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Tích phân hàm lượng giác

Chương 5c Tích phân chuong3a – nick yahoo, mail: chuong2a@gmail.com .2 .2 .3 .4 .5 .5 .6 .8 .8 4/ .9 .9 10 5/ Tích phân hàm lượng giác: .10 10 12 13 13 14 15 16 17 18 18 19 21 22 23 Check the result 23 25 27 27 28 30 31 32 32 dx 1/ I = ∫ x2 + a2 dx a/ I=∫ ⇒x= = x2 + a2 = ln x + x + a 2 t −a , dx = 2t 4t − 2t + 2a 4t 2 x + a = t − x ⇒ x + a = ( t − x ) = t − 2t.x + x , put ( ) dt = t2 + a2 2t t2 + a2 dx ⇒∫ x2 + a2 =∫ ( ' ' t − a 2t − ( 2t ) t − a 4t ) dt dt dt dt 2t = ∫ = ln t + C = ln x + x + a + C t t2 − a2 t− 2t   ⇒∫ =  ln x + x + a  = ln + + a − ln a 2  0 x +a 1b / b/ dx 1  i.a   = ln tan  arcsin     x  2 x2 + a2 dx ∫ ∫ dx x2 + a2 dx =∫ x2 − ' ⇒ dx = −i.a ( sin t ) dt = ( −1.a ) =∫ −i.a.cos t.dt dx x − ( i.a )  i.a  , t = arcsin    x  i.a ( i2 = −1) put x = sin t sin t sin t  i.a.cos t.dt   i.a.cos t.dt  1 ⇒ I = −∫  = −∫     sin t   i.a 2  sin t  −a 2 + a2   − ( i.a )  sin t  sin t  −i.a.cos t.dt  = ∫   sin t   −i.a.cos t.dt  = ∫   sin t  −1 + sin t a sin t − cos t a sin t  −i.a.cos t.dt  sin t −dt t 1  i.a   = ∫ =∫ = − ln tan = − ln tan  arcsin     sin t  x  2  sin t  i.a.cos t 1  a  = ln tan  arcsin    put a = i.b  x  2 x2 − a2 dx *∫ 1  i.b   = ln tan  arcsin     x  2 x + b2 dx ⇒∫ 1c / ∫ c/ ∫ dx x2 + a2 dx x2 + a2 = −i arcsin i.x +C a dx =∫ − ( −1.x ) + a2 =∫ dx a − ( i.x ) =∫ dx  i.x  a 1−    a  i.x i.dx a.du i.a.du = u ⇒ du = ⇒ dx = = = −i.a.du a a i i dx −1 i.a.du i.x ⇒∫ = ∫ = −i arcsin +C 2 a a a −x 1− u Put dx *I=∫ ⇒I=∫ Put x = a.sin t ⇒ dx = a.cos t.dt, t = arcsin a2 − x2 a.cos t.dt =∫ a.cos t.dt = ∫ dt = t = arcsin a − a sin t a cos t i.dy  i.y  Put x = i.y ⇒ I = ∫ = arcsin    a  a + y2 ⇒∫  i.y  = −i.arcsin    a  a + y2 dy x a x a ( i2 = −1) x = i.y ⇒ when x = 1, y = −i and when x = 2, y = −2i ⇒∫ dx x2 + a2 dx =∫ − ( −1.x ) = + a2 −2i ∫ −i dy a + y2 −2i i.y ⇒ ln x + x + a = −i.arcsin a −i i ( −i ) ⇔ ln + + a = −i.arcsin = −i.arcsin , arcsin z = −i ln  z ± z +    a a   2 eiw − e −iw w = arcsin z ⇒ z = sin w = ⇒ e2i.w − 2z.eiw − = ⇒ eiw = z ± z + ⇒ i.w = ln  z ± z +  ⇒ w = arcsin z = −i ln  z ± z +          x π 1 = ln tg  arctan +  + C a 4 2 x2 + a2 dx a.dt x I=∫ Put x = a.tan t ⇒ dx = , t = arctan a cos t x2 + a2 a.dt a.dt dt ⇒I=∫ =∫ =∫ cos t a tan t + a a.cos t tan t + cos t cos t *I=∫ dx  dt t π  ∫ cos t = ln  tan  +   ⇒ I = ∫    ⇒∫ x π 1 = ln tg  arctan +  a 4 2 x2 + a2 dx i.a  1 = ln x + x + a = i.ln tg  arcsin  x  2 x2 + a2 dx = −i arcsin i.x x π 1 = ln tg  arctan +  a a 4 2 so why dx ∫ x +a =∫ 2 dx ( x2 − −1.a ) dx =∫ − ( −1.x ) but + a2 2  i.a    1 2  ln x + x + a  ≠ i.ln tg  arcsin    x  1  1  2 2  i.x  x π  1 ≠  −i arcsin  ≠  ln tg  arctg +   a 1  a  1  2 dx 2a / I = ∫ a/ I=∫ a2 − x2 a.cos t.dt ⇒I=∫ * x a dat x = a.sin t ⇒ dx = a.cos t.dt, t = arcsin =∫ a.cos t.dt = ∫ dt = t = arcsin x a x a a − a sin t a cos t dx dx x dx =∫ Dat = u ⇒ du = ⇒ dx = a.du 2 a a a −x x a 1−   a ∫ dx ⇒∫ 2b / b/ a2 − x2 dx = arcsin a2 − x2 ∫ ∫ = dx a.du x = arcsin ∫ a − u2 a = −i.ln i.y + a − y + C a2 − x2 dx =∫ 2 a −x dx ( i.x ) +a Dat ( i.x ) + a = t − i.x ⇒ − x + a = ( t − i.x ) = t − 2t.ix − x ⇒x= = 2 t −a = 2t.i ( ) t2 − a2 i 2t.i −i4t + 2t i − 2i.a 4t dt = = ( ) − t2 − a2 i 2t ( −i 2t + 2a 4t , dx = ( ) = −i ( t + a ) dt 2t ) ( ' ' −i t − a 2t + ( 2t ) i t − a ( 2t ) ) dt ( −i t + a dx ⇒∫ ( 2t t − i.x =∫ a2 − x2 ) −i t + a dt =∫  ( ) =∫ ) dt =∫ ( (  ( dx *∫ ) −i t + a dt ( ) =∫ a − y2 dy ⇒∫ ∫ i.dy ∫ a − y2 dx a2 − x2 ( ) t ( t + a ) −i t + a dt = ln x + a + x + C put x = i.y, we have : x2 + a2 dx = i.dy, ) − t2 − a2 i   2t t − i   2t     2  i2 t − a  t2 − a2  2   2t t + 2t t −     2t 2t         −i.dt =∫ = −i.ln t + C = −i.ln i.x + a − x + C t 2 ) −i t + a dt = ln i.y + a − y + C = −i.ln i.y + a − y + C = arcsin x i.y = arcsin = −i.ln i.y + a − y2 a a i.y i.y  i.y  arcsin z = −i ln  z ± z +  , ⇒ arcsin = −i.ln +   +1   a a    a  i.y + a − y   = −i.ln = −i  ln i.y + a − y − ln a  = −i.ln i.y + a − y + C a   2c / ∫ a 1 = i.ln tg  arcsin  + C x 2 a2 − x2 dx dx 2c / ∫ =∫ a2 − x2 i.dx i2 dx =∫ ( x2 − a2 ) =∫ ( −1) ( x − a ) ( x2 − a2 ) ( i2 = −1) −i.dx =∫ dx ( x2 − a2 ) i ' −a ( sin t ) dt −a.cos t.dt a a dat x = ⇒ dx = = t = arcsin sin t x sin t sin t ⇒I=∫ − ( −i ) a.cos t.dt sin t a2 − a2 =∫ i.a.cos t.dt sin t = i.ln tg a sin t i.cos t.dt sin t dt =∫ = i.∫ sin t sin t cos t − sin t sin t t a 1 = i.ln tg  arcsin  + C x 2 Vay tai : ∫ dx a −x =∫ dx ( i.x ) +a dx =∫ ( −1) ( x −a 2  x a   1 2  arcsin a  ≠ i.ln tg  arcsin x   ≠ ln i.x + a − x  1    1  x2 − a2 −a.cos t.dt sin t ' −a ( sin t ) dt −a.cos t.dt a a dat x = ⇒ dx = = t = arcsin sin t x sin t sin t a2 sin t =∫   1 a 1 = − ln tg  arcsin  x 2 x2 − a2 dx ⇒I=∫ ) nhung dx 3a / I = ∫ a/∫ −a =∫ −a.cos t.dt sin t a − sin t sin t − cos t.dt sin t dt t a 1 = −∫ = − ln tg = − ln tg  arcsin  sin t x 2 sin t cos t 3b / 3b / dx ∫ 2 x −a dx ∫ = ln x + x + ( i.a ) dx =∫ x2 − a2 x2 + ( −1.a ) +C =∫ dx x + ( i.a ) Dat x + ( i.a ) = t − x ⇒ x − a = ( t − x ) ( dx = ) ' ' ( t + a 2t − ( 2t ) t + a ( 2t ) 2 t2 + a2 = t − 2t.x + x ⇒ x = , 2t 2 ) dt = 4t − 2t − 2a dt = t − a dt 4t 2t t2 − a2 dt dt 2t =∫ = ∫ = ln t + C = ln x + x + ( i.a ) + C 2 t t +a x2 + a2 t− 2t dx ⇒∫ 3c / ∫ 3c / ∫ =∫ dx x2 − a dx x2 − a2 i.dx i a −x = −i.arcsin Vay tai : ∫ dx =∫ =∫ x a =∫ dx −1 a − x i a − x −i.dx x = −i.arcsin a a2 − x2 dx x2 − a2 2 =∫ dx x2 + ( −1.a ) 2 =∫ dx −1 a − x 2 a x 1 − ln tg  arcsin  ≠ ln x + x − a ≠ −i.arcsin x1 a1 2 4/ I = ∫ dx −x − a dx I=∫ −x − a =∫ dx −i.dx =∫ i x2 + a2 x2 + a2 = −i.ln x + x + a i.a  i.x x π 1 1 = ln tg  arcsin  = − arcsin = −i.ln tg  arctg +  x  a a 4 2 2 i.a  1 = ln x + x + a = i.ln tg  arcsin  x  2 x2 + a2 dx ∫ = −i arcsin i.x x π 1 = ln tg  arctg +  a a 4 2 ∫ Vay tai : −x − a 2 dx −idx =∫ x2 + a2 dx =∫ − x + ( i.a )  =∫ nhung  −i.ln x + x + a 2  ( i.x ) − a dx 2  i.a    1  ≠ ln tg  arcsin x     1 1  2  i.x  x π  1 ≠  − arcsin  ≠  −i.ln tg  arctg +   a 1  a  1  2 x = arctan + C a x2 + a a dx x * I=∫ Dat x = a.tgt ⇒ dx = a tg t + dt, t = arctan a x2 + a2 5/ I=∫ dx ( dx Vay : ∫ x2 + a2 *I= ∫ dx *I= ∫ x2 + a2 dx x +a = =∫ ( ) a tan t + dt ( i x + i.a ln 2a x − i.a dx =∫ = ) a tan t + x2 − ( −1.a ) = =∫ ) t x ∫ dt = a = a arctan a + C a dx x − ( i.a ) 2 =∫ dx ( x − i.a ) ( x + i.a ) A B + ⇒ A ( x + i.a ) + B ( x − i.a ) = ( x − i.a ) ( x + i.a ) ( x − i.a ) ( x + i.a ) ⇔ x ( A + B ) + i.a ( A − B ) = ( i2 = −1) A + B = ⇔ A = −B  ⇔ i −i i i.a ( A − B ) = ⇒ A = = = , B=  2i.a 2i a 2.a 2.a  d ( x − i.a )  i ( ln x + i.a − ln x − i.a ) i  d ( x + i.a ) i x + i.a ⇒ I2 =  ∫ −∫ = ln = 2a  x + i.a x − i.a  2a 2a x − i.a 2 2 x  i x + i.a  1 so why : I = ∫ =∫ but  arctg  ≠  ln a 1  2a x − i.a 1 a  x + a x − ( i.a ) 6/ I=∫ dx dx dx x2 − a2 dx dx * I=∫ =∫ ( x − a) ( x + a) x2 − a2 A B = + ⇒ A ( x + a ) + B( x − a ) = ( x − a) ( x + a) ( x − a) ( x + a) A + B = ⇔ A = −B  ⇔ x ( A + B) + a ( A − B) = ⇔  1 a ( A − B ) = ⇒ A = 2a , B = − 2a  d( x + a)  d( x − a) −∫ ∫ 2a  x − a x+a ⇒I= *I= ∫ dx x2 − a2 =∫ dx x2 + ( −1.a )  ln x − a − ln x + a x−a = ln = 2a 2a x + a  =∫ dx x + ( i.a ) ( ) i.a ( tg t + 1) dt dx i.t −i.t −i x Vay : ∫ =∫ = ∫ dt = = = arctg + C a a i.a i a x + ( i.a ) ( i.a ) ( tg t + 1) i.a Dat x = i.a.tgt ⇒ dx = i.a tg t + dt, t = arctg 2 x i.a 2 1 x −a  x  −i Vay tai : ∫ =∫ nhung  ln ≠  arctg   2 2 i.a 1  2a x + a 1  a 1x −a x + ( i.a ) dx dx 5/ Tích phân hàm lượng giác: * I=∫ dx a sin x + b cos x + c 10 n 2π =0    −n.π + 2i.π  i =1 n a sin   + b  n     ∑ n →+∞ * lim * I= +π ∫ −π dx = ( sin x hàm le ) , a sin x + b 2π , x i ∈ [ −π, π] , ∆x = , a sin x i + b n dat f ( x i ) = n 2i.π − n.π + 2i.π 2π x i = −π + = , ⇒ lim I = lim ∑ n n  n →+∞ n →+∞ i =1   − n.π + 2i.π  n  a sin   + b n     n +π i =1 −π ∑ f ( x i ) ∆x = ∫ n →+∞ = lim * I=∫ dx sin x Put t = tan dx =0 a sin x + b x 2dt ⇒ x = arctan t, dx = , + t2 sin ( x/2 )  x x sin x = 2sin cos =  2 cos ( x/2 )  cos2 ( x/2 )   x cos x = cos2 − =  −  cos2 ( x/2 )  = ( ) = − t2     −1 tan ( x/2 ) =  .   cos2 ( x/2 )  tan ( x/2 ) +     = 2t + t2 −1 − tan ( x/2 ) + tan ( x/2 ) + ⇒I = ∫ + t2 dx 2dt 2t dt x =∫ ÷ = ∫ = ln t = ln tan + C sin x t + t2 + t2  at + 2t + a  sin x.dx 2t 2dt  2t 4t.dt  * I=∫ = ÷ +a = ÷    sin x + a ∫ t + t +  t +  ∫    t +1  t +1 ) ( =∫ 4t.dt , Put 4t = A + B ( t2 + 1) ( at2 + 2t + a ) ( t2 + 1) ( at2 + 2t + a ) t2 + at2 + 2t + a ⇒ A ( at + 2t + a ) + B ( + t ) = 4t ⇒ ( a.A + B ) t = 0, 2A.t = 4t, a.A + B = ⇒ A = 2, a.A + B = 2a + B = ⇒ B = −2a 21 ⇒I=∫ 4t.dt ( t2 + 1) ( at2 + 2t + a ) = 2.arctan t − ∫ * I=∫ =∫ 2dt t2 + − 2a ∫ dt at + 2t + a dt 2t t2 + + a  tan ( x/2 ) a +  sin x.dx a =x− arctan     2 sin x + a a −1 a −1   I1 = −2 ∫ ( where a > 1) dt dt dt = −2 ∫ = −2 ∫ 2 2t a2 −  1  t.a +  t2 + +  t +  +1−   + a  a  a  a a a2 − 1 a = M a2 a2 − dt  t.a +  ⇒ I1 = −2 ∫ = − arctan   M  a.M   t.a +    +M  a  If a > or a < −1 ⇒ a − > ⇒ = M2 ⇒  t.a +  t.a +  a  a =− arctan  arctan  =−   a    2 a −1 a −1  a −1   a −1  a  t.a +  dt a = 2.arctan t − arctan     2 ∫ t + 2t + a −1  a −1  a  tan ( x/2 ) a +  x a  = 2.arctan  tan  − arctan     2 2  a −1 a −1   ⇒ I = 2.arctan t − ⇒I=∫  tan ( x/2 ) a +  sin x.dx a =x− arctan     2 sin x + a a −1 a −1   π ( where a > 1)   tan ( π /2 ) a +   tan 0.a +   sin x.dx a = π−  arctan   − arctan       2 sin x + a a −1  a −1    a −    ⇒J=∫ π   = π−  − arctan     2 a −1   a −   a 22 sin x.dx a * I=∫ =x− sin x + a − a2 I1 = −2 ∫     ( where − < a < 1) dt dt dt = −2 ∫ = −2 ∫ 2 2t a2 −  1  t.a +  t2 + + t +  +1− +    a  a  a  a2 a2 If a < ⇒ a − < ⇒ ⇒ I1 = −2 ∫ =   ln tan ( x/2 ) a + + − a   tan ( x/2 ) a + − − a a2 − a2 dt  t.a +    −M  a  − a2 = −M ⇒ M = a =−    t.a +   t.a +   ln   − M − ln  +M  M   a   a      ln  t.a + + − a  a −1 − ln  t.a + − − a  a −1          − a2   a   ln t.a + + − a = − a  t.a + − − a  a  ,    dx dx  dx ∫ x − a = 2a  ∫ x − a − ∫ x + a     dt a  t.a + + − a ⇒ I = 2.arctan t − ∫ = 2.arctan t − ln  2t − a  t.a + − − a t + +1 a   x a  tan ( x/2 ) a + + − a   = 2.arctan  tan  − ln  2 2  − a  tan ( x/2 ) a + − − a  sin x.dx a ⇒I=∫ =x− sin x + a − a2 π sin x.dx a ⇒J=∫ = π− sin x + a − a2 = π+ a 1− a ln + − a2 1− 1− a   ln tan ( x/2 ) a + + − a   tan ( x/2 ) a + − − a   ln1 − ln + − a  − − a2  ( where a < 1) Check the result 23         ( where     − < a < 1) '   tan ( x/2 ) a +   x a   2.arctan  tan  − arctan     2 2   a −1 a −  x      x =  tan  + 1 2     −1 x   tan  2  '   tan ( x/2 ) a +  a  −  + 1    a −  a2 −     = ( cos ( x/2 ) ) ( cos ( x/2 ) ) −2 −1  tan ( x/2 ) a +      a −1   ( ( x/2 ) ) '   tan ( x/2 ) a  + tan ( x/2 ) a + + a −  a   −    a −1 a2 −       x  x    = − a a  tan  + 1 + 2a tan    2          −1 x   cos  2   2 x   a + 2a.sin ( x/2 ) cos ( x/2 )  = 1− a  cos   2    cos ( x/2 )      = − a  a + a.sin x    22 / I = ∫ =∫ =∫ = 1− dx dx =∫ cos x sinπ/2 − ( x d ( xπ/2 + sin ( xπ/2 + * I=∫ −1 ) ) ' −1 −2   x   tan      a2 −  a x   2 −1 a a + sin x − a sin x = = ( a + sin x ) ( a + sin x ) a + sin x ) =∫ = ln tgπ/2 + ( π/4 ) dx sinπ/2− (x − (π )) + C cos ( x/2 ) d ( x/2 ) dx dx =∫ =∫ sin x 2sin ( x/2 ) cos ( x/2 ) sin ( x/2 ) cos2 ( x/2 ) d ( tan ( x/2 ) ) tan ( x/2 ) = ln tan * I = ∫ tan x.dx = ∫ ' x +C −d ( cos x ) sin x.dx =∫ = − ln ( cos x ) cos x cos x 24 ' * I=∫  tan ( x/2 ) a −  cos x.dx 2a =x− arctan   where a < −1 or a >   cos x + a a +1 a −1   * I=∫ cos x.dx cos x + a Put t = tan x 2dt ⇒ x = 2arctan t, dx = , 2 1+ t  x cos x = cos2 − =  −  cos2 ( x/2 )  = ( ) = − t2 ⇒ I = − tan ( x/2 ) + tan ( x/2 ) + 1 + t2  .   cos2 ( x/2 )  −1  cos x.dx 2dt  − t   − t =∫  +a . ∫ cos x + a t +  t +   t +     ) ) ( (     ( ) −1 − t dt  − t + a.t + a −1 − t dt =∫   =∫   t +1 t + t ( a − 1) + a +   t +1 =∫ ( t + 1) ( t 2dt ( a − 1) + a + 1) Put )( ( −∫ = ( t + 1) ( t A.t + B + 2t dt ( a − 1) + a + 1) ) , C.t + D ( t2 + 1) ( t2 ( a − 1) + a − 1) t2 + t2 ( a − 1) + a + ⇒ ( A.t + B) ( t ( a − 1) + a + 1) + ( C.t + D ) ( t + 1) = ⇒ t ( a.A − A + C ) = 0, t ( a.A − A + C ) = 0, t ( a.B − B + D ) = 0, a.B + B + D = ⇒ A = C = 0, 2B = ⇒ B = 1, a + + D = ⇒ D = − a ⇒ I1 = ∫ =∫ 2.dt ( t2 + 1) ( t2 ( a − 1) + a + 1) dt t2 + +∫ ( − a ) dt = arctan t − dt ∫ a +1 t ( a − 1) + a + t + a −1 25 a +1 a +1  a +1  If > ⇔ a > or a < −1 ⇒ =  =M a −1 a −1  a −1  dt  t  ⇒ I1 = arctan t − ∫ = arctan t − arctan   M M t2 + M2 Put −2t ( t2 + 1) ( t2 ( a − 1) + a + 1) = A.t + B t2 + ) ( + C.t + D t ( a − 1) + a + ) ( ⇒ ( A.t + B) t ( a − 1) + a + + ( C.t + D ) t + = −2t ⇒ t ( a.A − A + C ) = 0, t ( a.A + A + C ) = 0, t ( a.B − B + D ) = −2t , a.B + B + D = ⇒ A = C = 0, a.B − B + D = −2, a.B + B + D = ⇒ B = 1, a + + D = ⇒ D = − ( a + 1) ⇒ I2 = ∫ −2t dt =∫ dt ( t2 + 1) ( t2 ( a − 1) + a + 1) ( t2 + 1) −∫ ( a + 1) dt ( t2 ( a − 1) + a + 1) ( a + 1)  t + a + −1 dt, = arctan t −  a −1 ( a − 1) ∫   a +1 a +1  a +1  If > ⇔ a < −1 or a > ⇒ =  =M a −1 a −1  a −1  ( a + 1) ( a + 1) arctan  t  dt ⇒ I = arctan t − = arctan t −   M ( a − 1) ( a − 1) ∫ t + M M ( a + 1) arctan  t  − arctan  t      M ( a − 1) M M M 1 2a  t   ( a + 1)  t  = 2arctan t − arctan    +  = 2arctan t − arctan   M ( a − 1)  M   M ( a − 1) M  M ⇒ I = I1 + I = arctan t −  tan ( x/2 )  x 2a a +1  = 2arctan  tan  − arctan   where M =  M ( a − 1) a −1   M  =x−  tan ( x/2 ) a −  2a a − arctan     a + ( a − 1) a +1   ⇒I=∫  tan ( x/2 ) a −  cos x.dx 2a =x− arctan   where a < −1 or a >   cos x + a a +1 a −1   26 π * cos x.dx π ∫ cos x + a = − * I=∫ I1 = ∫ π π a.π = − a2 − 2 a2 − 2a tan ( x/2 ) − a − a + cos x.dx a =x+ log where − < a < cos x + a tan ( x/2 ) − a + a + 1− a 2.dt ( t2 + 1) ( t2 ( a − 1) + a + 1) =∫ dt t2 + +∫ ( − a ) dt t ( a − 1) + a +  dt a +1 a +1 = arctan t − ∫ If < ⇔ −1 < a < ⇒ = −  a +1 a −1 a −1  t2 + a −1 dt t−M ⇒ I1 = arctan t − ∫ = arctan t − log   2M t+M t2 − M2 dt  dt dt  = −∫ ∫ ∫ t − M 2M  t − M t + M   I2 = ∫ −2t dt =∫ dt ( t2 + 1) ( t2 ( a − 1) + a + 1) ( t2 + 1) −∫ a +1 a −1   = −M   ( a + 1) dt ( t2 ( a − 1) + a + 1) ( a + 1)  t + a + −1 dt = arctan t − ( a + 1) t − M −1 dt = arctan t − )  a −1 ( a − 1) ∫  ( a − 1) ∫ (  ( a + 1) log t − M = arctan t − 2M ( a − 1) t+M ( a + 1) log t − M − log t − M 2M ( a − 1) t + M 2M t+M tan ( x/2 ) − M  ( a + 1) x   = 2arctan  tan  − log +   2 tan ( x/2 ) + M  2M ( a − 1) 2M   tan ( x/2 ) − M a a +1 a +1 =x− log , where M = = , M ( a − 1) tan ( x/2 ) + M a −1 1− a ⇒ I = I1 + I = arctan t − 27 −1 < a < =x− =x+ π * ∫ tan ( x/2 ) − M tan ( x/2 ) − M a 1− a a 1− a log =x+ log tan ( x/2 ) + M tan ( x/2 ) + M a + ( a − 1) a +1 1− a a − a2 log cos x.dx ( cos x + a ) ⇒I=∫ tan ( x/2 ) − a − a + tan ( x/2 ) − a + a + = where − < a < n tan ( x/2 ) − a − a + cos x.dx a =x+ log where − < a < cos x + a tan ( x/2 ) − a + a + 1− a π π 0 cos x.dx cos x.dx ⇒ I( a) = ∫ = π ⇒ I' ( a ) = ( −1) ∫ =0 cos x + a cos x + a ) ( π I ( a ) = ( −1) ∫ '' cos x.dx ( cos x + a ) π ⇒∫ cos x.dx ( cos x + a ) n = 0,  I ( a )    ( n −1) = ( −1) n −1 π ( n − 1) ! ∫ ( cos x + a ) = where − < a < Code Matlab: a = input('nhap vao can duoi a: '); b = input('nhap vao can tren b: '); x = a:10^(-3):b; y = cos(x).*(cos(x) + 1/2).^(-5); y1 = 0; y2 = 0; for i = : : length(x) - y1 = y1+y(i); end for j = : : length(x) - y2 = y2+y(j); end I = (b - a)/(3*length(x))*(4*y1 + 2*y2 + y(1) + y(length(x))) 28 cos x.dx n =0 a dx I=∫ 2 a −x +b , Put x = a.sin t ⇒ dx = a.cos t.dt, ) ( a − x = a − sin t = ( a.cos t ) π x = ⇒ t = 0, x = a ⇒ sin t = ⇒ t = ⇒ I = I=∫ π /2 ∫ a.cos t.dt = a.cos t + b π /2 ∫ cos t.dt , cos t + b/a tan ( x/2 ) − a − a + cos x.dx a =x+ log where − < a < cos x + a tan ( x/2 ) − a + a + 1− a If − < b/a < ⇒ −a < b < a π /2 ( b/a ) log tan ( t/2 ) − ( b/a ) − cos t.dt π ⇒ ∫ = + cos t + b/a tan ( t/2 ) − ( b/a ) + − ( b/a ) π /2 ( b/a ) + ( b/a ) + a+b π a−b tan   − π b a a 4 = + log a+b π a−b a − b2 tan   + a a a 4 a2 a ⇒I=∫ dx a2 − x2 + b dx I=∫    − x + 3   = = π b + log 2 a −b π + log 2 −3 a−b − a+b , where − a < b < a a−b + a+b 2− π = + log 2+ 2− 2+ Code Matlab, calculating by Simpson formula: a = input('nhap vao can duoi a: '); b = input('nhap vao can tren b: '); x = a - 10^(-3) : 10^(-3) : b - 10^(-3); y = ((25 - x.^2).^(1/2) + 3).^(-1); y1 = 0; y2 = 0; for i = : : length(x) - y1 = y1+y(i); end for j = : : length(x) - y2 = y2+y(j); end I = (b - a)/(3*length(x))*(4*y1 + 2*y2 + y(1) + y(length(x))) I = pi/2 + 3/4*log( ( 8^(1/2) - 2^(1/2) ) / ( 2^(1/2) + 8^(1/2) ) ) a = input('nhap vao can duoi a: '); b = input('nhap vao can tren b: '); x = a - 10^(-3) : 10^(-3) : b - 10^(-3); y = cos(x).*(cos(x) + 3/5).^(-1); y1 = 0; y2 = 0; for i = : : length(x) - y1 = y1+y(i); end for j = : : length(x) - y2 = y2+y(j); end I = (b - a)/(3*length(x))*(4*y1 + 2*y2 + y(1) + y(length(x))) 29 */ I = ∫ sin x = dx Put t = tan ( sin x.cos x ) 1/3 2t + t2 , cos x = ( − t2 + t2 )  + t2 2dt  ⇒I=∫  + t  t − t3   =∫ ( ) x 2dt ⇒ x = 2arctan t, dx = , + t2 ⇒ sin x.cos x = 1/3       =∫ ( 2dt ( t − t3 ( + t2 + t2 ) ) ) 2/3 22/3.dt =∫ 1/3 1/3 + t 21/3 t − t 1/3 + t2 t − t3 ( ) ( ) ( ) 22/3.dt ( t − t5 ) 1/3 Nếu R(– sinx, cosx) = – R(sinx, cosx) Khi đặt t = cosx (hàm lẻ theo sinx, chẵn theo cosx) Nếu R(sinx, – cosx) = – R(sinx, cosx) Khi đặt t = sinx (hàm lẻ theo cosx, chẵn theo sinx) Nếu R(– sinx, – cosx) = R(sinx, cosx) Khi đặt t = tanx (hàm chẵn theo sinx cosx) ( ) VD2 : Tính I = ∫ sin x.cos3 x + cos x dx Hàm duoi dau tich phan le theo cos x nên ta dat t = sin x ⇒ dt = cos x.dx ( ) ) ( ( ) ⇒ I = ∫ sin x.cos x + cos x.dx = ∫ t − t + dt t3 t5 sin x sin x = − +t= − + sin x + C 5 * I=∫ dx sin x + sin 2x − cos x dx VD3 : Tính I = ∫ sin x + sin 2x − cos x sin x + sin 2x − cos x = ( − sin x ) + ( − sin x ) ( − cos x ) − ( − cos x ) Hàm duoi dau tich phan chan theo sin x cos x nen ta dat t = tgx tgx t dt ⇒ sin x = = , cos x = , x = arctgt, dx = , 2 2 1+ t + tg x 1+ t 1+ t 30 sin 2x = 2sin x.cos x = 2t + t2  t + 2t −  d ( t + 1) dt dt ⇒I=∫ ÷ =∫ =∫ + t2  + t2  t + 2t − ( t + 1) −   = 2 t +1− ln = t +1+ 2 ln tgx + − dx ∫ x − a = 2a ln tgx + + x−a x+a π   x.dx π π π π    − arctan  = − arctan  =   2   2 2 a + sinx  a.M   a −1  a −1   a −1  * I=∫ where a > π π π 0 x.dx π−x π.dx * I=∫ =∫ ⇒ 2I = ∫ a + sinx a + sinx a + sinx substituting sinx = ⇒ dx = 2dt t2 + tan ( x/2 ) + tan ( x/2 ) , sin x = π π.dx π ⇒I= ∫ = + sinx +∞ 2t t2 + , x = ⇒ t = 0, t = +∞ 2dt  2t   a +  ∫ t2 +  t +1  +∞ π.dt , Put t = tan ( x/2 ) ⇒ x = 2arctan t −1 π ⇒ t = +∞ +∞ π.dt  t2 +  = ∫   t +  a.t + 2t + a    +∞ π dt π dt I= ∫ = ∫ = ∫ 2 a 2t +1 a  t +  + a −1 a.t + 2t + a t +   a  a a2 a −  a2 −  a  = M2 ⇒ = If a > ⇒ = M  a2 a2  a2 −   π ⇒I= a +∞ +∞ π   a.t +   = arctan   ∫ a.t +  a.M  0   a.M    +M  a  dt 31 π   x.dx π π π π    − arctan  = − arctan  =   2   2 2 a + sinx a.M     a −1  a −1   a −1  ⇒I=∫ where a > π * I=∫ x.dx = where a < a + sinx +∞ +∞ π.dt +∞ π dt π dt I= ∫ = ∫ = ∫ 2 a 2t +1 a  t +  + a −1 a.t + 2t + a t +   a  a a2  − a2  a −1 a  = −M2 ⇒ = If a < ⇒ = − M  a2 a2  − a2   π ⇒I= a π ⇒I=∫ π * I=∫ +∞ +∞ +∞ π  a.t + − a.M  =  log ∫ a.t + a.t + + a.M 0    2a.M    −M  a  dt =0 x.dx = where a < a + sinx x.dx =π + sinx π.dt +∞ +∞ π dt π dt I= ∫ = ∫ = ∫ 2 a 2t +1 a  t +  + a −1 a.t + 2t + a t +   a  a a2 If a = ⇒ a2 − a2 +∞ +∞   =0⇒I=π ∫ = −π   t + 10  ( t + 1) dt 32 π =π⇒I= ∫ x.dx =π + sinx 3π /4 π /4 π /4 π * I= ∫ = ∫ where y = x − ∫ 2 π /4 cos x + −π /4 2sin y + =2 π /4 ∫ dx dy 2sin y + =2 π /4 ∫ dy dy = − cos 2y π t  π /2 dt π   =  tan −1  tan   = − cos t 0   ∫ −2 π /2 2 − cos ( x/2 )  dx x x    I1 = ∫ , Put t = tan ⇒ cos x =  cos  − = −2 − cos x 2  cos ( x/2 )    = ( ) = − t2 , x = π ⇒ t = 1, x = ⇒ t = 0, x = arctan t ⇒ dx = − t2 + t2 + t2 + 2dt t2 + 1 2dt  − t2  2dt  2t + − − t  2dt π I1 = ∫  − = [ arctan t ] =  = ∫  =∫ 2   t2 +  t +  t2 +    t +1 t +1  33 ... ∫ =∫ nhung  ln ≠  arctg   2 2 i.a 1  2a x + a 1  a 1x −a x + ( i.a ) dx dx 5/ Tích phân hàm lượng giác: * I=∫ dx a sin x + b cos x + c 10 Put t = tg = x 2dt x x ⇒ x = 2arctgt, dx = , sin... cosx (hàm lẻ theo sinx, chẵn theo cosx) Nếu R(sinx, – cosx) = – R(sinx, cosx) Khi đặt t = sinx (hàm lẻ theo cosx, chẵn theo sinx) Nếu R(– sinx, – cosx) = R(sinx, cosx) Khi đặt t = tanx (hàm chẵn... =0    −n.π + 2i.π  i =1 n a sin   + b  n     ∑ n →+∞ * lim * I= +π ∫ −π dx = ( sin x hàm le ) , a sin x + b 2π , x i ∈ [ −π, π] , ∆x = , a sin x i + b n dat f ( x i ) = n 2i.π − n.π

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