# Integration;unsure method

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Feb 27th 2007, 12:10 PM
ChaosBlue
Integration;unsure method
Hi, I have another integration question (because they're just so fun :) ).

Int ( dx/(x^2-3x+2)^(1/2) )

The item giving me the biggest problem is the square rt. From what I know, I don't think u-sub will work (just complicate things more), integrating my parts seems like it would be very messy, partial fractions is out of the mix since its raised to a power and I don't think you can do complete the square for the same reason. I'm assuming I'm making this much more complicated than it really is. Can anybody help me start it out, thats all I need help with.

Thanks a ton.
• Feb 27th 2007, 12:42 PM
topsquark
Quote:

Originally Posted by ChaosBlue
Hi, I have another integration question (because they're just so fun :) ).

Int ( dx/(x^2-3x+2)^(1/2) )

The item giving me the biggest problem is the square rt. From what I know, I don't think u-sub will work (just complicate things more), integrating my parts seems like it would be very messy, partial fractions is out of the mix since its raised to a power and I don't think you can do complete the square for the same reason. I'm assuming I'm making this much more complicated than it really is. Can anybody help me start it out, thats all I need help with.

Thanks a ton.

Start this by looking under the square root sign. You want to do a "complete the square."

x^2 - 3x + 2 = (x^2 - 3x) + 2 = (x^2 - 3x + 9/4 - 9/4) + 2 = (x^2 - 3x + 9/4) - 9/4 + 2

= (x - 3/2)^2 - 1/4

Now let y = x - 3/2, dy = dx

Int[1/(x^2-3x+2)^{1/2}dx] = Int[1/(y^2 - 1/4)^{1/2}dy]

One more trick and we're done. Pull a 1/4 out from under the radical:
Int[1/((1/4)(2y)^2 - 1/4)^{1/2}dy] = 2*Int[1/((2y)^2 - 1)^{1/2}dy]

And now let z = 2y, dz = 2dy and your integral becomes:
Int[1/(x^2-3x+2)^{1/2}dx] = Int[1/(y^2 - 1/4)^{1/2}dy] = Int[1/(z^2 - 1)^{1/2}dz]

This is a more standard problem. Can you take it from here?

-Dan
• Feb 27th 2007, 12:48 PM
Jhevon
Quote:

Originally Posted by ChaosBlue
Hi, I have another integration question (because they're just so fun :) ).

Int ( dx/(x^2-3x+2)^(1/2) )

The item giving me the biggest problem is the square rt. From what I know, I don't think u-sub will work (just complicate things more), integrating my parts seems like it would be very messy, partial fractions is out of the mix since its raised to a power and I don't think you can do complete the square for the same reason. I'm assuming I'm making this much more complicated than it really is. Can anybody help me start it out, thats all I need help with.

Thanks a ton.

The way I see to do it is really long, so if anyone out there has an easier way, you are welcome to post.

We begin by completing the square.

int(1/sqrt(x^2 - 3x + 2))dx
= int(1/(sqrt( x^2 - 3x + (-3/2)^2 - 1/4))dx
= int(1/sqrt((x - 3/2)^2 - 1/4))dx
= int(1/sqrt((x - 3/2)^2 - (1/2)^2))dx
= int(1/sqrt[(1/2)^2*([(x - 3/2)/(1/2)]^2 - 1)])dx
= 2*int(1/sqrt[(2x - 3)^2 - 1])dx

let u= 2x - 3
=> du = 2 dx
so our integral becomes
int(1/sqrt(u^2 - 1))du

Now continue using trig substitution
• Feb 27th 2007, 12:50 PM
Jhevon
Quote:

Originally Posted by topsquark
Start this by looking under the square root sign. You want to do a "complete the square."

x^2 - 3x + 2 = (x^2 - 3x) + 2 = (x^2 - 3x + 9/4 - 9/4) + 2 = (x^2 - 3x + 9/4) - 9/4 + 2

= (x - 3/2)^2 - 1/4

Now let y = x - 3/2, dy = dx

Int[1/(x^2-3x+2)^{1/2}dx] = Int[1/(y^2 - 1/4)^{1/2}dy]

One more trick and we're done. Pull a 1/4 out from under the radical:
Int[1/((1/4)(2y)^2 - 1/4)^{1/2}dy] = 2*Int[1/((2y)^2 - 1)^{1/2}dy]

And now let z = 2y, dz = 2dy and your integral becomes:
Int[1/(x^2-3x+2)^{1/2}dx] = Int[1/(y^2 - 1/4)^{1/2}dy] = Int[1/(z^2 - 1)^{1/2}dz]

This is a more standard problem. Can you take it from here?

-Dan

Ha ha, i took forever typing this up only to see that u beat me to it once i clicked "submit reply"

Thanks a lot Dan
• Feb 27th 2007, 12:55 PM
topsquark
Quote:

Originally Posted by Jhevon
Ha ha, i took forever typing this up only to see that u beat me to it once i clicked "submit reply"

Thanks a lot Dan

It happens. You're welcome. :)

And besides, my variables are prettier anyway. :D

-Dan
• Feb 27th 2007, 12:59 PM
Jhevon
Quote:

Originally Posted by topsquark
It happens. You're welcome. :)

And besides, my variables are prettier anyway. :D

-Dan

Nu-uh, u is a much more suitable variable for substitution, most text books use u. You're just fighting the system.

Do you know what they do to non-conformists in the math field?!
• Feb 27th 2007, 02:39 PM
ChaosBlue
Thanks guys so much, you really helped me out. I do have one question though.

Quote:

Originally Posted by topsquark
One more trick and we're done. Pull a 1/4 out from under the radical:
Int[1/((1/4)(2y)^2 - 1/4)^{1/2}dy] = 2*Int[1/((2y)^2 - 1)^{1/2}dy]

And now let z = 2y, dz = 2dy and your integral becomes:
Int[1/(x^2-3x+2)^{1/2}dx] = Int[1/(y^2 - 1/4)^{1/2}dy] = Int[1/(z^2 - 1)^{1/2}dz]

topsquark: I understand we're essentially "manipulating" our equation so we can effectively use trig substitution, but I kind-of got lost in the process. I have trouble grasping it after we pull out the 1/4.? Is there anyway you (or Jhevon since you got it as well) could explain that part to me a bit more?

Thanks a ton.
• Feb 27th 2007, 02:42 PM
Jhevon
Quote:

Originally Posted by ChaosBlue
Thanks guys so much, you really helped me out. I do have one question though.

topsquark: I understand we're essentially "manipulating" our equation so we can effectively use trig substitution, but I kind-of got lost in the process. I have trouble grasping it after we pull out the 1/4.? Is there anyway you (or Jhevon since you got it as well) could explain that part to me a bit more?

Thanks a ton.

When he said pull out 1/4, he meant we factorized the function to obtain (1/4)*(something), he did that so he could get something of the form u^2 - 1 under the square root
• Feb 27th 2007, 03:08 PM
ChaosBlue
Quote:

Originally Posted by Jhevon
When he said pull out 1/4, he meant we factorized the function to obtain (1/4)*(something), he did that so he could get something of the form u^2 - 1 under the square root

I worded my question poorly (my bad :p ). What I mean is this.

I get where we use completing the square to transform:

int 1/sqrt(x^2-3x+2) dx--> int 1/sqrt( (x-3/2)^2 - 1/4 )dx

From there we have u = x-3/2 and therefore du = dx

Plug in and we get int 1/ (u^2 - 1/4)^1/2 du

Now, we pull out or factor out the 1/4; But I am struggling with when we do that, how

int 1/ (u^2 - 1/4)^1/2 du becomes

Int[1/((1/4)(2u)^2 - 1/4)^{1/2}du]
(we pull out 1/4, so how does u^2 become 2u^2 ?

and then

2*Int[1/((2u)^2 - 1)^{1/2}du]

Sorry for bothering you guys so much!
• Feb 27th 2007, 03:18 PM
Jhevon
Quote:

Originally Posted by ChaosBlue
I worded my question poorly (my bad :p ). What I mean is this.

I get where we use completing the square to transform:

int 1/sqrt(x^2-3x+2) dx--> int 1/sqrt( (x-3/2)^2 - 1/4 )dx

From there we have u = x-3/2 and therefore du = dx

Plug in and we get int 1/ (u^2 - 1/4)^1/2 du

Now, we pull out or factor out the 1/4; But I am struggling with when we do that, how

int 1/ (u^2 - 1/4)^1/2 du becomes

Int[1/((1/4)(2u)^2 - 1/4)^{1/2}du]
(we pull out 1/4, so how does u^2 become 2u^2 ?

and then

2*Int[1/((2u)^2 - 1)^{1/2}du]

Sorry for bothering you guys so much!

it's no bother, we're all here to learn.

ok, here it is. Say we had some function b, and we wanted to factor an a out of it, we could write is as a*(b/a).

similarly, when he factored 1/4 out of u^2, he ended up with (1/4)*(u^2/(1/4))

u^2/(1/4) = u^2/(1/2)^2 = [u/(1/2)]^2 = 2u^2
• Feb 27th 2007, 03:25 PM
Jhevon
Funny you should say thanks, we actually left the hard part to you :D
• Feb 27th 2007, 04:45 PM
topsquark
Quote:

Originally Posted by Jhevon
Nu-uh, u is a much more suitable variable for substitution, most text books use u. You're just fighting the system.

Do you know what they do to non-conformists in the math field?!

Good thing I'm not in the Math field, then. :p

-Dan
• Feb 27th 2007, 06:04 PM
Jhevon
Quote:

Originally Posted by topsquark
Good thing I'm not in the Math field, then. :p

-Dan

It's even worst in the physics field, which i suppose you're a part of
• Feb 27th 2007, 07:51 PM
topsquark
Quote:

Originally Posted by Jhevon
It's even worst in the physics field, which i suppose you're a part of

Yes, but as long as you define your variables, you can technically get away with any set of variables you wish.

I think I'm going to create a unit called the "Dan." It'll be defined as the amount of work needed to translate an equation from the variable z to the variable u...

-Dan
• Feb 27th 2007, 07:53 PM
Jhevon
Quote:

Originally Posted by topsquark
Yes, but as long as you define your variables, you can technically get away with any set of variables you wish.

I think I'm going to create a unit called the "Dan." It'll be defined as the amount of work needed to translate an equation from the variable z to the variable u...

-Dan

haha, bless your heart Dan, you need help
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last