Integrate functions of the form — Paper 2 · IGCSE Mathematics - Additional 0606

9 (a) Find e 2x + 1 dx . [2] y x d (b) (i) Given that y = , find y. [3] ln x d x J 1 1 1 N - + (ii) Hence find 2 2 OO d x . [3] y KK ln x ( ln x) x L P

8 marks

This question in 0606/22 Feb/March 2017

11 (a) Find y 6 dx . [3] x i - 5 ) d i . [2] (b) (i) Find y cos (4 2 i - 5 ) d i . [2] (ii) Hence evaluate y cos (4 .125

7 marks

Mark scheme: 11(a) x 2 ( x 6 + 1) 2 1 B1 = x + soi x 6 x 4 x 3 x −3 B2 B1 for any two out of three terms + + c oe, isw correct 3 −3 11(b)(i) k sin(4θ− 5) where M1 1 k > 0 or k = − 4 sin(4θ− 5) A1 ( +c ) 4 11(b)(ii) sin(4(2) − 5) sin(4(1.25) − 5) M1 FT their (b)(i), dep on M1 awarded in − (b)(i) 4 4 sin(3) sin(0) or − 4 4 0.0353 or 0.03528[…] oe, cao A1

This question in 0606/22 Feb/March 2019

dy r 8 r(b) A curve is such that = 3 - 2 cos 5x . The curve passes through the point , dx b 5 5 l. (i) Find the equation of the curve. [4] ydx and hence evaluate (ii) Find r ydx . [5] r y 2y

13 marks

Mark scheme: 12(a) 2e x B1 seen 2 2e a 1 M1 Uses limits correctly for their integral − = 50 and sets = 50 2 2 Rearranges and takes logs to base e: M1 Using their integral 2 a = ln101 oe 1 A1 Allow any exact equivalent a = ln101 or ln 101 final answer 2 12(b)(i) 2 B2 B1 for −k sin5 x where k > 0 [ y = ]3 x − sin5 x [ + c ] 5 8π 3π 2  π  M1 = − sin 5 × + c   5 5 5  5  2 A1 y = 3 x − sin5 x + π 5 12(b)(ii)   2   B3 2 y d x = 3 x − sin5 x + π d x B2 for cos5 x oe nfww     ∫ ∫  5    25 2 2 3 x 3 x 2 and B1FT for + … + πx [ + c ] = + cos5 x + πx [ + c ] 2 2 25  π  M1 their F(π) – their F    2  16[.0] or 15.95 to 15.96 or A1 13π 2 2 − 8 25

This question in 0606/21 May/June 2019

7 Giving your answer in its simplest form, find the exact value of 4 10 (a) dx, [4] y 0 5 x + 2 nl 2 2 4 x + 2 (b) e d.x [5] y 0 ` j

9 marks

Mark scheme: 7(a) 2ln(5x + 2) B2 B1 for kln (5x + 2) 2 ( ln(22) − ln(2) ) oe soi M1 2 A1 2ln11 or ln121 or ln11 e 8 x + 4 d x M17(b)  ln 2 M1  1 8 x+ 4  e    8  0 1 ln2 8 4 4 M2 1 ln2 ( e × e − e ) oe M1 for ( e 8 + 4 − e 4 ) 8 8 A1 255e 4 or exact equivalent 8

This question in 0606/22 May/June 2020

4 It is given that y = ln ( sin x + 3 cos x) for 0 1 x 1 r. 2 dy (a) Find . [3] dx dy 1 (b) Find the value of x for which = - . [3] dx 2

6 marks

Mark scheme: 4(a) d y cos x − 3sin x 3 M1 for attempt at chain rule must have = function in numerator and denominator d x sin x + 3cos x A1 for denominator A1 for numerator (b) –2 cos x – 3 cos x = sin x – 6 sin x M1 Expand and collect terms in sin x and cos x 1 = tan x M1 sinx Use = tanx cos x π A1 Must be radians x = 4

This question in 0606/21 Oct/Nov 2020

4 It is given that y = ln ( 1 + sin x) for 0 1 x 1 r. d y. [2] (a) Find d x d y r 1(b) Find the value of when x = , giving your answer in the form , where a is an integer. d x 6 a [2] d y (c) Find the values of x for which = tan x . [5] d x

9 marks

Mark scheme: 4(a) dy 1 M1 = dx 1 + sinx cosx A1 × cosx = 1 + sinx 4(b) π dy M1 insert into their 6 dx 1 A1 3 not 3 3 4(c) cos x sin x M1 sinx their = replace tan x with 1 + sin x cos x cos x use cos 2 x = 1 − sin 2 x M1 earned when equation reduced to a 2 quadratic in sinx 2sin x + sin x −=1 0 ( ) ( 2sin x − 1)( sin x + 1) = 0 M1 solve three term quadratic in sinx π A1 or 0.524 or better radians only x = 6 if M0 M0 M0 and (a) and (b) correct, allow SC2 for 1 π tanx = , x = 3 6 5 π A1 or 2.62 or better radians only x = A0 if extra solution(s) in range 6

This question in 0606/23 Oct/Nov 2020

11 (a) (i) Find 6 dx . [2] 10x - 1 e dd` j 2 c 2 x 3 + 5 (ii) Find ` j d x . [3] dd x e (b) (i) Differentiate y = tan ( 3 x + 1) with respect to x. [2] c 10r sec 2 ( 3x + 1) (ii) Hence find dd - sin x d x . [4] r e 2 o e12 Question 12 is printed on the next page.

11 marks

Mark scheme: 11(a)(i) (10 x − 1) −5 B2 (10 x − 1) −5 1 ( + c ) isw B1 for k ( + c ) , where k ≠ −×5 10 −5 10 11(a)(ii)  5 2 25  B1  4 x + 20 x +  d x   x  4 6 20 3 B2 B1 for any 3 terms correct x + x + 25ln x + c 6 3 11(b)(i) 3sec 2 (3 x + 1) B2 B1 for k sec 2 (3 x + 1) where k ≠ 3 11(b)(ii) sec 2 (3 x + 1) tan(3 x + 1) B1 dx =  2 6 oe, soi B1 − sin x d x = cos x oe   π   π  M1 F   − F   where  10   12  F(x) = k1 tan(3 x + 1) + k 2 cos x oe 0.322 or 0.3222[32...] rot to 4 figs A1

This question in 0606/22 Feb/March 2021

10 (a) Find ( e x + 1 ) 3 d x . [2] (b) (i) Differentiate, with respect to x, y = x sin 4x . [2] r 3 1 r 3 (ii) Hence show that 4x cos 4xdx = - . [4] yr 8 6 4

8 marks

Mark scheme: 10(a) 1 3 x + 3 1 3 3 x B2 3 x + 3 3 3 x 1 e + c or e × e + c nfww B1 for k e or ke × e where k ≠ or 0 3 3 3 10(b)(i) d(sin4x ) B1 = 4cos4 x soi dx Applies correct form of product rule: B1 FT their 4 cos 4x if possible 4x cos 4x + [1] sin 4x isw M1 FT use of their mx cos 4x + n sin 4x where m10(b)(ii)  (4 x cos4 x )d x =  x sin 4 x − sin 4 xd x     and n are constants 1 A1 x sin4 x + cos4 x [ + c ] soi 4 π  π  1  π  A1 sin  4 ×  + cos  4 ×  − 3  3  4  3   π  π  1  π    sin  4 ×  + cos  4 ×    4  4  4  4   Correct completion to given answer A1 1 π 3 − 8 6

This question in 0606/22 May/June 2021

3 (a) Differentiate ln ( x 3 + 3x 2 ) with respect to x, simplifying your answer. [2] x + 2 (b) Hence find dx . [2] y x ( x + 3)

4 marks

Mark scheme: 3(a) 3( x  2) 3 x  6 2 mark final answer or or simplified equivalent; x ( x  3) x 2  3 x 3 x 2  6 x B1 for oe x 3  3 x 2 23(b) 1ln( x 3  3 x 2 )  c 3  3 x 2 ) B1 for 1ln( x 3 3

This question in 0606/23 May/June 2023

3 (a) Find 4x + 5 - dx . [3] 2x + 3 3 1 (b) Hence find the exact value of 4x + 5 - dx , simplifying your answer. [3] + 3 o y1 e 2x

6 marks

Mark scheme: 3(a) 2 1 B3 2 1 2 x + 5 x − ln ( 2 x + 3 ) + c oe B2 for 2 x + 5 x − ln ( 2 x + 3 ) 2 2 2 1 or 2 x + 5 x − ln2 x + 3 + c 2 or 2 x 2 + 5 x + k ln ( 2 x + 3 ) + c with k ≠ 0 or B1 for 2 x 2 + 5 x +…+c 1 or ... − ln2 x + 3 2 or ... + k ln ( 2 x + 3 ) with k ≠ 0 3(b) Substitutes limits and subtracts in correct M1 FT their part (a) providing it includes a term order k ln ( 2 x + 3 ) with k ≠ 0  1   1  A1 18 + 15 − ln9 − 2 + 5 − ln5      2   2  1 9 1 5 A1 26 − ln or 26 + ln oe 2 5 2 9

This question in 0606/21 Oct/Nov 2023