1. anonymous

$\int\limits_{}^{} (4z + \sin(3z))e ^{6z ^{2}-\cos(3z)} dz$

2. anonymous

whew, that's a doozy.

3. anonymous

So what do you think you want to use for substitution?

4. anonymous

$\int\limits_{?}^{?} \sin(x) d x = -\cos(x) + constant$

5. anonymous

Not sure...possibly something to do with 3z? I have no idea to be honest

6. anonymous

I would let u = that yucky exponent on the e

7. anonymous

Ignore the two ?'s

8. anonymous

Okay but how does that help?

9. anonymous

Well. $u = 6x^2 - cos(3z)$ $\implies du = 12z +3sin(3z)dz = 3(4z + sin(3z))dz$

10. anonymous

Oh so then $du = 2z ^{3}-\sin(3z)dx$

11. anonymous

Close, but you integrated. You want to take the derivative

12. anonymous

oh

13. anonymous

Okay. Where do I plug u into?

14. anonymous

Well first rewrite dz in terms of du using that last equation I had.

15. anonymous

?

16. anonymous

$$du = 3(4z + sin(3z))dz \implies dz =\ ?$$

17. anonymous

I don't understand - what am I solving for?

18. anonymous

Well for this part you're solving for dz.

19. anonymous

Err...I divide 3(4z + sin(3z)) on both sides?

20. anonymous

Yes

21. anonymous

So $dz = \frac{1}{3(4z+sin(3z))} du$

22. anonymous

So now. Go back to your integral, and replace the exponent of e with u, and the dz with your expression for du, and you'll be pleasantly surprised.

23. anonymous

Wait a minute...Why do I replace dz?

24. anonymous

Because we are no longer integrating with respect to z, we are integrating with respect to u now. Because it turns out u is nicer.

25. anonymous

We are replacing our nasty expression with z for a nice expression with u.

26. anonymous

$(4z + \sin(3z)e ^{u}$du

27. anonymous

But remember that dz does not equal du.

28. anonymous

ButI thought we replaced dz

29. anonymous

$dz = \frac{1}{3(4z+sin(3z))} du$

30. anonymous

So you have to put that whole thing in for dz

31. anonymous

I'm so lost

32. anonymous

33. anonymous

$\int x*e^{x^2}dx$

34. anonymous

We let $$u = x^2$$

35. anonymous

So $du = (2x)dx \implies dx = \frac{1}{2x}du$

36. anonymous

With me so far?

37. anonymous

sure

38. anonymous

39. anonymous

yes

40. anonymous

Ok. So then $\int x*e^{x^2}dx = \int x*e^u(\frac{1}{2x} du)$

41. anonymous

And we can cancel the x in front with the x in the denominator from the du.

42. anonymous

to get $\frac{1}{2}\int e^udu$

43. anonymous

Which is nice and easy.

44. anonymous

$\frac{1}{2}\int e^udu = \frac{1}{2}e^u + c = \frac{1}{2}e^{x^2} + C$

45. anonymous

So I've got$1/3 e ^{u}$ for my problem when do I integrate?

46. anonymous

Now.

47. anonymous

Before or after I plug in u

48. anonymous

Don't switch back your u until after you integrate.

49. anonymous

How do I integrate that?

50. anonymous

$\int \frac{1}{3}e^u du = \frac{1}{3}\int e^udu$

51. anonymous

What is the integral of $$e^x$$?

52. anonymous

e^x

53. anonymous

Right.

54. anonymous

$1/3 *e ^{6z ^{2}-\cos(3z)}$

55. anonymous

+C

56. anonymous

oh right

57. anonymous

Nice job!

58. anonymous

Oh my goodness. Would you please walk me through one more?

59. anonymous

Sure, but you'll have to do more of it this time ;)

60. anonymous

How's it start?

61. anonymous

Okay I'll try:

62. anonymous

$\int\limits_{}^{}(4z ^{3} + 6\csc ^{2}(6z)(z ^{4}-\cot(6z))^{4} dz$

63. anonymous

Ugh

64. anonymous

Do I need to FOIL first?

65. anonymous

No.

66. anonymous

These are all ugly u substitution problems. Foiling will just make you cry.

67. anonymous

You want to find a nice thing to pick for a u substitution.

68. anonymous

Oh okay

69. anonymous

Do you notice anything interesting about these two things?

70. anonymous

Not really sure what to look for, actually.

71. anonymous

$$(4z^3 + 6csc^2(6z)) * (z^4 - cot(6z))^4$$ You are looking for something in the equation that looks like the derivative of another part of the equation.

72. anonymous

Err expression, not equation

73. anonymous

Oh! I guess the 4z^3 and z^4

74. anonymous

And what about the $$6csc^2(6z)$$ and the -cot(6z) ?

75. anonymous

AND the csc^2

76. anonymous

right.

77. anonymous

So what you want to do is pick the one that is not the derivative, and let that be u.

78. anonymous

So let$u = z ^{4} - \cot(6z)$

79. anonymous

Yep, then solve for dz in terms of z and du.

80. anonymous

So $\int\limits_{}^{}(4z ^{3}+6\csc ^{2}(6z))u ^{4}du$

81. anonymous

Nope. but close.

82. anonymous

$u = z^4 - cot(6z)$ So $du =\ ?$

83. anonymous

4z^3 + 6csc^2(6z)

84. anonymous

close.

85. anonymous

$du = (4z^3 + 6csc^2(6z))dz$

86. anonymous

So then what does dz = ?

87. anonymous

1 over all that mess

88. anonymous

yep.

89. anonymous

So plug in 1 over all that mess times du for dz in the original equation. and plug in u for what we said it equaled.

90. anonymous

$(4z ^{3} + 6\csc ^{2}(6z))u ^{4} * 1/4z ^{3} + 6\csc ^{2}(6z)$

91. anonymous

simplify!

92. anonymous

u^4?

93. anonymous

yes

94. anonymous

$(z ^{4} - \cot(6z))^{4}$

95. anonymous

No

96. anonymous

$\int(4z^3 + 6csc^2(6z)) * (z^4 - cot(6z))^4dz$ $= \int (4z^3 + 6csc^2(6z)) u^4(\frac{1}{4z^3 + 6csc^2(6z)})du$ $= \int u^4du =\ ?$

97. anonymous

? I don't know

98. anonymous

I can't pull anything out in front of the integral sign

99. anonymous

$\int x^2dx = \frac{x^3}{3} +C$ Right?

100. anonymous

Oh! Integrate!

101. anonymous

Yes. Once you have something nice, you just integrate.

102. anonymous

Then after you integrate you plug back in what u was.

103. anonymous

$(z ^{4} - \cot(6z)^{5}/ 5 +c$

104. anonymous

Yep.

105. anonymous

WOW. Good grief. Thanks so much. I am going to use these as examples for the next ones. Thanks again

106. anonymous

Very welcome =)