TMJC H2 Chapter 10 Integration Techniques Discussion Questions Solutions 2024

Chapter 10 Integration Techniques TMJC 2024 Page 1 of 24 H2 Mathematics (9758) Chapter 10 Integration Techniques Discussion Questions (Solutions) Level 1 Integration – Reverse of Differentiation 1 Find  d cosd   . Hence, find sin d   .  d cos cos sind        cos sin d cos cos d sin d cos sin d = cos d cos sin cos ', where ' C C C C C C                                         Integration of Standard Functions 2 Find the following integrals: (a)   5 32 e 3e dx x x (b) 1 2 21 d k xx       , 0k  (c)   2 5 2 dx x x x    Q2 Solutions (a) 5 3(2 e 3e ) dx x x 1 5 32 1 5 32 5 32 (2e 3e ) d 2e 3e 1 3 2 4e e x x x x x x x C C              Integration is reverse of differentiation

Chapter 10 Integration Techniques TMJC 2024 Page 2 of 24 (b)   2 1 1 2 1 2 4 41 d 1 d 44ln 4(1 4) 4ln 4 4ln 3 0 k k k x xx x x x x x k k k k k kk                                   (c) 2 3 1 1 2 2 2 (2 5)( 2) d 2 10 d 2 10 d x x x x x x x x x x x x            5 3 2 24 2 205 3x x x C    Need to expand first before integrating term by term.

Chapter 10 Integration Techniques TMJC 2024 Page 3 of 24 Integration involving the function and its derivative 3 Find the following integrals: (a) 23 7 dx x x (b) 2 d2 6 x xx  (c) 32 1e dxx x   Q3 Solution (a) 23 7 dx x x       1 2 2 3 2 2 3 2 2 1 14 3 7 d14 3 71 314 2 1 3 721 x x x x C x C              (b) 2 2 2 d2 6 1 4 d4 2 6 1 ln 2 64 x xx x xx x C         (c) 32 1e dxx x   3 3 2 1 1 1 3 e d3 1 e3 x x x x C       Formula to memorise (not in MF27) and apply   1 f ( ) f ( ) 1 f ( )(2) For , 1, f ( ) f ( ) d d ln f ( ) f ( ) (3) f ' ( ) e d e [f ( )](1) For , 1, f ' ( )[f ( )] d 1 x x n n xn n x x x x x C x x x C xn n x x x C n                     Note: 1. 22 6 x x  is a proper function 2.      f ' d ln f , f x x x C Cx     Remember to put modulus

Chapter 10 Integration Techniques TMJC 2024 Page 4 of 24 Integration of Rational Algebraic Functions (including MF27) 4 Find the following integrals: (a) 2 1 d3 4 tt (b)    1 d3 4 xx x  (c) 2 10 d 2 11 x x x  Q4 Solutions (a)     2 2 2 1 d3 4 1 d 3 2 1 1 3 2 ln22 3 3 2 3 3 2ln12 3 2 tt t t t C t t C t                      (b) 1 d( 3)( 4) xx x  1 1 d( 3) ( 4) ln 3 ln 4 3ln 4 xx x x x C x Cx              (c) 2 10 d 2 11 x x x    2 1 1 110 d 1 10 10 1tan 10 10 110 tan 10 x x x C x C                  

Chapter 10 Integration Techniques TMJC 2024 Page 5 of 24 Integration of Trigonometric Functions 5 (a) 3sin cos dx x x (b) 2sin dx x Q5 Solutions (a) 3sin cos dx x x = 4sin 4 x C (b)   2sin d 1 1 cos 2 d2 1 sin 2 2 2 x x x x xx C           2cos 2 1 2sinA A Use double angle formula f ( ) sin f ' ( ) cos x x x x  

Chapter 10 Integration Techniques TMJC 2024 Page 6 of 24 Integration by substitution 6 Using the substitution u x to find 1 d (1 ) x x x . Q6 Solutions (a) 2 d 2d u x u x x uu      2 2 1 d (1 ) 1 2 du(1 ) 2 du(1 ) 2 1ln2 1 1ln 1 x x x uu u u u Cu x C x               

Chapter 10 Integration Techniques TMJC 2024 Page 7 of 24 Integration by Parts 7 Find the following integrals: (a) ( 1)e d xx x  (b) sin 2 dx x x (c) e 1 ln dx x x Q7 Solutions (a) ( 1)e d xx x  e ( 1) e d e ( 1) e 2 e x x x x x x x x C x C               (b) sin 2 d 1 1cos2 cos 2 d2 2 1 1cos2 sin 22 4 1 1sin 2 cos 24 2 x x x x x x x x x x C x x x C               (c) 2 2 2 2 2 1ln d ln d 2 2 ln d2 2 ln 2 4 x xx x x x x x x x x x x x x C                 e2 2e 1 1 2 2 2 2 2 2 ln d ln 2 4 e e 1 1ln e ln12 4 2 4 e e 1 02 4 4 e 1 4 4 1 e 14 x xx x x x                                   Strategy: Consider solving the question without limits first. Strategy: Apply the limits only after solving the question.

Chapter 10 Integration Techniques TMJC 2024 Page 8 of 24 Level 2 Integration – Reverse of Differentiation 8 Find   2 1d ed xxx  . Hence, find  e 2 dxx x x  . Q8 Solutions     2 1 1 2 1 1 d e 2 e ed e 2 x x x x x x xx x x                 1 2 1 2 1 2 1 2 e 2 d e e e 2 d e 1e 2 d e e e e where e x x x x x x x x x x x C x x x x C Cx x x x Cx B B                    Alternative     1 2 1 2 1e 2 d e 2 d e 1 ee e x x x x x x x x x x x C x C              Integration is reverse of differentiation 1ex 1e exRewrite as and factorise the econstant .

Chapter 10 Integration Techniques TMJC 2024 Page 9 of 24 Integration involving the function and its derivative 9 Find the following integrals: (a) sin 2 cos 2 d   (b) cos 6e sin d 6 x x x (c) 1 2 sin d 1 x x x     (d)   1 d1 ln 3 xx x  (e)   3 33 e d 2 e x x x     (f) 2 7 3 d7 6 x xx x   Q9 Solution (a) Method 1: 2 2 sin 2 cos 2 d 1 2cos 2 sin 2 d2 1 sin 2 2 2 1 sin 24 C C                Method 2: sin 2 cos 2 d 1 2sin 2 cos 2 d2         1 sin 4 d2 1 cos 48 C        Formula to memorise (not in MF27) and apply   1 f ( ) f ( ) 1 f ( )(2) For , 1, f ( ) f ( ) d d ln f ( ) f ( ) (3) f ' ( ) e d e [f ( )](1) For , 1, f ' ( )[f ( )] d 1 x x n n xn n x x x x x C x x x C xn n x x x C n                         f sin 2 f ' 2cos 2 x x     sin 2 2sin cosA A AUse double angle formula

Chapter 10 Integration Techniques TMJC 2024 Page 10 of 24 (b) cos 6e sin d 6 x x x cos 6 cos 6 16 sin e d6 6 6e x x x x C          (c)     1 2 1 2 21 sin d 1 1sin d 1 sin 2 x x x x x x x C                  (d)   1 d1 ln 3 1 d1 ln 3 ln 1 ln 3 xx x x xx x C         (e)           3 33 3 33 33 3 23 23 e d e 2 e d 2 e 1 3e 2 e d3 2 e1 3 2 1 2 e6 x x x x x x x x x x x C C                         (f) 2 2 2 7 3 d7 6 1 14 6 d2 7 6 1 ln 7 62 x xx x x xx x x x C          f ( ) cos 6 1f ' ( ) sin 6 6 xx xx   1 2 f ( ) sin 1f ' ( ) 1 x x x x      f ( ) 1 ln 3 1 1f ' ( ) 3 3 x x x x x     3 3 f ( ) 2 e f ' ( ) 3e x x x x      2f ( ) 7 6 f ' ( ) 14 6 x x x x x    