Math help needed...

2008-01-28 20:11:45

#1

S.Lythberg: Mastermind; +429|6932|Chicago, IL

Ok BF2s land, I've done math for several of you, now it's your turn:

I need to find the integral of (x^2)/((x^2 + 9)^(1/2))dx

its a trig substitution problem, so x becomes 3sec(u), and dx becomes 3sec(u)tan(u), but I cant rework the equation to bring it back into terms of x that I can integrate...

anyone?

2008-01-28 20:14:05

#2

phil-12-12: Banned; +21|6668|c-c-c-Canada

Since no one is helping you. ill kind of will be a help

www.google.ca

2008-01-28 20:14:20

#3

Ender2309: has joined the GOP; +470|7057|USA

there's no trig here...

2008-01-28 20:14:48

#4

cowami: OY, BITCHTITS!; +1,106|6775|Noo Yawk, Noo Yawk

Sorry, I start integrals next term.

2008-01-28 20:17:03

#5

S.Lythberg: Mastermind; +429|6932|Chicago, IL

Ender2309 wrote:
there's no trig here...

its the only way to eliminate the radical, replace x with a trigonometric term that can be rewritten without a radical using an identity...

2008-01-28 20:21:32

#6

Ender2309: has joined the GOP; +470|7057|USA

S.Lythberg wrote:
Ender2309 wrote:
there's no trig here...
its the only way to eliminate the radical, replace x with a trigonometric term that can be rewritten without a radical using an identity...

ah.

detail your steps so far and i'll be better able to help.

2008-01-28 20:38:22

#7

S.Lythberg: Mastermind; +429|6932|Chicago, IL

Ender2309 wrote:
S.Lythberg wrote:
Ender2309 wrote:
there's no trig here...
its the only way to eliminate the radical, replace x with a trigonometric term that can be rewritten without a radical using an identity...
ah.

detail your steps so far and i'll be better able to help.

ok

trig substitution: x=3sec(u)

radicand becomes: 9sec²(u) - 9, which is equal to 9tan²(u)

that term becomes 3tan(u), leaving me with: (9sec²(u))/(3tan(u)) dx

dx = 3 sec(u)tan(u)

so, when that is plugged in for dx, the equation becomes Integral [9sec³u du], which I cant seem to bring back to terms of x

Last edited by S.Lythberg (2008-01-28 20:39:17)

2008-01-28 20:46:55

#8

Ender2309: has joined the GOP; +470|7057|USA

i think your problem is that you're trying to use a dx without solving for du.

edit: wait, i might have found a way around trig, gimme a sec.

Last edited by Ender2309 (2008-01-28 20:48:15)

2008-01-28 20:47:05

#9

some_random_panda: Flamesuit essential; +454|6876

I did integrals last year. I'm not digging it up again until I need to.

2008-01-28 20:49:15

#10

Ryan: Member; +1,230|7329|Alberta, Canada

What grade is this?

2008-01-28 20:49:42

#11

S.Lythberg: Mastermind; +429|6932|Chicago, IL

Ender2309 wrote:
i think your problem is that you're trying to use a dx without solving for du.

dx=3sec(u)tan(u) du, which got me to the 9sec³u du, but I cant bring that back to terms of x dx, theres always a u lingering somewhere.

2008-01-28 20:50:46

#12

S.Lythberg: Mastermind; +429|6932|Chicago, IL

Ryan wrote:
What grade is this?

calc 231

2008-01-28 20:51:04

#13

nukchebi0: Пушкин, наше всё; +387|6809|New Haven, CT

Use uk substitution on the trig portion of the equation.

Make it 9sec^2(u)/3k, with k=3tan(u)

dk=3sec^2(u) du

dk/3sec^2(u)=du

Sub in for du.

Get 3/k.

You have the integral of 1/k multiplied by three.

This gets 3ln(k)

k=3tanu

Edit: I need to read better.

Last edited by nukchebi0 (2008-01-28 20:59:55)

2008-01-28 20:51:56

#14

Ender2309: has joined the GOP; +470|7057|USA

lyth do you have an answer for the antideriv part of this?

edit: or give me A and B and i'll plug it in.

Last edited by Ender2309 (2008-01-28 20:53:37)

2008-01-28 20:55:48

#15

S.Lythberg: Mastermind; +429|6932|Chicago, IL

int 9sec³u = (9/2)secutanu + (9/2)ln{abs value}(secu + tanu) + C

2008-01-28 20:58:00

#16

Ender2309: has joined the GOP; +470|7057|USA

S.Lythberg wrote:
int 9sec³u = (9/2)secutanu + (9/2)ln{abs value}(secu + tanu) + C

...did you pull that off a calculator?

2008-01-28 20:58:55

#17

S.Lythberg: Mastermind; +429|6932|Chicago, IL

Ender2309 wrote:
S.Lythberg wrote:
int 9sec³u = (9/2)secutanu + (9/2)ln{abs value}(secu + tanu) + C
...did you pull that off a calculator?

no, thats the integral, but I cant replace the u's with x's

2008-01-28 21:00:41

#18

nukchebi0: Пушкин, наше всё; +387|6809|New Haven, CT

You can do abs value like this.

ln|sin|

Shift+backward slash.

2008-01-28 21:02:27

#19

Ender2309: has joined the GOP; +470|7057|USA

S.Lythberg wrote:
Ender2309 wrote:
S.Lythberg wrote:
int 9sec³u = (9/2)secutanu + (9/2)ln{abs value}(secu + tanu) + C
...did you pull that off a calculator?
no, thats the integral, but I cant replace the u's with x's

my 89 pumps that out as the answer, which i can't for the life of me do by hand for some reason. sorry.

2008-01-28 21:04:36

#20

S.Lythberg: Mastermind; +429|6932|Chicago, IL

Ender2309 wrote:
S.Lythberg wrote:
Ender2309 wrote:

...did you pull that off a calculator?
no, thats the integral, but I cant replace the u's with x's
my 89 pumps that out as the answer, which i can't for the life of me do by hand for some reason. sorry.

I already did it last week, and its in the table of identities, I just cant get it back in terms of x

2008-01-28 21:05:35

#21

Ender2309: has joined the GOP; +470|7057|USA

S.Lythberg wrote:
Ender2309 wrote:
S.Lythberg wrote:

no, thats the integral, but I cant replace the u's with x's
my 89 pumps that out as the answer, which i can't for the life of me do by hand for some reason. sorry.
I already did it last week, and its in the table of identities, I just cant get it back in terms of x

what about just leaving it really messy?

2008-01-28 21:09:22

#22

nukchebi0: Пушкин, наше всё; +387|6809|New Haven, CT

9sec³u du

Doesn't 3secu=x?

In this case, its 3x³

Thus, 3/4x⁴.

I think this works.

Although it seems way too simple.

3(3)secusecusecu
3(3secu3secu3secu)
3(x*x*x)
3x³

Unless I messed up the algebra by assuming something wrong, it appears to work here.

k=secu
dk=secutanu du
dk/secutanu=du

9k³dk

9/4k⁴=Integral

9/4(secutanu)²(secutanu)²

secu

Last edited by nukchebi0 (2008-01-28 21:16:36)

2008-01-28 21:14:18

#23

S.Lythberg: Mastermind; +429|6932|Chicago, IL

Ender2309 wrote:
S.Lythberg wrote:
Ender2309 wrote:
my 89 pumps that out as the answer, which i can't for the life of me do by hand for some reason. sorry.
I already did it last week, and its in the table of identities, I just cant get it back in terms of x
what about just leaving it really messy?

well, its gonna be messy, but not inverse secant messy, which is what i'm coming up with

books answer:

(3x/2)[(x/3)²-1]^1/2 + (9/2)ln | (x/3) + [(x/3)² - 1]^1/2 | + C

2008-01-28 21:18:42

#24

nukchebi0: Пушкин, наше всё; +387|6809|New Haven, CT

This is when I would just put a question mark next to it and save for questions in class tomorrow.

2008-01-28 21:19:36

#25

S.Lythberg: Mastermind; +429|6932|Chicago, IL

nukchebi0 wrote:
9sec³u du

Doesn't 3secu=x?

In this case, its 3x³

Thus, 3/4x⁴.

I think this works.

Although it seems way too simple.

3(3)secusecusecu
3(3secu3secu3secu)
3(x*x*x)
3x³

Unless I messed up the algebra by assuming something wrong, it appears to work here.

k=secu
dk=secutanu du
dk/secutanu=du

9k³dk

9/4k⁴=Integral

9/4(secutanu)²(secutanu)²

secu

you forgot about du, which has to be returned to dx in order to integrate

BF2S Forums Math help needed...

Ender2309 wrote:

S.Lythberg wrote:

Ender2309 wrote:

Ender2309 wrote:

S.Lythberg wrote:

Ender2309 wrote:

Ender2309 wrote:

Ryan wrote:

S.Lythberg wrote:

Ender2309 wrote:

S.Lythberg wrote:

S.Lythberg wrote:

Ender2309 wrote:

S.Lythberg wrote:

Ender2309 wrote:

S.Lythberg wrote:

Ender2309 wrote:

S.Lythberg wrote:

Ender2309 wrote:

S.Lythberg wrote:

Ender2309 wrote:

S.Lythberg wrote:

Ender2309 wrote:

nukchebi0 wrote:

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