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fuk

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

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

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.

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)

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)

forget that square root

no its Integral (3 cos^2 x)

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.

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.

;)

its rather. I hate that stupid part of english.

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.

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

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

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

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

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.

oh yes, it's fun! =)

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

just takes time to digest

Good luck on the test!