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?

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

fuk

- anonymous

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

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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

- 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

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

- anonymous

You need to solve for dx in terms of du.

- anonymous

plug u back in

- anonymous

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

- anonymous

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

- 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

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

oh wait, no.

- anonymous

It's a tough one XD

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

so z = sin u dz = cos x dx

- anonymous

i mean dz = cos u du

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

other way round bob.

- anonymous

no

- anonymous

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

- anonymous

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

- anonymous

so that'll be negative sin(u)

- 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

negative sin*

- anonymous

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

- anonymous

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

- anonymous

Yessir.

- anonymous

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

- anonymous

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

- anonymous

yeah, np

- anonymous

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

- anonymous

dx = 3cosu du

- anonymous

Right.

- anonymous

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

- anonymous

What do you have?

- anonymous

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

- anonymous

you didn't swap out dx.

- anonymous

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

- anonymous

dx = 3cosu du
You said it yourself.

- anonymous

Agreed, but if you ....
OH!

- anonymous

3cosu * 3 cosu ?

- anonymous

Indeed (along with a du)

- 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

forget that square root

- anonymous

no its Integral (3 cos^2 x)

- 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

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

- 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

\[\cos^2 x = (1 + \cos 2x )/2\]

- anonymous

that's the one.

- anonymous

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

##### 1 Attachment

- anonymous

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

- anonymous

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

- anonymous

particularly a pdf that is a screenshot of wolframalpha.

- 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

its rather. I hate that stupid part of english.

- 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

That's not ugly at all!

- anonymous

Break it into two integrals.

- anonymous

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

- anonymous

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

- 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

Nah it's just division by 2

- anonymous

indeed.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yes. Me too.

- anonymous

With a 2 in the denominator that is correct.

- anonymous

oh, wait

- anonymous

shouldn't be a -sin(2u)

- anonymous

\[\int cos(u)du = sin(u) +C \]

- anonymous

should be positive.

- anonymous

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

- anonymous

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

- anonymous

yeah, so that works nicely given what u is.

- anonymous

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

- anonymous

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

- 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

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

- anonymous

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

- anonymous

all kinds of fun trig subs going on.

- anonymous

We can work through it if you want

- 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

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

- anonymous

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

- anonymous

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

- anonymous

What happened to the two?

- anonymous

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

- 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

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

- anonymous

Very sweet, I just need a second to absorb.

- 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

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

oh yes, it's fun! =)

- 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

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

- 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

just takes time to digest

- 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

Good luck on the test!

Looking for something else?

Not the answer you are looking for? Search for more explanations.