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 x x x x x x x xx xx      =+ = + − =− − =− +       Alternatively, 2 0 2 2 0 0 2 2 0 0 22 0 22 0 e cos 2 d 11e sin 2 e sin 2 d22 1 1 1e sin 2 e cos 2 e cos 2 d2 4 4 51 e cos 2 d e 144 1 e cos 2 d e 1 5 x xx x x x x x xx x x x x x x x xx xx        =− = + − =− +  =− +       Qn 6 2015/ACJC Prelim/1/1 22 d3 , 3 , 2 . d uu x x u x x= − = − =− ( ) ( ) ( ) 1 32 2 13 22 53 22 53 22 22 13 d 3 d 2 1 3 d2 1 5 1 3 3 .5 x x x u u u u u u u u c x x c − =− − =− − = − + = − − − +   Combine 2 0 4 cos2xe x dx   with LHS to become 2 0 5 cos2xe x dx  

Chapter 10 Integration Techniques TMJC 2024 Page 5 of 8 Qn 7 2015/NJC Prelim/2/1 (a) 2 3tan d 3secd x x   = = When 3,x= tan 1 4 =  = When 3x= , 3tan 36 =  = ( ) ( )( )( ) 3 223 24 22 6 24 2 6 4 2 6 4 6 1 d 9 1 3sec d 9 tan 9 tan 9 11 3sec d9 tan 3sec 1 cos d9 sin 11 9 sin 1 1 1 9 sin sin46 22 9 x xx                 + = + = = −=    −=+    −=     (b) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 21 ln 4 d 1 ln 4 d 2ln 4 d 4 2 4 8 ln 4 d 4 8ln 4 2 d 4 ln 4 2 4 tan 2 xx xx xx x x x x x x x x x x x x x xx x x c − + =  + = + −  +  +−= + −  + = + − −  + = + − + +       Change limits to  Let 2ln( 4)ux=+ 1v= 2 d2 d4 ux xx= + dv x x=

Chapter 10 Integration Techniques TMJC 2024 Page 6 of 8 Qn8 2015/PJC Prelim/1/9 (a)(i) Consider ( ) 22d e 2 ed xx xx = Therefore, 22 1e d e 2 xxx x C =+ (a)(ii) 23edxxx ( ) 22 edxx x x= = 2221 ed2 xxx xe x− = 22211 ee22 xxxC −+ = 22211 ee22 xxxC −+ (b) 1 du uu −  = ( ) 2 2 1 sin 2sin cos dsin x x x xx −  = ( )cos 2sin cos dsin x x x xx = 22cos dxx = cos2 1dxx+ = sin 2 2 x xC++ = sin cosx x x C++ = 11 sinu u u C −− + + = 21 sinu u u C −− + + Let 2ux= 2 exvx= d 2d u xx = 21de 2 xvx = 2 1 sin sin sin u x x u xu − =  = = x 1 sin 1 ux= 1cos 1 ux −=

Chapter 10 Integration Techniques TMJC 2024 Page 7 of 8 Qn 9 2017/JJC Prelim/1/2 (a) ( ) ( )sin 3 cos 3 1 2sin 3 cos32 1 sin 62 1 cos 61 d d 2 d C         = = =− +    (b) xx   =  =  = 2 2 2 xx   =  =  = x =  d1 d 2x x  = . ( ) 32 2 cos d      ( ) 2 1 d 2 o csxx x x x   =  2 1 cos2 dx xx  =    ( ) 2 2 sin s i1 1d2 n xxxx    = −      2 0 cos1 22 x   =−  + ( )11 2 02 = − − − +  1 24 =− − Qn 10 2017/NYJC Prelim/1/4 (i) 2 1 3tan d 3secd x x   −= = Let ux= sinvx= d 1d u x = d cosv x x=−

Chapter 10 Integration Techniques TMJC 2024 Page 8 of 8 Qn 10 2017/NYJC Prelim/1/4 ( ) ( ) 22 2 2 2 2 2 2 11 dd 2 10 13 1 3sec d 3tan 3 1 3sec d3sec sec d ln sec tan 2 10 1ln 33 xx xx x C x x x C      = −+ −+ = + = = = + + − + −= + +     (ii) ( )13 2 2 42xx+ = − + ( ) ( ) ( ) 2 1 2 2 22 2 2 1 2 22 2 2 3 d 2 10 2 2 4 d 2 10 1 2 2 4 dd2 2 10 13 1 2 10 1 4d2 13 2 10 12 10 4ln 33 x x xx x x xx x xx xx x xx x x x x xx x C + −+ −+= −+ −=+ −+ −+ −+=+ −+ − + −= − + + + +    