## anonymous 5 years ago How would you integrate the following function? Particularly, what would you do after the last following step? integral( (9-x^2)^(1/2) dx) x = 3 sin u u = invsin(x/3) integral( 3 (cos u) d?

1. anonymous

fuk

2. anonymous

I have the answer sheet, and wolfram gives a decent explanation, but I can't understand what the next step is.

3. anonymous

pull out 3 get 3*integral(cos u du) = 3sin u + C

4. anonymous

You just have to know the integral of cos(u)

5. anonymous

Nah, it's still in the form dx, not du. My worksheet says the answer is. (1/2)x (9-x^2)^(1/2) + 9 invsin(x/3))

6. anonymous

Once you integrate it you can plug back in what you substituted for u

7. anonymous

You need to solve for dx in terms of du.

8. anonymous

plug u back in

9. anonymous

Yeah, that's what I'm having issues with. I've never been strong with trig functions.

10. anonymous

When you plug u in, there's a sin and cosine function, which indicates integration by parts, that seem right?

11. anonymous

$x = 3 sin(u) \implies dx = 3cos(u)du$ $(9-x^2)^{1/2} dx= [9-9cos^2(u)]^{1/2} \cdot (3cos(u)\ du)$

12. anonymous

Then you can factor out a 3 from the sqrted factor and a 3 from the du part and have a 9 out front and rewrite it as $$(1-cos^2(u))^{1/2}cos(u) du$$. This solves nicely with another substitution.

13. anonymous

oh wait, no.

14. anonymous

It's a tough one XD

15. anonymous

u use sin(x)^2 + cos(x)^2 = 1

16. anonymous

yeah, 1-cos^2(u) = sin^2(u)

17. anonymous

which you can take the square root of easily and also u-sub easily.

18. anonymous

so z = sin u dz = cos x dx

19. anonymous

i mean dz = cos u du

20. anonymous

integral(z) = z^2/2 then replace with u and then replace with x

21. anonymous

I have ... $3\cos(\sin^-(x/3))$ Where would I go from here?

22. anonymous

that's not right. How did you get there?

23. anonymous

$\cos x =(1- \sin^2 x)^{?}$

24. anonymous

other way round bob.

25. anonymous

no

26. anonymous

x = 3sin(u) x/3 = sin(u) invsin(x/3) = u integral ( 3cos(u)) integral (3cos(invsin(x/3)))

27. anonymous

you don't want to sub back in your u until after you integrate.

28. anonymous

so that'll be negative sin(u)

29. anonymous

because the integral of cosine is sin, right? See, I'm confused because it's still technically in the form dx, not du.

30. anonymous

negative sin*

31. anonymous

It shouldn't be. When you do the u-sub you need to replace your dx with some function of u du

32. anonymous

This is what you're starting with right? $\int (9-x^2)^{1/2} dx$

33. anonymous

Yessir.

34. anonymous

Ok, and we let $$x = 3sin(u)$$.

35. anonymous

So you know, I'm studying for a test tomorrow, this isn't homework.

36. anonymous

yeah, np

37. anonymous

Ok, so now what is dx in terms of du?

38. anonymous

dx = 3cosu du

39. anonymous

Right.

40. anonymous

So now lets swap out $$x^2$$ and $$dx$$ with the appropriate expressions of u and du.

41. anonymous

What do you have?

42. anonymous

(9-9sin^2u)^(1/2) dx (9(1-sin^2u))^(1/2) dx 3 cos u dx

43. anonymous

you didn't swap out dx.

44. anonymous

I don't know how the hell to do that my friend :P

45. anonymous

dx = 3cosu du You said it yourself.

46. anonymous

Agreed, but if you .... OH!

47. anonymous

3cosu * 3 cosu ?

48. anonymous

Indeed (along with a du)

49. anonymous

okay, so we have ... $\int\limits_{}^{} 3 \cos (u)cos(u) du$ which is ... $\int\limits_{}^{} 9 \cos ^2 (u) ^ {1/2} du$

50. anonymous

forget that square root

51. anonymous

no its Integral (3 cos^2 x)

52. anonymous

Right. So then you can use the double/half angle equation to swap the $$cos^2$$ for an expression involving the cos(2u)

53. anonymous

She said she would be providing those equations on the test.

54. anonymous

Should be 9 cause there's a 3 from the 9 under the sqrt and a 3 from the du replacement of dx.

55. anonymous

$\cos^2 x = (1 + \cos 2x )/2$

56. anonymous

that's the one.

57. anonymous

I think polpak's solution is about the same as the attachment.

58. anonymous

Yeah but polpak is a better teacher than any pdf :p

59. anonymous

I'm working out the cos squared substitution on paper as we speak.

60. anonymous

particularly a pdf that is a screenshot of wolframalpha.

61. anonymous

;)

62. anonymous

Though alpha is good for checking your answers and finding where you made a mistake, but sometimes it's process is a little cumbersome.

63. anonymous

its rather. I hate that stupid part of english.

64. anonymous

Tell me about it, I checked wolfram and couldn't find how they went from 3cos(u) to 9cos^2(u). This is getting uglier and uglier as we go along. I have... $\int\limits_{}^{} 9 {( 1+\cos2u)}/2 du$

65. anonymous

That's not ugly at all!

66. anonymous

Break it into two integrals.

67. anonymous

and pull the 9/2 all the way out front.

68. anonymous

Oh wow. You're right. Hah! I need to practice these more, for sure.

69. anonymous

Then the only thing you have to do is another sub for z = 2u on the second integral if you don't feel confident doing it in your head.

70. anonymous

Nah it's just division by 2

71. anonymous

indeed.

72. anonymous

$9/2 ( u - \sin(2u)/u)$ Does this look alright?

73. anonymous

why is that u under the sin(2u) ?

74. anonymous

Oh, sorry that's supposed to be a 2. I have dyslexia, it's a pain in the rear end sometimes.

75. anonymous

yes. Me too.

76. anonymous

With a 2 in the denominator that is correct.

77. anonymous

oh, wait

78. anonymous

shouldn't be a -sin(2u)

79. anonymous

$\int cos(u)du = sin(u) +C$

80. anonymous

should be positive.

81. anonymous

the derivative of cos is -sin, but the integral of cos is just sin

82. anonymous

$u = \sin^{-1}(x/3)$ You're right, I got that mixed up with the rules for sin integration.

83. anonymous

yeah, so that works nicely given what u is.

84. anonymous

So if I were to substitute u, then I would be finished?

85. anonymous

you can simplify probably, but it's up to you

86. anonymous

One quick note, in my packet, the answer is 1/2 (x * (9-x^2)^(1/2) + 9 sin -1 (x/3)) How can it be so different?

87. anonymous

because they plug back in for sin(2u) and use the formula for $$sin(2\theta)$$

88. anonymous

to just get a sin or cos squared which they then swap out for 1-cos or sin squared

89. anonymous

all kinds of fun trig subs going on.

90. anonymous

We can work through it if you want

91. anonymous

well, I'm happy leaving it in the form inverse sine. I'm just curious what would happen when you substitute sin^-1(x/3) into sin(2u)/2

92. anonymous

because I believe she would want it back in the form of x

93. anonymous

I would imagine black holes are created, or something of that variety (when sin and inverse sin collide)

94. anonymous

No, you just get back what you started with. $sin(sin^{-1}(x/3)) = x/3$

95. anonymous

What happened to the two?

96. anonymous

it's still there, I'm showing a different example

97. anonymous

Ok, so here it is. $\frac{9}{2}sin(2u) =\frac{ 9(2)}{2}cos(u)sin(u)$$=9(\sqrt{1-sin^2(u)})sin(u)$

98. anonymous

and when you plug your u back in your sin will get nuked by the inverse

99. anonymous

Very sweet, I just need a second to absorb.

100. anonymous

Thanks again for your help though, I wish there was some way I could repay you. I hope a challenging problem can suffice since that's all I have to offer.

101. anonymous

the first step comes from: $sin(2\theta) = cos(\theta)sin(\theta)$ the second step comes from: $cos\theta = \sqrt{1-sin^2\theta}$

102. anonymous

oh yes, it's fun! =)

103. anonymous

Maybe somebody will post a linear algebra problem :p I know I enjoy helping people with Calc 1 and Algebra problems, so I imagine it's the same for you, except more advanced.

104. anonymous

indeed =) I'm taking Linear algebra now ;)

105. anonymous

Have with with that! :D I enjoy the idea of precision in math, however the process of learning about it kills me. Especially when there's a concept I can't comprehend.

106. anonymous

just takes time to digest

107. anonymous

Well, I'm off to bed, thanks again for your help. She said that a problem to the one you helped me solve just now will the be the hardest on the exam. Besides that, it's mostly partial fractions, integrals by parts. 1 step integrals. Have a nice night!

108. anonymous

Good luck on the test!