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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have the answer sheet, and wolfram gives a decent explanation, but I can't understand what the next step is.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0pull out 3 get 3*integral(cos u du) = 3sin u + C

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You just have to know the integral of cos(u)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Nah, it's still in the form dx, not du. My worksheet says the answer is. (1/2)x (9x^2)^(1/2) + 9 invsin(x/3))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Once you integrate it you can plug back in what you substituted for u

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You need to solve for dx in terms of du.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, that's what I'm having issues with. I've never been strong with trig functions.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When you plug u in, there's a sin and cosine function, which indicates integration by parts, that seem right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x = 3 sin(u) \implies dx = 3cos(u)du\] \[(9x^2)^{1/2} dx= [99cos^2(u)]^{1/2} \cdot (3cos(u)\ du)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then 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 \((1cos^2(u))^{1/2}cos(u) du\). This solves nicely with another substitution.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u use sin(x)^2 + cos(x)^2 = 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, 1cos^2(u) = sin^2(u)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which you can take the square root of easily and also usub easily.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so z = sin u dz = cos x dx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0integral(z) = z^2/2 then replace with u and then replace with x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have ... \[3\cos(\sin^(x/3))\] Where would I go from here?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that's not right. How did you get there?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\cos x =(1 \sin^2 x)^{?}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x = 3sin(u) x/3 = sin(u) invsin(x/3) = u integral ( 3cos(u)) integral (3cos(invsin(x/3)))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you don't want to sub back in your u until after you integrate.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so that'll be negative sin(u)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because the integral of cosine is sin, right? See, I'm confused because it's still technically in the form dx, not du.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It shouldn't be. When you do the usub you need to replace your dx with some function of u du

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is what you're starting with right? \[\int (9x^2)^{1/2} dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, and we let \(x = 3sin(u)\).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So you know, I'm studying for a test tomorrow, this isn't homework.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, so now what is dx in terms of du?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So now lets swap out \(x^2\) and \(dx\) with the appropriate expressions of u and du.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(99sin^2u)^(1/2) dx (9(1sin^2u))^(1/2) dx 3 cos u dx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you didn't swap out dx.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't know how the hell to do that my friend :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dx = 3cosu du You said it yourself.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Agreed, but if you .... OH!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Indeed (along with a du)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay, so we have ... \[\int\limits_{}^{} 3 \cos (u)cos(u) du\] which is ... \[\int\limits_{}^{} 9 \cos ^2 (u) ^ {1/2} du\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0forget that square root

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no its Integral (3 cos^2 x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Right. So then you can use the double/half angle equation to swap the \(cos^2\) for an expression involving the cos(2u)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0She said she would be providing those equations on the test.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Should be 9 cause there's a 3 from the 9 under the sqrt and a 3 from the du replacement of dx.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\cos^2 x = (1 + \cos 2x )/2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think polpak's solution is about the same as the attachment.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah but polpak is a better teacher than any pdf :p

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm working out the cos squared substitution on paper as we speak.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0particularly a pdf that is a screenshot of wolframalpha.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Though alpha is good for checking your answers and finding where you made a mistake, but sometimes it's process is a little cumbersome.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its rather. I hate that stupid part of english.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Tell 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... \[\int\limits_{}^{} 9 {( 1+\cos2u)}/2 du\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's not ugly at all!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Break it into two integrals.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and pull the 9/2 all the way out front.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh wow. You're right. Hah! I need to practice these more, for sure.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Nah it's just division by 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[9/2 ( u  \sin(2u)/u)\] Does this look alright?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why is that u under the sin(2u) ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, sorry that's supposed to be a 2. I have dyslexia, it's a pain in the rear end sometimes.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0With a 2 in the denominator that is correct.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0shouldn't be a sin(2u)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int cos(u)du = sin(u) +C \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the derivative of cos is sin, but the integral of cos is just sin

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[u = \sin^{1}(x/3)\] You're right, I got that mixed up with the rules for sin integration.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, so that works nicely given what u is.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So if I were to substitute u, then I would be finished?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can simplify probably, but it's up to you

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0One quick note, in my packet, the answer is 1/2 (x * (9x^2)^(1/2) + 9 sin 1 (x/3)) How can it be so different?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because they plug back in for sin(2u) and use the formula for \(sin(2\theta)\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0to just get a sin or cos squared which they then swap out for 1cos or sin squared

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0all kinds of fun trig subs going on.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We can work through it if you want

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because I believe she would want it back in the form of x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I would imagine black holes are created, or something of that variety (when sin and inverse sin collide)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, you just get back what you started with. \[sin(sin^{1}(x/3)) = x/3\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What happened to the two?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it's still there, I'm showing a different example

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, so here it is. \[\frac{9}{2}sin(2u) =\frac{ 9(2)}{2}cos(u)sin(u)\]\[ =9(\sqrt{1sin^2(u)})sin(u) \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and when you plug your u back in your sin will get nuked by the inverse

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Very sweet, I just need a second to absorb.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the first step comes from: \[sin(2\theta) = cos(\theta)sin(\theta)\] the second step comes from: \[cos\theta = \sqrt{1sin^2\theta}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Maybe 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0indeed =) I'm taking Linear algebra now ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Have 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just takes time to digest

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, 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!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Good luck on the test!
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