ln(2 x − 5) . Apply second logarithm property correctly and obtain ln25 A1 4 3(b) 1 2 x− 5 B1 Integrate to obtain e 2 1 15 1 3 B1FT or exact equivalent, FT on their ke 2 x −.5 Obtain final answer e − e 2 2 2
7 (a) Prove that 2 sin i cosec 2 i / sec i. [2] … … … … … … … … … … … … … … … … … (b) Solve the equation tan 2i + 7 sin i cosec 2 i = 8 for - r 1 i 1 r . [5] … … … … … … … … … … … … … … … … … … (c) Find 8 sin 2 12 x cosec 2 x d x . [3] y … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …
10 marks
Mark scheme: 7(a) 1 cosec2 sin 2 M1 Obtain 1 cos and confirm sec A1 Answer given – necessary detail needed. 2 7(b) Attempt to obtain quadratic equation in sec or cos only *M1 Obtain 2 7 2 sec 1 sec 8 involving one trigonometric ratio A1 Or equivalent, may be unsimplified, but reduce to 2 2sec 7sec 18 0 2 18cos 7cos 2 0 . Attempt to solve 3-term quadratic equation for sec, using a correct method, to find at least one value of DM1 Or equivalent using cos. Obtain any two of the four correct solutions 0.952, 1.76 A1 Or greater accuracy. Obtain remaining two correct solutions A1 Or greater accuracy; and no others between π and π . 5 7(c) Identify integrand as 2 1 2 2sec x B1 Integrate 2 1 2 sec k x to obtain 1 2 2 tan k x M1 Obtain correct 1 2 4tan x A1 Condone omission of ...c . 3
7 (a) Prove that 2 sin i cosec 2 i / sec i. [2] … … … … … … … … … … … … … … … … … (b) Solve the equation tan 2i + 7 sin i cosec 2 i = 8 for - r 1 i 1 r . [5] … … … … … … … … … … … … … … … … … … (c) Find 8 sin 2 1 x cosec 2 x d x . [3] y 2 … … … … … … … … … … … … … … … … … …
10 marks
Mark scheme: 7(a) 1 cosec2 sin 2 M1 Obtain 1 cos and confirm sec A1 Answer given – necessary detail needed. 2 7(b) Attempt to obtain quadratic equation in sec or cos only *M1 Obtain 2 7 2 sec 1 sec 8 involving one trigonometric ratio A1 Or equivalent, may be unsimplified, but reduce to 2 2sec 7sec 18 0 2 18cos 7cos 2 0 . Attempt to solve 3-term quadratic equation for sec, using a correct method, to find at least one value of *DM1 Or equivalent using cos. Obtain any two of the four correct solutions 0.952, 1.76 A1 Or greater accuracy. Obtain remaining two correct solutions A1 Or greater accuracy; and no others between π and π . 5 7(c) Identify integrand as 2 1 2 2sec x B1 Integrate 2 1 2 sec k x to obtain 1 2 2 tan k x M1 Obtain correct 1 2 4tan x A1 Condone omission of ...c . 3
a 10 5 It is given that d x = 7 , where a is a constant greater than 1. y 2x + 1 a (a) Show that a = 3 0.5e 1 .4 ( 2 a + 1 ) - 0. 5 . [5] … … … … … … … … … … … … … … … … … … … … … … … … … (b) Use an iterative formula, based on the equation in part (a), to find the value of a correct to 3 significant figures. Use an initial value of 2 and give the result of each iteration to 5 significant figures. [3] … … … … … … … … … … … … … … … … … … … … … … … … … …
8 marks
Mark scheme: 5(a) Obtain integral of form k ln(2 x + 1) *M1 Obtain correct 5ln(2 x + 1) A1 Apply limits correctly and equate to 7 DM1 Apply appropriate logarithm property to reach at least a 3 = ... DM1 3 1.4 A1 AG – necessary detail needed. Confirm a = 0.5e (2a + 1) − 0.5 5 5(b) Use iterative process correctly at least once M1 Obtain final answer 2.18 A1 Answer required to exactly 3 sf. Show sufficient iterations to 5 sf to justify answer or show a sign change in the A1 interval [2.175, 2.185] 3
6 (a) Find the quotient and remainder when 18x 3 - 6 x 2 - 30x + 4 is divided by ( 3x - 1 ) . [3] … … … … … … 5 3 2 18x - 6x - 30x + 4 (b) Hence find d x . Give your answer in the form a - ln b , where a and b are y 1 3x - 1 integers. [5] … … … … … … … … … … … … … … … … … … …
8 marks
Mark scheme: 6(a) Carry out division at least as far as 6 x 2 + k1 M1 Obtain quotient 6 x 2 − 10 A1 Obtain remainder − 6 A1 3 6(b) 2 6 B1 FT Following their quotient and remainder. Identify integrand as 6 x − 10 − 3 x − 1 Integrate to obtain at least 2x 3 and term of the form k 2 ln(3 x − 1) *M1 Obtain 2 x 3 − 10 x − 2ln(3 x − 1) A1 FT Following their quotient and remainder. Apply limits and appropriate logarithm properties DM1 Obtain 208 − ln49 A1 5