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?

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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?

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fuk
I have the answer sheet, and wolfram gives a decent explanation, but I can't understand what the next step is.
pull out 3 get 3*integral(cos u du) = 3sin u + C

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

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