DHS Integration & Its Applications (9758) (Revision Solutions)

Y5 Topical Revision Package 1 7. Integration and its Applications (solutions) 1 7. Integration and its Applications (solutions) 1 (a) (b) (c) (d) CJC PROMO 2010/QN10 ∫ x 3 + x dx = ∫ 3 + x – 3 3 + x dx = ∫ 1 – 3 3 + x dx = x – 3 ln | 3 + x | + c ∫x2 ln x dx = [ x3 3 ln x ] – ∫x3 3 1 x dx = [ x3 3 ln x ] – 1 3 ∫x2 dx = [ x3 3 ln x ] – x3 9 + C ∫ x + 3 x2 + 4x + 7 dx = 1 2 ∫ 2x + 6 x2 + 4x + 7 dx = 1 2 ∫ 2x + 4 + 2 x2 + 4x + 7 dx = 1 2 [∫ 2x + 4 x2 + 4x + 7 dx + ∫ 2 x2 + 4x + 7 dx] = 1 2 [ ln (x2 + 4x + 7) + ∫ 2 (x + 2)2 + 3dx ] = 1 2 [ ln (x2 + 4x + 7) + 2 3 tan–1 x + 2 3 ] + C Let x = 2 sin θ dx = 2 cos θ dθ ∫ 2x – 1 4 – x2 dx = ∫ 4 sin θ – 1 2 cos θ 2 cos θ dθ = ∫ 4 sin θ – 1 dθ = – 4 cos θ – θ + c = – 4 4 – x2 2 – sin–1 x 2 + c = – 2 4 – x2 – sin–1 x 2 + c

Y5 Topical Revision Package 1 7. Integration and its Applications (solutions) 2 2 (i) (ii) (iii) DHS PROMO 2009/QN9 d (sin 2 ) 2cos 2d xxx  ( ) ( ) ( ) ( ) ( ) 2 2 2 1 sin cos d (sin cos ) cos sin d cos sin ( sin cos ) c os sin d cos sin 1 1 cos sin xx x x x x xx xx x x x xx xx C xx − − − + = +− − = −− − − −=−+ − = +− ∫∫ ∫ C ( ) 6 2 0 6 6 00 22 6 0 sin sin2 cos sin2 d cos sin 11sin2 . .2cos2 dcos sin cos sin 1 cos sinsin . 2 d3 cos sincos sin66 xx xx x xx x xxxx xx xx xxx π π π π π ππ + − = − −− −= − −− ∫ ∫ ∫ ( ) [ ] 6 0 6 0 3 2 cos sin d 31 33 2 sin cos2 33 3 22 x xx xx π π = −+ − +=−− = − ∫ 3 (a) (b) ACJC PROMO 2010/QN9 2cos3 cos 3 1 ln sin 3 cot 3sin 3 cot 3 3 x ec x dx x x cxx − = +++∫ ( ) ( ) ( ) 2 22 1 16 1 2 2 2 21 4 11 1 16 4 1 16 11 1 32 1 164 32 xx dx dx dx x xx dx x x dx x − − = − − −− = +− − − ∫ ∫∫ ∫∫ ( ) 12 12 12sin 1 1614 324 11sin 4 1 164 16 x xC x xC − −  = + −+ = +−+

Y5 Topical Revision Package 1 7. Integration and its Applications (solutions) 3 (c) ( ) ( ) ( )∫∫ −−   −=− − dxx x xx dx x x 1 1 1 1lnln1 2 = ∫    +−−− dxx x x x 1 1 1 1 ln = ln ln 1 ln1 x x xCx+ −− +− 4 TJC PROMO 2009/QN2 (a) 2 11 1 11 ln lnln 2 ln 2 ln 2 xxdx dx dx x Cxx x x = = = +∫∫∫ (b) 21 21 211 21 21 x xxe dx e dx e C xx − −−= = + −−∫∫ 5 4 22 31 3(1 tansec sec tan ) tand dCθ θ θθ θ θ+ + += =∫∫ 6 (a) JJC PROMO 2010/QN12 (2 6) 2 6A x b Ax A b+ += + + Comparing coefficients of x , 121 2AA= ⇒= Comparing coefficients of constant term, 34 1BB+=⇒= 14 (2 6) 12xx∴+= + + 2 4 d 6 13 x x xx + ++∫ 2 1(2 6) 12 d 6 13 x x xx ++ = ++∫ 22 1 26 1 d d2 6 13 6 13 x xx xx xx += + ++ ++∫∫ 2 22 1 26 1 d d2 6 13 ( 3) 2 x xx xx x += + ++ ++∫∫ ( ) 211 13ln 6 13 tan2 22 xxx c− += +++ + 

Y5 Topical Revision Package 1 7. Integration and its Applications (solutions) 4 (b) Using 1x u= , 2 1ddxu u = − When 2x= , 1 2u= . When 4x= , 1 4u= . 1 4 3 2 1 dxex x∫ 1 4 3 21 2 1 duue u u = − ∫ 1 4 1 2 duue u= −∫ 1 2 1 4 duue u=∫ (shown) 1 4 3 2 1 dxex x∫ 1 2 1 4 duue u=∫ 1 1 22 1 14 4 duuue e u= − ∫ 11 1224 14 11 24 uee e  =−−  1 1 11 2 4 2411 24e e ee      = − −−     11 4231 42ee= −

Y5 Topical Revision Package 1 7. Integration and its Applications (solutions) 5 (c) 2 if 2 2 ( 2) if 2 xxx xx + ≥−+= − + <− 0 3 -3 2 dxx+∫ [ ] -2 0 3 3 -3 -2 ( 2) d ( 2) dx x xx= −+ + +∫∫ 2 044 32 ( 2) ( 2) 44 xx − −−  ++=−+    411(0 1) (2 0)44= − −+ − 17 4= 7 (a) (b) RVHS PROMO 2010/QN11 xx duue e u dx= ⇒= = Then ( ) 2 2 2 11 22 1 2 1 2 xx dudxee uu u duu du u − = + + = + = + ∫∫ ∫ ∫ -1 112 tan tan 222 2 xue cc − = += +  ( ) 1 0 2 1 0 2 11 00 22 11 00 22 45 32 2(2 2 ) 1 32 22 1 2 32 32 22 1 2 32 41 x dx xx x dx xx x dx dx xx xx x dx dx xx x − +− −−−= +− −= −− +− +− −= −− +− −− ∫ ∫ ∫∫ ∫∫ 11 2 -1 0 0 1 -1 1 0 1 2 2 3 2 sin 2 11 01 2 2 4 2 3 sin sin 22 xxx −  − = − +− −    − − = −−− −    

Y5 Topical Revision Package 1 7. Integration and its Applications (solutions) 6 4 3 8 0 6 24 3 48 6 π π = −−− −−= 8 VJCPROMO2013/QN5 ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 22 2 2 22 2 22 2 2222 2 22 d 1 (2 )(a) i d1 1 1 1 21 1 21 11 21 11 x x xx xx x x x x x x xx xx +− = + + −= + −−= + += − ++ = − ++ ( ) ( ) ( ) ( ) ( ) 1 1 2 222 00 1 11 2 02 0 1 22 0 1 22 0 21ii d111 112 d tan 21 112 d 241 11 d 481 xxxxx xx x x x x x π π −   −= ++ + −=+ = + + = + + ∫ ∫ ∫ ∫ 2 2 22 2 e(b) RHS = + 1e ee 1e x x xx x A AA − −+= − Comparing the numerator to that of the LHS, 22ee1 1 xxAA A − += ⇒= 2 22 2 1e d1 d1e 1e 1ln1 e2 x xx x xx xC = +−−  = − −+ ∫∫ If 2x≥ , then |x − 2| = (x − 2) If x >2, then |x − 2| = −(x − 2)

Y5 Topical Revision Package 1 7. Integration and its Applications (solutions) 7 9 HCI/2020Prelim/I/6 a. 3e d5 0.3e x x x−∫ e d5 0.3e 10ln 0 5 0.3 .1 e 30 x x x x C = − = −− −− + ∫ b. Let ( )cos ln dI xx=∫ ( ) ( ) cos ln 1' sin ln ux ux x = =− , '1v vx = = ( ) ( ) ( ) ( ) 1cos ln sin ln d cos nl sin nl d I x xx x x x xx xx = −−  = + ∫ ∫ ( ) ( ) sin ln 1' cos ln ux ux x = = , '1v vx = = ( ) ( ) ( ) ( ) ( ) ( )o n 1 s cs c cos ln cos ln in l ssin ln o n ld d n l I xxxx xx x x xx x x xx − = + =+− ∫ ∫ ( ) ( ) ( ) ( ) 2 cos ln sin ln cos ln sin ln2 Ix x x x xI x xC = + = ++ c. When 2e 5 0, ln2.5x x−= = ( ) . 2e 5 ln 2 2 2e , e5 5 .5 2, ln 5 x x x x x −   −< −≥ −

Y5 Topical Revision Package 1 7. Integration and its Applications (solutions) 8 3 0 2e 5 dx x−∫ ln2.5 3 0 l n2.5 2e 5 d 2e 5 dxx xx= − −+ −∫∫ ln2.5 3 0 ln 2.5 5 2e 2e 5xxxx = − +−  ( ) ( ) ( ) ln 2.5 0 3 ln 2.5 5ln 2.5 2e 2e 2e 15 2e 5ln 2.5 = − ++  −− − ln2.5 310ln2.5 4e 2e 13= − +− ( ) 310ln 2.5 4 2.5 2e 13= − +− 310ln2.5 2e 23= +− 10 NJC/2020Promo/6 (i) ( ) ( ) ( ) ( ) 22 1 22 2 1 2 22 2 2 1 122 2 2 1 22 2 2 d 1 1d 1 21 d2 11 12 2 1 1 1 x x kx x kx x kx kx xk kx k kx Ck C − − −+ − = − −=−− −−= − − −+ + + = ⌠ ⌡ ⌠⌡ ⌠⌡ (ii) ( ) 1 22 sin d 1 xkx x kx − − ⌠ ⌡ ( ) 1 22 dsin d 1 , vxu kx x kx − == − ( ) 1 22 2 222 d1 ,1d 1 uk v kxxk kx −= −= −

Y5 Topical Revision Package 1 7. Integration and its Applications (solutions) 9 ( ) ( ) ( ) ( ) ( )( ) ( )( ) 1 22 11 1 22 2222 22 22 1 1 22 2 2 1 1 22 2 2 sin d 1 11sin 1 1 d 1 sin 1 1 d sin 1 D xkx x kx kk x kx kx xkk kx kx k x xkk kx k x x kk − − − − − −−= −−− − −− = −− += + + ⌠ ⌡ ⌠⌡ ⌠⌡ (iii) ( ) ( ) ( ) ( )( ) 1 22 1 2 2 1 2 1 1 212 2 0 1 1 1 0 2 When 1, sin d becomes 1 sin d so 1 sin d 1 sin 1 11s 2 1 22 π 2 in 1 11 2 π 2 4 1 1 4 xk k x x kx xxx x xxx x xxx − − − − − = − − − = − =−  −+     = −+ =   −+ −  ⌠ ⌡ ⌠ ⌡ ⌠ ⌡ (iv) Method 1 ( ) ( ) 1 2 1 2 sin d 2 1 1 1 π 4 m m xmxm x xm + −   −− −− =  − ⌠  ⌡ Both integrand and limits of integration underwent a translation of m units in the positive or negative x-direction so the area under the curve is preserved. Or

Y5 Topical Revision Package 1 7. Integration and its Applications (solutions) 10 Method 2 Let u xm= − d 1d u x = xm= , 0u mm=−= 1 2 xm= + , 11 22 u mm= +−= ( ) ( ) ( ) 2 1 2 2 1 2 1 1 0 sin d 1 sin d 1 1 42 1 π m m xmxm x xm uuu u + − − −− −− = − = −    ⌠  ⌡ ⌠ ⌡ 11(a)