Question

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?

Answer 1

fuk

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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))

Answer 6

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

Answer 7

You need to solve for dx in terms of du.

Answer 8

plug u back in

Answer 9

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

Answer 10

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

Answer 11

\[x = 3 sin(u) \implies dx = 3cos(u)du\]
\[(9-x^2)^{1/2} dx= [9-9cos^2(u)]^{1/2} \cdot (3cos(u)\ du)\]

Answer 12

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.

Answer 13

oh wait, no.

Answer 14

It's a tough one XD

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

so z = sin u dz = cos x dx

Answer 19

i mean dz = cos u du

Answer 20

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

Answer 21

I have ...

\[3\cos(\sin^-(x/3))\]

Where would I go from here?

Answer 22

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

Answer 23

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

Answer 24

other way round bob.

Answer 25

no

Answer 26

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

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

Answer 27

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

Answer 28

so that'll be negative sin(u)

Answer 29

because the integral of cosine is sin, right?

See, I'm confused because it's still technically in the form dx, not du.

Answer 30

negative sin*

Answer 31

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

Answer 32

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

Answer 33

Yessir.

Answer 34

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

Answer 35

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

Answer 36

yeah, np

Answer 37

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

Answer 38

dx = 3cosu du

Answer 39

Right.

Answer 40

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

Answer 41

What do you have?

Answer 42

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

Answer 43

you didn't swap out dx.

Answer 44

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

Answer 45

dx = 3cosu du

You said it yourself.

Answer 46

Agreed, but if you ....

OH!

Answer 47

3cosu * 3 cosu ?

Answer 48

Indeed (along with a du)

Answer 49

okay, so we have ...
\[\int\limits_{}^{} 3 \cos (u)cos(u) du\]

which is ...


\[\int\limits_{}^{} 9 \cos ^2 (u) ^ {1/2} du\]

Answer 50

forget that square root

Answer 51

no its Integral (3 cos^2 x)

Answer 52

Right. So then you can use the double/half angle equation to swap the \(cos^2\) for an expression involving the cos(2u)

Answer 53

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

Answer 54

Should be 9 cause there's a 3 from the 9 under the sqrt and a 3 from the du replacement of dx.

Answer 55

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

Answer 56

that's the one.

Answer 57

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

Answer 58

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

Answer 59

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

Answer 60

particularly a pdf that is a screenshot of wolframalpha.

Answer 61

;)

Answer 62

Though alpha is good for checking your answers and finding where you made a mistake, but sometimes it's process is a little cumbersome.

Answer 63

its rather. I hate that stupid part of english.

Answer 64

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...
\[\int\limits_{}^{} 9 {( 1+\cos2u)}/2 du\]

Answer 65

That's not ugly at all!

Answer 66

Break it into two integrals.

Answer 67

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

Answer 68

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

Answer 69

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.

Answer 70

Nah it's just division by 2

Answer 71

indeed.

Answer 72

\[9/2 ( u - \sin(2u)/u)\]

Does this look alright?

Answer 73

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

Answer 74

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

Answer 75

yes. Me too.

Answer 76

With a 2 in the denominator that is correct.

Answer 77

oh, wait

Answer 78

shouldn't be a -sin(2u)

Answer 79

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

Answer 80

should be positive.

Answer 81

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

Answer 82

\[u = \sin^{-1}(x/3)\]

You're right, I got that mixed up with the rules for sin integration.

Answer 83

yeah, so that works nicely given what u is.

Answer 84

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

Answer 85

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

Answer 86

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?

Answer 87

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

Answer 88

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

Answer 89

all kinds of fun trig subs going on.

Answer 90

We can work through it if you want

Answer 91

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

Answer 92

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

Answer 93

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

Answer 94

No, you just get back what you started with.

\[sin(sin^{-1}(x/3)) = x/3\]

Answer 95

What happened to the two?

Answer 96

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

Answer 97

Ok, so here it is.

\[\frac{9}{2}sin(2u) =\frac{ 9(2)}{2}cos(u)sin(u)\]\[ =9(\sqrt{1-sin^2(u)})sin(u) \]