Integral Calculus PDF

Integral Calculus Indefinite Integral Def: 1. Integration is the inverse of differentiation 2. Integration is summation Methods of Integration 1. Integration by substitution 2. Integration by parts 3. Integration by partial fraction 4. Integ...

Integral Calculus Indefinite Integral Def: 1. Integration is the inverse of differentiation 2. Integration is summation Methods of Integration 1. Integration by substitution 2. Integration by parts 3. Integration by partial fraction 4. Integration by successive reduction Integration is the inverse of differentiation dx7 = 7x 6  7 x dx = x 6 7 dx d ( x 7 + c) = 7 x6  7 x dx = x +c 6 7 dx S=Summation FORMULA 1.  2 dx 2 = 1 tan −1 x +c, x=atan  , acot  x +a a a 2.  2 dx 2 = 1 ln x − a +c, partial fraction x −a 2a x + a 3.  2 dx 2 = 1 ln a + x +c , partial fraction a −x 2a a − x  Type  2 dx ax + bx + c Example 1.1: Workout  2 dx 2 x + 3x + 1 dx  2x dx = 1 dx =1  dx + 3x + 1   2 = 3 1 9 3 1 3 1 2 2 2( x 2 + x + ) ( x + )2 + − (x + ) 2 − ( ) 2 2 2 4 2 16 4 4 3 1 x+ − = 1 ln 34 14 + c = ln 4 x + 2 + c 1 2. x+ + 4x + 4 2 4 4 px + q Type  dx ax + bx + c 2 Example 1.2: Workout  23x + 4 dx 2 x + 3x + 1 Differentiation of deno. 2x + 3x + 1 is 4x+3 2 3 9 (4 x + 3) + 4 − 3x + 4  2 x 2 + 3x + 1dx =  2 x 2 + 3x + 1 dx 4 4 = 3  24 x + 3 dx + 7  2 dx 4 2 x + 3x + 1 4 2 x + 3x + 1 = 3  dz + 7  2 dx = 3 lnz+ 7  2 dx +C 4 z 4 2 x + 3x + 1 4 4 2 x + 3x + 1 FORMULA dx 4.  =ln(x+ x2 + a2 )+c x=atan  , acot  x +a 2 2 dx 5.  =ln(x+ x2 − a2 )+c x=asec  , acosec  x −a 2 2 2  dx 6.  = sin −1 x +c x=asin  , acos  a2 − x2 a dx  x +a 2 2 x=atan  dx= a sec 2  d  a sec2 d  a 2 tan2  + a 2 =  sec d =ln(sec  +tan  )+ c1 x2 x2 + a2 sec  = 1 + tan  = 1 + 2 = 2 2 a a2 x2 + a2 sec = a x2 + a2 x = ln( + )+ c1 a a x2 + a2 + x =ln( )+ c1 ==ln( x 2 + a 2 + x )-lna+ c1 ==ln( x + x 2 + a 2 )+c a dx Type  ax 2 + bx + c dx Example 1.3: Workout  2 x + 3x + 1 2 dx dx 1 dx  2 x 2 + 3x + 1 =  3 1 = 2  3 1 9 2( x 2 + x + ) ( x + )2 + − 2 2 4 2 16 1 dx   =  = 1 ln  x + 3 + ( x + 3 )2 − ( 1 )2  +c 2 3 1 2  4 4 4  (x + ) 2 − ( ) 2 4 4 dx Type  put ax+b= z 2 (ax + b)(cx + d ) dx Example 1.4: Workout  3x-4= z 2 3dx=2zdz (2 x − 3)(3x − 4) 1 2 2 2z 2 + 8 − 9 2z 2 − 1 x = ( z 2 + 4) 2 x − 3 = ( z + 4) − 3 = = 3 3 3 3 3  dx = =2 3 zdz 2 dz 6 = (2 x − 3)(3x − 4) 3 (2 z − 1) z 2 2  1  2 z − 2   2 2 2  1  = ln( z + z 2 −   +c 6  2 dx Example 1.5: Workout  6 + 11x − 10 x 2 dx 1 dx  6 + 11x − 10 x 2 = 10  3 11 + x − x2 5 10 1 dx = 10  3 11 121 − (x − )2 + 5 20 400 11 x− = 1  dx = 1 sin −1 19 20 +c= 1 sin −1 20 x − 11 +c 10  19  2 11 2 10 10 19   − (x − ) 20  20  20 px + q Type  dx ax 2 + bx + c 3x + 4 Example 1.6: Workout  dx 2 x 2 + 3x + 1 Differentiation of 2 x 2 + 3x + 1 is 4x+3 3 9 (4 x + 3) + 4 − 3x + 4  2 x 2 + 3x + 1dx =  4 2 x 2 + 3x + 1 4 dx = 3  42x + 3 dx + 7  dx 4 2 x + 3x + 1 4 2 x + 3x + 1 2 3  dz +7 dx =3 z+ 7  dx +c =…… 4 z 4 2 x 2 + 3x + 1 2 4 2 x 2 + 3x + 1 4  ax + b Type  dx cx + d 2x + 4 Example 1.7: Workout  dx 3x + 3 2x + 4 2x + 4 2x + 4  dx = dx =  dx =……. 3x + 3 (3x + 3)(2 x + 4) (6 x 2 + 18 x + 12) dx Type  put cx + d = z 2 (ax + b) cx + d Example 1.8: Workout  dx (1 − x) x dx  (1 − x) x x= z 2 dx=2zdz = ln 1 + z +c= ln 1 + x 2 +c 2 2 zdz 2dz = = (1 − z 2 ) z 1− z2 1− z 1− x dx 1 Type  put ax + b = (ax + b) px + qx + r2 z dx Example 1.9: Workout  (1 + x) 1+ x − x2 dx  (1 + x) 1+x = 1 , dx = − 12 dz x= 1 −1 1+ x − x 2 z z z 3z − z 2 − 1 2 1  21 1 + x − x = 1 + − 1 -  − 1 =1 1 2 3 1 - 2 + −1= − 2 −1 = z z  z z z z z z2 1 −dz 2 dz dx = z =−  (same as  ) 1 3z − z 2 − 1 3z − z − 1 2 6 + 11x − 10 x 2 z z2 dz dz =−  =−  − 1 + 3z − z 2 3 9 −1 − (z − )2 + 2 4 5  dz =−  2 =……..  5 3   − (z − )2  2  2   1 x 2 dx Example 1.10 :  1 x = z L C M of the deno min ator of the fractional power 1+ x 3 x = z , dx = 6 z dz 6 5 3 5 8 =  z 6 z 2dz =6  z2 dz 1+ z z +1 =6  z ( z + 1) − z ( z + 1) 2+ z ( z + 1) − ( z + 1) + 1 dz 6 2 4 2 2 2 2 z +1 =6  ( z 6 − z 4 + z 2 − 1)dz +6  21 dz z +1 7 5 3 1 =6( z − z + z -z)+ 6 tan −1 z +c z = x 6 7 5 3 x dx Example 1.11:  3 do yourself 1+ x 4 Integration by parts  du  FORMULA  uvdx = u  vdx −   dx  vdx dx d(uw)=udw+wdu uw =  udw +  wdu  dx = x uw=  u dwdx +  w du dx let dw =v w =  vdx dx dx dx u  vdx =  uvdx +   du  vdx dx  dx   du   uvdx = u  vdx −   dx  vdx dx Example 1.12.  xe x dx = x  e x dx −   dx  e x dx dx = xe x −  e x dx = xe x − e x + c  dx  Example 1.13.  x 2 e x dx = x 2  e x dx −  (2 x  e x dx )dx = x 2 e x − 2[ xe x −  e x dx ] = x2e x − 2 xex + 2ex +c=( x2 − 2 x + 2)e x +c 6 x e dx = ( x3 − 3 x2 + 6 x − 6 ) e x +c 3 x  ln xdx  sin  cos −1 −1 xdx xdx 1  ln x1dx = ln x.x −  x xdx = xlnx-x+c LIATE L=Logarithm I=Inverse A=Algebraic T=Trigonometric E=Exponential Example 1.14: Workout  x sin xdx  x sin xdx , u=x v=sinx  x sin xdx u=sin x v=x −1 −1  e sin xdx u= e or sin x or, v=sinx or x x ex Example 1.15 Workout  e ax sin bxdx ,  e ax cos bxdx I=  e ax cos bxdx = e ax sin bx −  ae ax sin bx dx b b ax e sin bx a = b − b  e ax sin bxdx = e sin bx − a e ax  − cos bx  −  ae ax  − cos bx dx ax b b   b   b   e ax sin bx a ax a2 = − 2 e cos bx − 2  e ax cos bxdx b b b ax e sin bx a ax a2 = − 2 e cos bx − 2 I b b b  a2  ae sin bx + be cos bx ax ax 1 + 2  I=  b  b2  a2 + b2  ae ax sin bx + be ax cos bx  2   I=  b  b2 ae ax sin bx + be ax cos bx I= +c a2 + b2 7 FORMULA x x2 + a2 a2 7.  x 2 + a 2 dx = + ln(x+ x2 + a2 ) +c x=atan  , acot  2 2 x x2 − a2 a2 8.  x − a dx = 2 2 − ln(x+ x 2 − a 2 )x+c x=asec  , acosec  2 2 x a −x 2 2 a2 x 9.  a − x dx = 2 2 − sin −1 +c x=asin  , acos  2 2 a The results can be obtained by two methods. Integration by substitution and Integration by parts I=  x 2 + a 2 1 dx = x 2 + a 2. x - 1  22 x 2. x dx 2 x +a x2 =x x2 + a2 - dx x2 + a2 x2 + a2 − a2 =x x2 + a2 - dx x2 + a2 1 =x x2 + a2 - x 2 + a 2 dx + a 2  x2 + a2 dx 1 2I= x x2 + a2 + a2  dx x2 + a2 x2 + a2 2 I= x + a ln(x+ x2 + a2 )+c 2 2 Type  ax 2 + bx + c dx Example 1.16: Workout  2 x 2 + 3x + 1 dx 3 1 3 1 9  2 x 2 + 3x + 1 dx =  2( x 2 + x + )dx = 2  ( x + ) 2 + − dx 2 2 4 2 16 3 1 = 2  ( x + ) 2 − ( ) 2 dx 4 4  3 3 2 1 2 1  x +  (x + ) − ( )  1  2[ 4 = 4 4 + ln  x + 3 + 4 3 ( x + ) 2 − ( ) 2  ]+c 2 2  4 4 4  8 Type  ( px + q) ax 2 + bx + c dx Example 1.17: Workout  (3x + 4) 2 x 2 + 3 x + 1 dx Differentiation of 2x2 + 3x + 1 is 4x+3 3 9  (3x + 4) 2 x 2 + 3 x + 1 dx =   (4 x + 3) + 4 −  2 x 2 + 3x + 1 dx 4 4 = 3  (4 x + 3) 2 x 2 + 3x + 1 dx + 7  2 x 2 + 3x + 1 dx 4 4 =3  z dz + 7  2 x 2 + 3x + 1 dx 4 4 =1 z z +7 2 x 2 + 3x + 1 dx 2 4 = …….. Formula e (af ( x) + f ( x))dx = e ax f ( x) ax e (af ( x) + f ( x))dx ax = a  e ax f ( x)dx +  e ax f ( x))dx = a  f ( x)e ax dx +  e ax f ( x))dx e ax e ax = a[ f ( x) −  f ( x) dx] +  e ax f ( x))dx a a = e f ( x) −  e f ( x)dx +  e ax f ( x))dx ax ax = e ax f (x) For a=1  e ( f ( x) + f ( x))dx = e x x f ( x) Example 1.18: Workout  e x x + 12 dx 2 ( x + 1) x2 +1 x ( x − 1) + 2 2 e =  ( x + 1)2 dx x dx e ( x + 1) 2  2  =  ex  x − 1 + 2 dx =  e x ( f ( x) + f ( x))dx = e x f ( x) = e x x −1 +c  x +1 ( x + 1)  x +1 9  x −1 ( x + 1)1 − ( x − 1)1 2 f ( x) = then f ( x) = = x +1 ( x + 1) 2 ( x + 1) 2 dx Type  a + b sin x + c cos x Examples  dx ,  dx , dx dx  2 sin x + 3 cos x ,  6 + 3 sin x + 4 cos x 5 + 4 sin x 3 + 5 cos x dx Example 1.19: Workout  6 + 3 sin x + 4 cos x dx  6 + 3 sin x + 4 cos x sinx=2sin x cos x cosx=cos 2 x -sin 2 x 2 2 2 2 dx = x x x x 6 + 3.2 sin cos + 4(cos 2 − sin 2 ) 2 2 2 2 2 x sec dx x 1 x = x 2 x x z = tan dz = sec 2 dx 2 2 2 6(1 + tan 2 ) + 6 tan + 4(1 − tan 2 ) 2 2 2 2dz = 2dz = 2 =  2 dz =  dz 6(1 + z ) + 6 z + 4(1 − z ) 2 2 2 z + 6 z + 10 z + 3z + 5 3 9 (z + )2 + 5 − 2 4 3 z+ dz 2 = 2 = tan −1 2 +c 3  11  11 11 (z + ) 2 +    2 2  2  Type  a sin x + b cos x + c dx d sin x + e cos x + f 2 sin x + 3 cos x Examples  sin x dx , cos x  sin x + cos x dx ,  dx , sin x + cos x 4 sin x + cos x 2 sin x + 3 cos x + 5  sin x + 2 cos x + 6 dx Example 1.20: Workout  3 sin x + 14 cos x + 6 dx 4 sin x + 5 cos x + 3 3 sin x + 14 cos x + 6  4 sin x + 5 cos x + 3 dx 10 Nu. =l.Deno.+m. Diff. Coeff.of deno. +n 3 sin x + 14 cos x + 6 = l( 4 sin x + 5 cos x + 3 ).+m(4cosx-5sinx)+n =(4l-5m)sinx+(5l+4m)cosx +(3l+n) 4l-5m=3 20l-25m=15 m=1 l=2 n=0 5l+4m=14 20l+16m=56 3l+n=6 3 sin x + 14 cos x + 6 = 2( 4 sin x + 5 cos x + 3 ).+(4cosx-5sinx)+0 3 sin x + 14 cos x + 6 4sinx + 5cosx + 3 4 cos x − 5 sin x  4 sin x + 5 cos x + 3 dx = 2 dx +  4 sin x + 5 cos x + 3 dx 4 sin x + 5 cos x + 3 =2x +ln( 4 sin x + 5 cos x + 3 ) +c Integration by partial fraction dx Example 1.21: Workout  ( x − 1)( x − 2) dx  ( x − 1)( x − 2) 1 = A + B ( x − 1)( x − 2) ( x − 1) ( x − 2) 1=A(x-2)+B(x-1) Putting x=1 A=-1 x=2 B=1 1 =− 1 + 1 ( x − 1)( x − 2) ( x − 1) ( x − 2) Alternative method We can do the partial fraction easily without writing the constants A and B 1 1 1 = + ( x − 1)( x − 2) ( x − 1)(1 − 2) (2 − 1)( x − 2) =− 1 + 1 ( x − 1) ( x − 2) dx dx dx x−2  ( x − 1)( x − 2) =  (x − 2) -  (x − 1) =ln(x-2)-ln(x-1)+c= ln x − 1 +c 11  dx Example 1.22: Workout  ( x − a)( x − b)( x − c) 1 1 1 1 = + + ( x − a)( x − b)( x − c) ( x − a)(a − b)(a − c) (b − a)( x − b)(b − c) (c − a)(c − b)( x − c) dx 1 dx 1 dx  ( x − a)( x − b)( x − c) = (a − b)(a − c)  x − a + (b − a)(b − c)  x − b + 1 dx  (c − a)(c − b) x − c 1 1 1 = ln( x − a) + ln( x − b) + ln( x − c) +c (a − b)(a − c) (b − a)(b − c) (c − a)(c − b) Example 1.23: Workout  2 2dx 2 2 ( x + a )( x + b ) 1 1 1 = 2 + ( x + a )( x + b ) ( x + a )(−a + b ) (−b + a )( x 2 + b 2 ) 2 2 2 2 2 2 2 2 2 dx 1 dx 1 dx  ( x 2 + a 2 )( x 2 + b 2 ) = (b 2 − a 2 )  ( x 2 + a 2 ) + (a 2 − b 2 )  ( x 2 + b 2 ) = 2 1 2 1 tan −1 x + 2 1 2 1 tan −1 x +c (b − a ) a a (a − b ) b b xdx Example:1.24:  z = x2 dz = 2xdx ( x + 4)( x 2 + 6) 2 =1  dz =…….. 2 ( z + 4)( z + 6) dx Type  put ( x − a) = z( x − b) ( x − a ) ( x − b) n m dx Example 1.25: Workout  ( x + 1) ( x + 2) 3 2 put ( x + 1) = z( x + 2) x(1 − z) = 2 z − 1 12  2z − 1 x= 1− z (1 − z )2 − (2 z − 1)(−1) 1 dx = dz = dz (1 − z ) 2 (1 − z ) 2 2z − 1 1 x+2= +2= 1− z 1− z 1 1 dz dz (1 − z ) 2 (1 − z ) 2 =  (1 − z2) 3 dx dz  ( x + 1) 2 ( x + 2) 3 =  z 2 ( x + 2) 2 ( x + 2) 3 = 1 z z2 (1 − z ) 5 1 − 3z + 3z 2 − z 3 1 1 = 2 dz =  2 dz -3  dz +3  dz -  zdz z z z =- 1 -3lnz+3z- z +c, z = x + 1 2 z 2 x+2 dx Example 1.26: Workout  cos x(5 + 3 cos x) 1 1 3 1 1 = − [partial fraction of ] cos x(5 + 3 cos x) 5 cos x 5 5 + 3 cos x z (5 + 3z ) dx 1 3 1  cos x(5 + 3 cos x) =  [ 5 cos x − 5 5 + 3 cos x ]dx = 1  sec xdx − 3  1 dx 5 5 5 + 3 cos x 1 3 1 = ln(sec x + tan x) - same as type  dx 5 5 a + b sin x + c cos x dx Example 1.27: Workout  sin x(3 + 2 cos x) dx sin xdx sin xdx  sin x(3 + 2 cos x) =  sin 2 x(3 + 2 cos x) = (1 + cos x)(1 − cos x)(3 + 2 cos x) dz =−  =….. (1 + z )(1 − z )(3 + 2 z ) 13 Example 1.28: Workout sin x cos x  sin x + cos x  ( tan x + cot x ) dx=  ( + )dx =   dx cos x sin x  sin x cos x   sin x + cos x  = 2  dx let z=sinx-cosx, dz=(cosx+sinx)dx  1 − (sin x − cos x) 2    dz = 2 = 2 sin −1 z + c = 2 sin −1 (sin x − cos x) + c 1− z 2 Type  sin m x cos n xdx i) m or n or both odd ii) m and n is even iii) m+n is an even negative integer Examples i)  sin 2 x cos 5 xdx ,  sin 5 x cos 4 xdx ,  sin 5 x cos 3 xdx ,  sin 5 xdx ,  cos 3 xdx ii)  sin 4 x cos 2 xdx ,  sin 4 x ,  cos 4 xdx 3 3 2 iii)  sin 5 x dx ,  sin 7 x dx cos x cos 2 x Example 1.29: Workout  sin 2 x cos 5 xdx  sin x cos xdx =  sin x cos x cos xdx =  sin x(1 − sin x) cos xdx =……. 2 5 2 4 2 2 2 Example 1.30: Workout  sin 4 x cos 2 xdx  sin x cos xdx =  sin x sin x cos xdx =  sin x(sin x cos x) dx 4 2 2 2 2 2 2 =  sin 2 x( 1 2 sin x cos x) 2 dx = 1  sin 2 x sin 2 2 xdx = 1  (1 − cos 2 x)(1 − cos 4 x)dx 2 4 16 1 16  = (1 − cos 2 x − cos 4 x − cos 2 x cos 4 x)dx 2cosAcosB=cos(A+B)+ cos(A-B) 1 1 = 16  (1 − cos 2 x − cos 4 x − (cos 6 x + cos 2 x))dx 2 =……. 14 Alternative method Example: Workout  sin 4 x cos 2 xdx By De Moivre’s theorem (cos x + i sin x) n = cos nx + i sin nx ( cos x + i sin x)1 = cos x + i sin x (cos x + i sin x)2 = cos2 x + 2i sin x cos x − sin 2 x = cos 2 x + i sin 2 x (cos x + i sin x) n = cos nx + i sin nx Let z = cos x + i sin x 1 = z −1 = (cos x + i sin x) −1 = cos x − i sin x z z n = (cos x + i sin x)n = cos nx + i sin nx 1 n = z − n = (cos x + i sin x) − n = cos nx − i sin nx z 1 1 z+ = 2 cos x , z − = 2i sin x , z z 1 1 z n + n = 2 cos nx , z n − n = 2i sin nx z z 4 2 2 2  1  1  1  1  24 i 4 sin 4 x 22 cos2 x =  z −   z +  =  z −   z 2 − 2   z  z  z  z  =  z 2 − 2 + 12   z 4 − 2 + 14   z  z  = z 6 − 2 z 2 + 12 − 2 z 4 + 4 − 24 + z 2 − 22 + 16 z z z z =  z 6 + 16  -2  z 4 + 14  -  z 2 + 12  +4  z   z   z  =2cos6x-2.2cos4x-2cos2x+4 1  sin x cos 2 xdx =  (cos6x-2cos4x-cos2x+2)dx 4 64  sin do it 8 xdx 3 2 Example 1.31: Workout  sin 7 x dx 3 m = ,n = − 7 m + n = −2 2 2 2 cos x 15  3 2 3 3 5 5 sin x 2 2  7 dx =  tan x sec xdx =  z dz = z 2 + c = tan 2 x + c 2 2 2 5 5 cos 2 x =  sec1 x dx =  (1 + tan 4 2 dx x) sec2 x  1 7 1 dx put tanx=z 2 2 2 sin x cos x 2 tan x tan x Integration by successive reduction Example 1.32: Find the reduction formula for I n =  x n e ax dx hence find  x 3 e ax dx I n =  x n e ax dx ax e ax = xn e −  nx n −1 dx a a ax = x n e − n  x n−1e ax dx a a ax = x n e − n I n −1 a a ax n e n In = x − I n −1 a a I 3 =  x e dx 3 ax e ax 3 I3 = x3 − I2 a a ax ax = x 3 e − 3 ( x 2 e − 2 I1 ) a a a a 3 ax 2 ax = x e − 3x e2 + 62 I 1 a a a 3 ax 3 x 2 e ax e ax 1 e ax =x e − + 6 ( x − I0 ) I 0 =  e ax dx = a a2 a2 a a a 3 ax 2 ax ax ax = x e − 3x e2 + 6 xe3 − 6e4 + c a a a a 16 Example 1.33: Find the reduction formula for I n =  tan n xdx hence find  tan 6 xdx I n =  tan n xdx =  tan n−2 x tan 2 xdx =  tan n−2 x(sec 2 x − 1)dx =  tan n−2 x sec 2 xdx -  tan n−2 xdx n −1 = tan x − I n−2 n −1 tan n −1 x In = − I n−2 n −1 I 6 =  tan 6 xdx tan 5 x I6 = − I4 5 5  3  = tan x −  tan x − I 2  5  3  tan 5 x tan 3 x = − + tan x − I 0 5 3 5 3 = tan x − tan x + tan x − x + c 5 3  4 Example 1.34: Find the reduction formula for I n =  tan n xdx hence 0  4 find  tan 4 xdx 0 tan n −1 x In = − I n−2 n −1   tan n −1 x  4 1 In =   − I n−2 = − I n−2  n −1 0 n −1 1 1 1   2 I 4 = − I 2 = − (1 − I 0 ) = − 1 + = − 3 3 3 4 4 3 17 Example 1.35: Find the reduction formula for I n =  sin n xdx hence find  sin 4 xdx I n =  sin n xdx =  sin n−1 x sin xdx = sin n−1 x(− cos x) −  (n − 1) sin n−2 x cos x(− cos x)dx = − sin n −1 x cos x + (n − 1)  sin n − 2 x(1 − sin 2 x)dx = − sin n −1 x cos x + (n − 1)  sin n − 2 xdx − (n − 1)  sin n xdx = − sin n −1 x cos x + (n − 1) I n − 2 − (n − 1) I n I n + (n − 1) I n = − sin n −1 x cos x + (n − 1) I n − 2 nI n = − sin n −1 x cos x + (n − 1) I n − 2 sin n −1 x cos x n − 1 In = − + I n−2 n n Find I4 ………………  2 Example 1.36: Find the reduction formula for I n =  sin n xdx hence 0  2 find  sin 4 xdx 0   sin n −1 x cos x  2 n − 1 n −1 I n = −  + I n−2 = I n−2  n 0 n n   3 31 3 1  3 2 2  I4 = I2 = I0 = = I 0 =  sin 0 xdx =  dx = 4 42 4 2 2 16 0 0 2 Example 1.37: Find the reduction formula for I n =  e ax cos n xdx n I n =  e ax cos xdx cosn xeax e ax = −  n cosn −1 x(− sin x) dx a a 18  cosn xeax n = +  (cosn −1 x sin x)e ax dx a a n ax  ax ax  = cos xe + n cosn −1 x sin x e −  [(n − 1) cosn − 2 x(− sin x) sin x + cosn −1 x cos x] e dx a a  a a    n ax n −1 ax = cos xe + n cos x2sin xe − n2  eax [(n − 1) cosn − 2 x(cos2 x − 1) + cosn x dx a a a xeax n cosn −1 x sin xeax n   n = cos + 2 − 2  e ax [(n − 1) cosn x − (n − 1) cosn − 2 x + cosn x dx a a a   n ax n −1 ax = cos xe + n cos x2sin xe − n2  eax [(n cosn x − cosn x − (n − 1) cosn − 2 x + cosn x dx a a a ax cosn −1 x{a cos x + n sin x} n n e ax cosn xdx − (n − 1) e ax cosn − 2 xdx =e − 2     a2 a cosn −1 x{a cos x + n sin x} n ax =e − 2 nI n − (n − 1) I n − 2  a2 a n −1 e cos x{a cos x + n sin x} n 2 ax n(n − 1) = 2 − 2 In + In−2 a a a2  n2  e ax cosn −1 x{a cos x + n sin x} n(n − 1) 1 + 2  I n = + I n−2  a  a2 a2 e ax cos n −1 x{a cos x + n sin x} n(n − 1) In = + 2 I n−2 n2 + a2 n + a2 n Example 1.38: Find the reduction formula for I m ,n =  x m (ln x) dx n x m+1 n −1 1 x m +1 =  x (ln x) dx = (ln x) m +1  I m,n m n − n(ln x) dx x m +1 x m +1 n m +1 m +1  = (ln x) n − x m (ln x) n −1 dx x m +1 n = (ln x) n − I m,n −1 m +1 m +1 x m+1 n I m,n = (ln x) n − I m,n −1 m +1 m +1 19 Example 1.39: Find the reduction formula for I m ,n =  cos m x cos nx dx hence find  cos 3 x cos 3x dx I m ,n =  cos m x cos nx dx sin nx sin nx = cos m x −  m cos m −1 x(− sin x) dx n n sin nx m = cos m x +  cos m−1 x sin nx sin x dx n n cos(nx-x)=cosnx cosx+sinnx sinx sinnx sinx =cos(n-1)x-cosnxcosx sin nx m = cos m x +  cos m −1 x[cos(n - 1)x - cosnxcosx]dx n n sin nx m m = cos m x +  cos m−1 xcos(n - 1)x -  cos m xcosnx dx n n n sin nx m m = cos m x + I m −1,n −1 − I m,n n n n  m m cos x sin nx m 1 +  I m , n = + I m−1,n −1  n n n m+n cos m x sin nx m   I m,n = + I m−1,n −1  n  n n cos m x sin nx m I m,n = + I m −1,n −1 m+n m+n cos 3 x sin 3x 3 cos3 x sin 3x 3  cos 2 x sin 2 x 2  I 3, 3 = + I 2, 2 = +  + I 1,1  6 6 6 6 4 4  cos 3 x sin 3x cos 2 x sin 2 x 1 = + + I 1,1 6 8 4 cos x sin 3x cos x sin 2 x 1  cos x sin x 1 3 2  = + +  + I 0, 0  6 8 4 2 2  cos3 x sin 3x cos2 x sin 2 x cos x sin x 1 = + + + x +c 6 8 8 8 I 0, 0 =  cos x cos 0 x dx =  dx = x 0 20