ejercicios primer departamental

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Page 1: EJERCICIOS Primer Departamental

EJERCICIOS DEL CUADERNO

1) ∫ x4 d= x 4+1 +c 4+1

= x 5 +c = d [x5/5 +c] = dx 5 + dc = 1 dx5 +dc = 5x 4 = 5 dx dx 5 dx 5 dx 5

=x 4 dx

2) ∫ 5 x4 dx = 5∫x4 dx = 5 (x 5 ) +c = 5

=x 5 +c

3) ∫7dx = 7∫dx = 7∫x -3 dx = 7x -2 +c = X3 X3 -2

=-7 +c2x2

__ 4) ∫3√ x4 dx = ∫(x4)⅓ dx =∫ (x 4 ) ⅓ 4 dx =1 ∫ u ⅓ +1 du = 1 u 4/3 + c = 4 4 ⅓ +1 4 4/3

=1 3√x4

4

5) ∫ ( 5ax⅔ √ 3 ) dx = ∫5ax⅔ dx - ∫ √ 3 dx = 5a∫ x⅔ dx - √ 3 ∫dx = X2 X 2 __X2 =5a x 5/3 +c - √ 3 ∫ x2 dx = 15 a x5/3 -√ 3 x -1 +c

5/3___ __ 5 -1

= 3 a 3 √ x 5 - √ 3 x +c

__ __ __ __6) ∫(3 √ x +1) (√ x -2)] dx = ∫[3x -5√ x -2] dx = ∫3x dx - ∫5√ x

dx - ∫ 2 dx = 3 ∫x dx -5 ∫x1/2 dx -2∫dx = 3x 2 - 5x 3/2 -2x +c __ 2 3/2

= 3 x2 -10 √ x3 -2x +c 2 3

Page 2: EJERCICIOS Primer Departamental

7) ∫(2x+1) (x-3) dx = ∫(2x2 -5x -3) dx = ∫2x2 dx – x dx - ∫3dx = 2∫x2 dx -5∫x dx - 3 ∫dx

= 2x 3 – 5x 2 - 3x +c

3 2 ___

8) ∫(2a - b + 3c 3√ x2 ) dx = 2a ∫x1/2 dx - b∫x-2 +3c∫x2/3 +c √ X x2

= 2a x 1/2 – b x -1 +3c – x 5/3 +c ½ -1 __5/3

=4a √x + b + 9 c 3√ x5 + c

X 5 _ __9) ∫(t 2 – 5t – 2t 3 √ t ) dt = ∫t 2 dt - ∫5t dt ∫2t 3 √ t dt = ∫dt - ∫5 dt -∫2

t -2/3 dt t 2 t2 t2 t2 t t2

=∫dt – 5 ∫ dt - ∫2t-2/3 dt = t – 5 Ln l t l – 2 ∫t-

2/3 dt t

= t – 5 Ln l t l – 2t 1/3 + c 1/3

10) ∫( x 3 + 5x 2 -4 ) dx = ∫x 3 dx - ∫5x 2 dx - ∫4 dx = ∫xdx + 5∫ dx - 4∫x-2 dx

X2 x2 x2 x2

= x 2 + 5x -4 (x -1 ) +c = x 2 +5x +4 +c 2 -1 2 x

11) ∫(3cos x +4x8) dx = ∫3cos x dx + ∫4x 9 dx = 3∫cos x dx + 4∫x 9 dx

9 9 = 3 sen x +4x 9 +c

9

12) ∫(tanǿ + cotǿ)2 dǿ= ∫(tan2ǿ +2tanǿ cotǿ +cot2 ǿ) dx = ∫ (tan2ǿ+2 [(sen ǿ ) (cos ǿ )] +cot2 ǿ) dx

[(Cosǿ) (senǿ)] = ∫ (tan2ǿ dx + ∫ csc2ǿ) dx

=∫(tan2ǿ+1+cot2ǿ+1) dx = ∫ (sec2 ǿ + csc2ǿ) dx = ∫sec2ǿ dx + ∫csc2ǿ dx =

= tanǿ - cotǿ +c

13) ∫(sen x + 1/x) dx = ∫ sen x dx + ∫ dx = X

=cos x + Ln l x l +c

Page 3: EJERCICIOS Primer Departamental

14) ∫cos x dx = √ 1-cos2 x

15) ∫ (x + 4)4 dx = ∫u4 du = u 5 + c = (x+4) 5 +c 5 5

U= x+4du= dx

16) ∫4(4x+1)3 dx = ∫ (4x+1)3 dx = ∫u3 du = u 4 +c = (4x+1)4

+c 4 4

U= x+4du= dx

17) ∫3x2 (x3-5)2 dx = ∫ (x3-5)2 3x2dx= ∫u2du = u 3 +c = (x 3 -5 )3

+c 3 3

U= x3-5 du= 3x2dx

18) ∫(x2-4)5 2x dx = ∫ (x2-4)5 2x dx = ∫u5du = u 6 +c = (x 2 -4 )6

+c 6 6

U= x2-5 du= 2x dx

19) ∫ x (x2-3)3 dx = ∫ (x2-3)3x dx = ∫ (x2-3)3 2x dx = 1 ∫ (x2-3)3

2x dx 2 1

U= x2+3 = 1 ∫u3du = 1 [u 4 +c] = u 4 +c = (x 2 -3 )4 +c du= 2x dx 2 2 4 8 8

20) ∫ (x-1) dx =∫ 2(x-1)dx = 1 ∫2(x-1)dx = 1 ∫du = 1 Ln l u l + c

X2 -2x 2(x2-2x) 2 (x2-2x) 2 u 2

U= x2 -2x = 1 Ln l x2-2x l +cdu= 2(x-1)dx 2

21) ∫_2x dx_ = 2x dx (6-5x2)-1/3 = ∫(-5) _2x dx__ = ∫dx (6-5x2)-1/3

3√6-5x (-5) (6-5x2)-1/3

= -1 ∫ (6-5x2)-1/3 (-10x dx) = -1 ∫u-1/3 du = -1 u 2/3 +c

Page 4: EJERCICIOS Primer Departamental

U=6-5x2 5 5 5 2/3 du=-10xdx = -1/5 u2/3 +c = -3 u2/3 +c = -3 (6-5x2)2/3 +c

2/3 10 10

22) ∫ e ǿ d ǿ = ∫ be ǿ d ǿ = 1 ∫ be ǿ d ǿ = 1 ∫du =1 Ln l u l +c =1 Ln l a+beǿ l +c

a+beǿ b (a+beǿ) b a+beǿ b u b b u= a+beǿ du= beǿ dǿ

23) ∫ x ex2 dx = ∫e x2 x dx= ∫ e x2 2 x dx = 1 ∫ e x2 2x dx = 1 e

x2 +c 2 2 2 U= x2

du= 2xdx

24) ∫ e x2+3 x dx = ∫ e x2+3 2 xdx = 1 ∫ e x2+3 2xdx = 1 e x2+3 +c 2 2 2

U= x2+3 Du= 2xdx

25) ∫ (e3x – 4)2 dx = ∫ (e6x-8e3x+16) dx = ∫ e6xdx - ∫8e3xdx+∫16dx

= ∫ e6x 6 dx - ∫8e3x 3 dx + ∫16dx = ∫ 6 3 U= 6x = 1 ∫ e6x 6dx -8 ∫ e3x 3dx +16 ∫dx

Du= 6dx 6 3 U= 3x = 1 e6x -8 e3x +16x +c Du= 3dx 6 3 _ _ _ __

26) ∫ e √x = ∫ e √x 2 dx = 2∫ e √x _dx_ = 2 e√x +c

√ X √ X 2 √ X 2√ x __ U= √ x du= dx_ 2√ x __

27) ∫ 3 dx = ∫3dx = ∫x-1/2 3dx = 3∫x-1/2 dx = 3 x -1/2 +c = 6x1/2

+c = 6√ x +c

Page 5: EJERCICIOS Primer Departamental

√ X x1/2 1/2 __ U= √ x du= 3dx

28) ∫ 3 dx = ∫3dx = ∫(ex)-1/2 3dx = 3∫(ex)-1/2 dx = 3 (e x ) 1/2 +c = 6x1/2 +c

√ ex (ex)1/2 __ 1/2

__ = 6(ex) ½ +c= = 6√ ex +c U= √ ex

du= 3dx

29) ∫ tdt_ = ∫ 6tdt_ = 1 ∫ du = 1 ∫ u-4 du = 1 u 3 = u 3 +c = (3t2+1) 6(3t+1)4 6 u4 6 6 du -18

= ___1___ + c U=3t2+1 -18(3t+1)3

du=6tdt

30) ∫z sen (7z5-3) dz = ∫ 35z 4 dz sen (7z2-3) = 1_ ∫sen dz 1 -cos u +c

35 35 35 U= 7z5-3 = -cos (7z 5 - 1) +c Du= 35z4dz 35

31) ∫ 3dq__ = 3 ∫ dq___ = 3 ∫u-4 du = 3 u -3 = 3 u -3 = 3___+c5q (Lnq +2) 5 q (Lnq +2)4 5 5 -3 -15 -15 (Lnlql+2)3

U= Ln q+2 Du= dq/q

32) ∫ [e1-3y +2 cos (y/q)] dy = ∫ e1-3y dy + ∫2cos y dy =∫e1-3y -3dy +∫cos(y)

9 -3 (9) U=e1-3y = 1 ∫ eu du +2 ∫cos (y) dy du=1-3y 3 (9) = -e 1-3y + 2sen (y) +c

3 (9)

33) ∫ sen 3x dx = ∫sen 3x 3 dx = 1 ∫sen 3x 3dx = -1 cos 3x +c

3 3 3 U= 3x

Page 6: EJERCICIOS Primer Departamental

du=3dx

34) tan 9x dx = ∫ tan 9x 9 dx = 1 ∫sen 9x 9dx = 1 Ln lsec 9xl +c

9 9 9 U= 9x du=9dx

35) tan 3x dx = ∫cos 3 3/5dx = 5 ∫ cos 3 x 3 dx = 5 sen 3 x +c

5 5 3/5 3 5 5 3 5 U= 3/5x du=3/5dx

36) ∫ dx__ = ∫sec2 2x 2dx = 1 tan 2x +cCos2 2x 2

U= 2x Du= 2dx

37) ∫ sec 5x tan 5x dx = ∫sec 5x 5 dx = tan 5x 5 dx = 1 ∫ sec 5x +c

5 5 5 U= 5x Du= 5dx __

38) ∫ sen √ x dx = ∫ sen (x) 1/2 dx = ∫sen(x)1/2 (x)-1/2 dx= ∫sen(x)1/2 ½ (x) -1/2 dx

√ X (x)1/2 ½ = 2∫sen(x) 1/2 ½ (x)-1/2 dx= 2∫sen(x) 1/2 dx = -2 cos (x) 1/2 +c = -2 cos √ x +c

39) ∫xsenx2dx= ∫sen x2 2 x dx = 1 ∫ sen x2 2x dx = -1 cos (x2) +c

2 2 2 U= x2

du= 2xdx

INTEGRACION DE FUNCIONES RACIONALES (INMEDIATAS)

Page 7: EJERCICIOS Primer Departamental

_____ _____

40) ∫ dx__ = ∫ du__= Ln l u+√ u2+a2 l +c = dx__= 1 Ln l 3x+√ 9x2+16 l +c

√9X2-16 √ u2+a2 √ 9x2+16 3

u2= 9x2 u=3x a2= 16 a= 4 du= 3dx

41) ∫ dx__ = ∫ du__ = 1 Ln l u-a l +c = 1 Ln l x-3 l +c X2-9 √ u2+a2 2a u+a 6 x+3

u2= x2 u=x a2= 9 a= 3 du= dx

42) ∫ dx__ = ∫ 2 dx__ = 1 ∫ 2dx_ = 1 arc sen 2x +c √1 – 4 x2 2√1-4x2 2 √ 1-4x2 2

u2=x2 u=2x a2= 1 a= 1 du= 2dx

43) ∫ 5dx_ = ∫ 5 dx__ = ∫(5) 3dx_ = 5 ∫ 3dx__ = 5 arc sen 3x +c

√1 – 9 x2 5√1-9x2 3 √ 1-9x2 3 √ 1-9x2 3 u2=x2 u=2x a2= 1 a= 1 du= 2dx _____

44) ∫ dx__ = ∫ dx__ = 1 ∫ dx_ = 1 ∫ 3dx__ = 1 Ln l 3x √ 9x2-3 +c

√9X2-3 3√9x2-3 3 √9x2-3 3 √9x2 -3 3 u2=9x2 u=3x a2= 3 a= 1.5 du= 3dx

45) ∫ dx__ = ∫ du__ = 1 1 Ln l u-a l +c = 1 Ln l 5x-3 l +c

25 X2-9 √ u2+a2 2(3) 5 u+a 30 5 x+3

u2=25x2 u=5x a2= 9 a= 3 du= 5dx

Page 8: EJERCICIOS Primer Departamental

INTEGRACION DE FUNCIONES TRIGONOMETRICAS (PRODUCTOS Y POTENCIAS)

46) ∫ sen 3x cos3x dx= ∫sen 3x cos3x 3 dx = 1 ∫ sen 3x cos 3x 3 dx = 1 ∫ udu

3 3 3 U= sen 3x = 1 u 2 +c = 1 u2 +c = sen 2 3x +c Du=cos3x3dx 3 2 6 6

47) sen ax cosax dx= ∫sen ax cosax a dx = 1 ∫ sen ax cos ax a dx = 1 ∫ udu

a a a U= sen ax = 1 udu = 1 u 2 +c = 1 u2 +c = 1 (sen ax)2 +c Du=cosaxadx a a 2 2a 2ª

48) ∫ sen x/3 = ∫ sen x/3 (-1/3) dx =-3∫du =-3∫Ln l u l+c=-3 Ln l 1+cosx/3 l+c1+cos x/3 u (-1/3) u

U= 1+cos x/3 Du=-sen x/3(1/3) dx

49) Sec ax tan ax dx = ∫sec ax tan ax a dx = 1 ∫ seca z tan ax a dx

a a u= sec ax = 1 sec ax +c du= tan ax adx a

50) ∫cos x dx = ∫du = ∫ u-2 du = u -1 +c = -1 +c = - 1_ +c = -csc x +c

Sen2 x u2 -1 u sen x U= sen x du = cos x dx

51) ∫sen3xdx= ∫sen2x senx dx=∫(1-cos x) sen x dx=∫sen x dx -∫cos2x sen x dx

= ∫u2 du = -cos x +u 3 +c = -cos x + cos 3 +c u= cos x 3 3

Page 9: EJERCICIOS Primer Departamental

du= -sen x dx

52) ∫sen2 x dx = ∫ 1 (1-cos 2x) dx = ∫ (1 -cos 2x ) dx = ∫1 dx - ∫cos 2x dx

2 2 2 2 2 U= 2x = 1 ∫dx – 1 ∫cos 2x dx = 1 ∫ dx –sen2x +c Du= 2dx 2 2 2 = x – 1 sen 2x +c 2 4

53) ∫cos2 x dx=∫cos2 x cos x dx -∫(1-sen2x)cos x dx =∫cosx dx – sen 2 cosxdx

U U= sen x = ∫cos x dx - ∫u2 du = -sen x –u 3 +c Du= cos x dx 3 = sen x –sen 3 x +c

354) ∫cos2 2x dx= ∫1 (1+cos 2x)dx = ∫1 (1+cos2x (2x)) dx =

1 ∫1+cos 4x dx 2 2 2 = 1 ∫ dx +1 ∫cos 4x dx = x + 1 (sen 4x) +c U= 4x 2 2 2 8

Du= 4dx

55) ∫cos5x dx= ∫(cos2)2 x cos x dx =∫(1-sen2 x)2 cosx dx=∫(1-2sen2x +sen4x)

Cos x dx= ∫ cos xdx -∫2sen2x cos x dx +∫sen4x cosx dx = sen x-2 sen 3 + sen 5 +c

3 54

56) ∫ sen5x dx =∫(sen x2)2sen x dx =∫(1-cos2x)2senx dx=∫(1-2 cos2x+cos4 x)

Sen x dx= ∫ sen x dx -∫ 2cos2x senxdx +∫cos4xsenxdx U= cos x u= cos x Du= -senx dx du= -sen xdx = ∫ sen x dx -2 ∫cos2x -1 sen x dx+ ∫cos4x -1 sen x dx -1 -1 = ∫ sen x dx +2 ∫u2 du - ∫ u 4 du = cos 4 x –sen x dx

Page 10: EJERCICIOS Primer Departamental

4 4

57) ∫cos 3 x dx = ∫ cos3x (senx)-1/2 dx= ∫cos2x cos x (senx)-1/2

√ sen x =∫ (1-sen2x) cosx (sen x)-1/2 dx=∫sen-1/2x cosx dx-∫cosx senx3/2 dx U= sen x u=sen x Du= cos x dx du=cos x dx =∫ u-1/2 du -∫u3/2 du= u 1/2 - u 5/2 = ½ 5/2 _____ _____ = 2 sen x1/2 -2 sen x 5/2 = 2√ sen x - 2 √sen 3 x +c 5 5

57) ∫sen2x cos3x dx= ∫ sen2x cos2x cosx dx= ∫sen2x (1-sen2x) cos xdx =

= ∫sen2x cosx dx - ∫ sen2x sen2x 6cosx dx U= sen x = ∫ sen2x cos x dx - ∫ sen4 x cos x dx Du= cosxdx = ∫ u2 du - ∫u4 du = u 3 - u 5 +c U= sen x 3 5 Du=cosxdx = sen4 x cos x dx sen 3 x - sen3x +c 3 5

58) ∫ sen33x cos53x dx=∫cos53x sen23x sen3x dx=∫cos53x(1-cos23x)sen3xdx

U=cos 3x =∫cos53x sen3x dx= ∫cos23x sen 3x dx Du=sen3x3dx =∫cos53x sen3x -3 dx -∫cos23x sen3x -3 dx -3 -3 = -1 ∫u5du + 1 ∫ u7du = -1 u 6 + 1 u 8 +c 3 3 3 6 3 8 = -1 cos63x + 1 cos8 3x +c 18 29

59) ∫sen33x cos53x dx=∫cos53x sen23x sen3x dx=∫cos53x(1-cos23x)sen3xdx

=∫ cos53x sen3x dx -∫cos7 3x sen3x dx =∫ cos53x sen3x -3 dx -∫ cos7 3x sen3x -3 dx -3 -3 U=cos3x u=cos3x Du= -sen3x3dx du= -sen 3x3dx = -1 ∫u5du + 1 ∫u7du = -1 u 6 +1 u 8 +c 3 3 3 6 3 8 = -1 cos6 3x +1 cos83x +c 18 24

60) ∫sen2x cos4x dx= ∫sen2x cos2x cos2x dx=∫1 sen2x (1+cos2x)

2 2 = ∫ 1 sen2 2x (1+cos2x) 4 2 =1 ∫ sen2 2x dx + 1 ∫1+cos 2x sen2 2x 2dx 8 8

Page 11: EJERCICIOS Primer Departamental

U=2x u=sen2x Du=2dx du= cos2x 2dx =∫sen2 2x 1 (1-cos 4x)= 2 = 1 [2x - 1 sen4x] +sen3 2x +c 16 2 16 48 = 2x - sen4x + sen 3 2x +c 32 64 48

61) ∫cos4x sen3x dx= ∫cos4 x sen2x senx dx= ∫cos4x (1-cosx2) sen x dx

= ∫cos4 x senx dx - ∫ cos6 x senx dx= = ∫cos4 x -1 sen x dx - ∫cos6x -1 sen x dx -1 -1 = ∫-1 ∫ u4du +1 ∫ u6 du = -1 u 5 + u 7 +c 5 7 = -cos 5 x + cos 7 +c 5 7

62) ∫ sen52x sen32x dx= ∫sen8 2x dx =∫sen82x 2∫sennudu= -1 senn-1 u cos n-1

2 n n U= 2x = 1 ∫ sen82x dx = 1 ∫ senn u du = Du=2dx 2 2 = 1 [1 sen8-1 2x cos2x 8-1 ∫ sen8-2 2x 2dx] 2 8 8 =1 [1 sen72x cos2x 7 ∫ sen6 2x 2 8 8

63) ∫ sen52x scos22x dx= ∫(sen22x)2 sen 2x cos22x dx = ∫ (1-cos22x) 2 sen2x cos22x dx = ∫ (1-2cos22x +cos42x) sen2x cos22x dx= = ∫ sen 2x cos22x dx -∫2 sen 2x cos42x dx+ ∫sen2xcos22xdx U=cos2x du=-sen2x2dx = -1 ∫u2du + 2 ∫ u4 du -1 ∫u6 du= -1 cos 3 2x +1 cos 5 2x-1 cos 7 +c 2 2 2 2 3 2 5 2 7

64) ∫sec23x sec23x dx = ∫sec2 3x (1 tan23x) dx= ∫sec23xdx+∫sec23x tan23xdx

U=tan 3x du= sec23x3dx = ∫sec23x 3 dx + ∫tan23x sec23x dx 3 = 1 ∫ sec2 3x 3dx + 1 ∫u2du= 1 ∫tan3x + 1 u 3 +c 3 3 3 3 3 = 1 tan 3x + 1 tan3 3x +c 3 9

Page 12: EJERCICIOS Primer Departamental

65) ∫tan32x dx=∫tan22x tan2xdx=∫(sec2-1)tan2xdx=∫tan2xdx-∫sec22xtan2x2dx

=∫sec2xsec2xtan2x2dx-∫tan2x2dx U=sec2x u=2x Du=sec2x tan2x 2dx du=2x 2dx =∫u du -∫ tan2dx = u 2 -Ln lsec u l +c 2 = 1 sec2x 2x - Ln l sec2x l +c 4 2

66) ∫sec64x dx=∫ sec44x sec24x dx= ∫sec44x (1+tan24x)dx =∫sec44x dx +∫sec44x tan24x dx U= tan24x du=sec44x dx

67) ∫tan53x dx= ∫(tan23x)2tan3xdx= ∫(sec23x -1)2 tan3xdx = ∫(sec43x – 2sec23x+1)tan3x dx=∫sec45xtan3x dx-∫2sec23x tan3x+∫tan3x dx = ∫sec33x sec3x tan3x 3 dx -2∫3 sec 3x sec3x tan3x dx +∫tan3x dx 3 3 U=sec33x u=3/5 sec Du=sec3x tan3x dx du= sec3x tan3x dx = 1 ∫u3du -2∫ u du +∫3 tan 3x dx= 1 u 4 - 2 u 2 + 1 Ln lsec 3xl +c 3 -2 3 3 4 3 2 3 = sec 4 3x - 2 sec23x + 1 Ln l sec 3x l +c 12 6 3

68) ∫ tan34x dx= ∫tan24x tan4x dx= ∫ tan2 4xdx - ∫sec24x tan 4x 4dx

= ∫sec24x sec 4x tan4x 4 dx - ∫tan24x dx 4 U=sec4x u=4x u=tan4x Du= sec4x tan4x 4dx du=4x4dx du=sec24x4dx = ∫sec24x tan4x dx -∫tan4x dx = ∫sec24x tan4x 4 dx- ∫tan4x 4 dx 4 4 = 1 ∫u du - 1∫ tan 4x 4dx= 1 u 2 - 1 Ln lsec ul +c 4 4 4 2 4 = 1 sec4x + 1 Ln l cos 4x l +c 8 4

69) ∫tan5x sec4x dx=

Si tan²x = sec² x-1 4

= ∫((tan²x))² tanx sec x dx 4

=∫ (sec²x-1) tanx sec x dx 4 4

=∫(sec x-2 sec²x-1) tanx sec x dx

Page 13: EJERCICIOS Primer Departamental

8 6 4

=∫sec x tanx dx- ∫2 sec x Tan x dx + ∫ tan x sec x dxu= sec xdu = secx tanx dx 7 5

=∫sec x secx tan x dx – 2 ∫sec x sec x tanx dx + ∫sec³ sex x tanx dx 7 5 =∫u du - 2∫u du + ∫u³ du 8 6 4

= u - 2 u + u + C 8 6 4

70) ∫tan3x sec2x dx= ∫u3du = u 4 +c = tan 4 x +c 4 4 U=tan x du=sec2x

71) ∫sec 2 2 ǿ tan 2 ǿ dǿ = 1 ∫sec2ǿ tan 2 ǿ 2dǿ= 1 sec 2ǿ +c 3sec 2ǿ 3 2 6 U=sec x du=sec x

72) ∫tan44x sec24x dx= ∫tan44x sec24x 4 dx= 1 ∫tan44x sec24x 4dx

4 2 =1 ∫ u4du – 1 u 5 +c = 1 tan54x +c 4 4 5 20

73) ∫sec3xtan3x dx= ∫ tan2x tan x sec3x dx= ∫(1-sec2x) tan x sec3x dx

= ∫sec3x tanx dx - ∫sec5x tanx dx = ∫sec x tanx sec2x - ∫sec x tanx sec4x dx U=sec x du= secx tanx dx =∫ u 3 du - ∫ u4du = u 3 - u 5 +c= sec 2 x – sec 3 x +c 2 3 5 3 5

74) ∫tan3x sec7x dx= ∫tan2x tan x sec7 x dx= ∫(1-sec2x) tanx sec7x dx

= ∫sec7x tanx dx - ∫ sec9x tanx dx

Page 14: EJERCICIOS Primer Departamental

U= secx du= tanx secx dx =∫ u8du -∫ u6 du= ∫ u 9 + u 7 +c= sec 9 x + sec 7 x +c 9 7 9 7

INTEGRACION DE LAS EXPRESIONES COT, CSC

75) ∫csc4dx= ∫csc2x csc2x dx= ∫(1+cot2x) csc2x dx= csc2dx +∫cot2x csc2x dx

U=cot x du= -csc2x dx = ∫csc2 dx - ∫u2du = -cot x –u 3 +c 3 = -cot x –cot 3 x +c 3

76) ∫cot2x dx= ∫-cot u –u +c = - cot x –x +c

77) ∫cot32x dx= ∫cot22x cot2x dx= ∫(csc22x -1) cot 2x dx= ∫csc22x cot2x dx

= ∫ csc22x cot 2x dx -∫cot 2x dx = 1 ∫u du -1 ∫cot 2xdx 2 2 U=cot2x u= 2x Du= -csc22x 2dx dx= 2dx = 1 u 2 - 1 Ln l sen 2x l +c = 1 cot2 2x - 1 Ln l sen 2x l +c 2 2 2 4 2

78) ∫ csc62x dx = ∫(csc22x)2 csc22x dx= ∫(1+cot22x)2 csc22x dx

= ∫ (1+2cot22x+cot42x) csc22x dx = ∫ csc22x dx +∫2cot22x xsx22x dx + ∫cot42x csc22x2dx U= 2x u=cot2x u=cot 2x Du=2dx du= -csc2x du= -csc 2x 2dx = 1 ∫csc22x2dx -1 ∫ 2cot22x csc2 2x 2dx -1 ∫cot42x-csc2 2xdx 2 2 2 = 1 ∫csc2u du -1 ∫ 2u2 du -1 ∫u4 du 2 2 2 =1 –cot u -1 2u 3 -1 ∫u 5 = -1 cot 2x -1 cot32x -1 cot52x +c 2 2 3 2 5 2 3 10

Page 15: EJERCICIOS Primer Departamental

79) ∫csc2(9-6x)dx= ∫ csc (9-6x) -6 dx= -1 ∫ csc2udu = 1 cot u +c=U=9-6x -6 6 6

Du=6dx = 1 cot (9-6x) +c 2

INTEGRACION DE LAS EXPRESIONES COT, CSC

80) ∫ csc bx cot bx dx= ∫csc bx cot bc b dx= 1 ∫csc bx cot bx dx=

U=bx b b Du=bxdx = -1 csc bx +c b

81) ∫cot43x csc23x dx= ∫cot23x cot23x csc23x dx=∫(csc23x-1)cot23x csc23x dx

= ∫csc43x cot23x dx- ∫cot23x csc2 3x dx U=3x = ∫csc23x csc23x cot23x dx- ∫cot23x csc23x dx Du= 3dx = ∫ (1+cot23x) csc23x cot2 3xdx - ∫ cot23x csc23x dx = ∫ csc23x cot23x dx + ∫ cot43x csc23x dx-∫cot23x csc23xdx U=cot3x du= -csc23x3dx = -1 ∫ u2du -1 ∫ u4du + 1 ∫ u2 du 3 3 3 = -1 cot3 3x -1 cot 5 3x +1 cot33x 3 3 3 5 3 3 = -1 cot53x +c 15 = ∫u2 du = -cos x +u 3 +c = -cos x + cos 3 +c

3 3

82.- ∫x cos 3x dx = x cos 3x x - ∫ - x sen 3x 3 dx = x 1/3 sen 3x - ∫1/3 sen 3x dx

= x sen 3x / 3 – 1/3 ∫sen 3x 3 dx = x sen 3x / 3 – 1/9 ∫sen u du

= x sen 3x / 3 + 1/9 cos 3x + c

83.- ∫x sen 2x dx = - x ½ cos 2x + ½ ∫ cos 2x dx = - x cos 2x / 2 + ¼ sen 2x + c