TMJC H2 Chapter 10 Integration Techniques Extra Practice Solutions 2024

Chapter 10 Integration Techniques TMJC 2024 Page 1 of 8 H2 Mathematics (9758) Chapter 10 Integration Techniques Extra Practice Solutions Qn 1 2018/ACJC Prelim/1/6 (i) ( ) ( ) ( ) 1 12 2 2 11 22 22 2 2 1 2 Let sin 2 , 1 4 14 21 , 1 4 8 1 4 814 1 1 4 4 sin 2 d 14 dv xu x x x dx x du v x x x x dxdx x xC xxx x −− −− − = = = − − = = − =− − − − =− − + −  ( ) ( ) ( ) 1 2 2 2 12 12 1 1 2sin 2 1 4 1 4 d44 14 11sin 2 1 4 d 42 11sin 2 1 442 x x x x x x x x x x x C − − −     = − − − − −         − = − − + = − − + +     (ii) ( ) 2 22 21 1 d26 1 2 2 2 d d2 2 6 15 1 2 1ln 2 6 tan2 55 x xxx x xxxx x xx x C − − ++ +=− ++ ++ += + + − +   

Chapter 10 Integration Techniques TMJC 2024 Page 2 of 8 Qn 2 2011/CJC Prelim/2/2 22 3131)31)(31( 102 x CBx x A xx x + +++= ++ + By cover-up rule, when 1 3 1 −=−= Ax )31)(()31( 1102 2 xCBxxx ++++−=+ ( )( ) ( ) ( ) 22 2 10x A Bx C 1 3x1 3x 1 3x 1 3x ++ =+ ++ + + When x = 0, C = 3 When x = 1, B = 1 x x x xx xx x d 31 3 31 1d )31)(31( 102 1 0 2 1 0 2  + +++−= ++ + ( ) ( ) 1 21 0 1 11ln 1 3 ln 1 3 3 tan 336 11ln 4 ln 4 3 tan 336 13ln 463 x x x  − − = − + + + + =− + + =− + Qn3 2011/DHS Prelim/1/8 (a) ( ) ( ) 22 2 12 1 11 1 1 xx xx x x x =−+ − −=+− − ( ) ( ) 22 11 d d1 2 1 1 1 ln 1 1 x xxx x x x xC x −=+− + − − = − + + −  (b) (i) ( )( ) 11 2 1 12 2 12 1sin d sin . d 1 1 sin 2 1 d 2 sin 1 x x x x x x x x x x x x x x x C −− −− − =− − = + − − = + − +   (b) (ii) ( ) 2 2 2 22 21 d23 231 d 23 2 2 11 d 23 12 11ln 2 3 tan 22 x xxx x xxx x xxx x xx x x C − −+ −=+ −+ −= + − −+ −+ −= + − + − +    Express ( )11xx=− − + and split the fraction Let 1sinux −= 1v= 2 d1 d 1 u x x = − d1v x dx x== Long division for improper fraction

Chapter 10 Integration Techniques TMJC 2024 Page 3 of 8 Qn 4 2011/IJC Prelim/1/3 ( ) ( ) 2 2 ln 2 d 25 2 ln 2 x x xx   −   ( ) 2 2 21 ed2e 25 2 u u u u u = −  2 2 d25 2 u uu= − 2 2 2 1 2 25 25 d2 25 2 1 251d2 25 2 u uu uu − + −=− − =− − −   ( ) ( )( ) ( ) ( ) ( ) 1 1 5 2 25 ln2 2 2 5 5 2 1 5 5 2 ln2 2 2 5 2 5 2 ln 215ln 2 ln2 2 2 5 2 ln 2 uuc u uuc u xxc x   +=− − +  −   +=− − + −  +=− − + − 1 d 1ee2 d 2 uu xx u=  = Visualise 1d e d 2 uxu =

Chapter 10 Integration Techniques TMJC 2024 Page 4 of 8 Qn 5 2015/MI Prelim/1/2 (i) sin d1 2cos 1 2sin d2 1 2cos 1 ln 1 2cos2 x xx x xx xC + −=− + =− + +   (ii) 2 0 22 0 0 22 0 0 22 0 22 0 e cos 2 d e cos 2 2e sin 2 d e cos 2 2e sin 2 4e cos 2 d 5 e cos 2 d e 1 1 e cos 2 d e 1 5 x xx x x x x x xx