anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
fuk
anonymous
  • anonymous
I have the answer sheet, and wolfram gives a decent explanation, but I can't understand what the next step is.
anonymous
  • anonymous
pull out 3 get 3*integral(cos u du) = 3sin u + C

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anonymous
  • anonymous
You just have to know the integral of cos(u)
anonymous
  • 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))
anonymous
  • anonymous
Once you integrate it you can plug back in what you substituted for u
anonymous
  • anonymous
You need to solve for dx in terms of du.
anonymous
  • anonymous
plug u back in
anonymous
  • anonymous
Yeah, that's what I'm having issues with. I've never been strong with trig functions.
anonymous
  • anonymous
When you plug u in, there's a sin and cosine function, which indicates integration by parts, that seem right?
anonymous
  • anonymous
\[x = 3 sin(u) \implies dx = 3cos(u)du\] \[(9-x^2)^{1/2} dx= [9-9cos^2(u)]^{1/2} \cdot (3cos(u)\ du)\]
anonymous
  • 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.
anonymous
  • anonymous
oh wait, no.
anonymous
  • anonymous
It's a tough one XD
anonymous
  • anonymous
u use sin(x)^2 + cos(x)^2 = 1
anonymous
  • anonymous
yeah, 1-cos^2(u) = sin^2(u)
anonymous
  • anonymous
which you can take the square root of easily and also u-sub easily.
anonymous
  • anonymous
so z = sin u dz = cos x dx
anonymous
  • anonymous
i mean dz = cos u du
anonymous
  • anonymous
integral(z) = z^2/2 then replace with u and then replace with x
anonymous
  • anonymous
I have ... \[3\cos(\sin^-(x/3))\] Where would I go from here?
anonymous
  • anonymous
that's not right. How did you get there?
anonymous
  • anonymous
\[\cos x =(1- \sin^2 x)^{?}\]
anonymous
  • anonymous
other way round bob.
anonymous
  • anonymous
no
anonymous
  • anonymous
x = 3sin(u) x/3 = sin(u) invsin(x/3) = u integral ( 3cos(u)) integral (3cos(invsin(x/3)))
anonymous
  • anonymous
you don't want to sub back in your u until after you integrate.
anonymous
  • anonymous
so that'll be negative sin(u)
anonymous
  • anonymous
because the integral of cosine is sin, right? See, I'm confused because it's still technically in the form dx, not du.
anonymous
  • anonymous
negative sin*
anonymous
  • anonymous
It shouldn't be. When you do the u-sub you need to replace your dx with some function of u du
anonymous
  • anonymous
This is what you're starting with right? \[\int (9-x^2)^{1/2} dx\]
anonymous
  • anonymous
Yessir.
anonymous
  • anonymous
Ok, and we let \(x = 3sin(u)\).
anonymous
  • anonymous
So you know, I'm studying for a test tomorrow, this isn't homework.
anonymous
  • anonymous
yeah, np
anonymous
  • anonymous
Ok, so now what is dx in terms of du?
anonymous
  • anonymous
dx = 3cosu du
anonymous
  • anonymous
Right.
anonymous
  • anonymous
So now lets swap out \(x^2\) and \(dx\) with the appropriate expressions of u and du.
anonymous
  • anonymous
What do you have?
anonymous
  • anonymous
(9-9sin^2u)^(1/2) dx (9(1-sin^2u))^(1/2) dx 3 cos u dx
anonymous
  • anonymous
you didn't swap out dx.
anonymous
  • anonymous
I don't know how the hell to do that my friend :P
anonymous
  • anonymous
dx = 3cosu du You said it yourself.
anonymous
  • anonymous
Agreed, but if you .... OH!
anonymous
  • anonymous
3cosu * 3 cosu ?
anonymous
  • anonymous
Indeed (along with a du)
anonymous
  • anonymous
okay, so we have ... \[\int\limits_{}^{} 3 \cos (u)cos(u) du\] which is ... \[\int\limits_{}^{} 9 \cos ^2 (u) ^ {1/2} du\]
anonymous
  • anonymous
forget that square root
anonymous
  • anonymous
no its Integral (3 cos^2 x)
anonymous
  • anonymous
Right. So then you can use the double/half angle equation to swap the \(cos^2\) for an expression involving the cos(2u)
anonymous
  • anonymous
She said she would be providing those equations on the test.
anonymous
  • anonymous
Should be 9 cause there's a 3 from the 9 under the sqrt and a 3 from the du replacement of dx.
anonymous
  • anonymous
\[\cos^2 x = (1 + \cos 2x )/2\]
anonymous
  • anonymous
that's the one.
anonymous
  • anonymous
I think polpak's solution is about the same as the attachment.
1 Attachment
anonymous
  • anonymous
Yeah but polpak is a better teacher than any pdf :p
anonymous
  • anonymous
I'm working out the cos squared substitution on paper as we speak.
anonymous
  • anonymous
particularly a pdf that is a screenshot of wolframalpha.
anonymous
  • anonymous
;)
anonymous
  • 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.
anonymous
  • anonymous
its rather. I hate that stupid part of english.
anonymous
  • 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... \[\int\limits_{}^{} 9 {( 1+\cos2u)}/2 du\]
anonymous
  • anonymous
That's not ugly at all!
anonymous
  • anonymous
Break it into two integrals.
anonymous
  • anonymous
and pull the 9/2 all the way out front.
anonymous
  • anonymous
Oh wow. You're right. Hah! I need to practice these more, for sure.
anonymous
  • 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.
anonymous
  • anonymous
Nah it's just division by 2
anonymous
  • anonymous
indeed.
anonymous
  • anonymous
\[9/2 ( u - \sin(2u)/u)\] Does this look alright?
anonymous
  • anonymous
why is that u under the sin(2u) ?
anonymous
  • anonymous
Oh, sorry that's supposed to be a 2. I have dyslexia, it's a pain in the rear end sometimes.
anonymous
  • anonymous
yes. Me too.
anonymous
  • anonymous
With a 2 in the denominator that is correct.
anonymous
  • anonymous
oh, wait
anonymous
  • anonymous
shouldn't be a -sin(2u)
anonymous
  • anonymous
\[\int cos(u)du = sin(u) +C \]
anonymous
  • anonymous
should be positive.
anonymous
  • anonymous
the derivative of cos is -sin, but the integral of cos is just sin
anonymous
  • anonymous
\[u = \sin^{-1}(x/3)\] You're right, I got that mixed up with the rules for sin integration.
anonymous
  • anonymous
yeah, so that works nicely given what u is.
anonymous
  • anonymous
So if I were to substitute u, then I would be finished?
anonymous
  • anonymous
you can simplify probably, but it's up to you
anonymous
  • 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?
anonymous
  • anonymous
because they plug back in for sin(2u) and use the formula for \(sin(2\theta)\)
anonymous
  • anonymous
to just get a sin or cos squared which they then swap out for 1-cos or sin squared
anonymous
  • anonymous
all kinds of fun trig subs going on.
anonymous
  • anonymous
We can work through it if you want
anonymous
  • 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
anonymous
  • anonymous
because I believe she would want it back in the form of x
anonymous
  • anonymous
I would imagine black holes are created, or something of that variety (when sin and inverse sin collide)
anonymous
  • anonymous
No, you just get back what you started with. \[sin(sin^{-1}(x/3)) = x/3\]
anonymous
  • anonymous
What happened to the two?
anonymous
  • anonymous
it's still there, I'm showing a different example
anonymous
  • anonymous
Ok, so here it is. \[\frac{9}{2}sin(2u) =\frac{ 9(2)}{2}cos(u)sin(u)\]\[ =9(\sqrt{1-sin^2(u)})sin(u) \]
anonymous
  • anonymous
and when you plug your u back in your sin will get nuked by the inverse
anonymous
  • anonymous
Very sweet, I just need a second to absorb.
anonymous
  • 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.
anonymous
  • anonymous
the first step comes from: \[sin(2\theta) = cos(\theta)sin(\theta)\] the second step comes from: \[cos\theta = \sqrt{1-sin^2\theta}\]
anonymous
  • anonymous
oh yes, it's fun! =)
anonymous
  • 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.
anonymous
  • anonymous
indeed =) I'm taking Linear algebra now ;)
anonymous
  • 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.
anonymous
  • anonymous
just takes time to digest
anonymous
  • 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!
anonymous
  • anonymous
Good luck on the test!

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