Integrate: xsin(pix^2)dx I'm going U sub with u equaling to pix^2 but I get stuck trying to compensate for the du = 2pixdx
then dx = du/(2pix)

- Mini

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- amistre64

u = pi x^2
du = 2pi x dx
dx = du/2pi x

- amistre64

x sin(u) sin(u)
----- = ----- du
2pi x 2pi

- Mini

so i can pull out 1/2pi in front

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## More answers

- amistre64

yep; it aint gonna change any :)

- Mini

if thats the case then the answer he has listed is wrong, the bounds are from 0 to 1
his answer listed is 1

- amistre64

you need to have a 2pi up top to counter your integration

- Mini

So when I set something dx = blah blah blah, i have to account for what im plugging in

- anonymous

(the answer isn't 1)

- Mini

for instance dx = 2xdu, i need to replace dx with the 2xdu but also put in a 1/2x?

- amistre64

whatch....
D(-cos(2pix^2)) = 2pix sin(2pix^2) right?

- amistre64

4pix lol...srry

- Mini

lol newton his answers are rarely ever what they seem to be, its also a double integration though

- amistre64

post the entire questionthen; its rather difficult to step in half way and try to determine where something messed up at

- amistre64

4pi x sin(2pix^2)
-------------- dx is doable right?
4pi

- Mini

|0to1|cube root y to 1 (sin(pix^2))/x^2 dxdy

- Mini

hard to type a double integral with this crappy window!

- amistre64

your u was ok but we messed up the derivative of it lol
u = 2pix^2
du = 4pi x dx
dx = du/4pi x

- Mini

uh why is u = 2pi...?

- Mini

u = pix^2

- anonymous

\[\iint \]
\.[\iint \]

- amistre64

0

- amistre64

you think changeing them would make it easier?

- Mini

hold on phone lol

- amistre64

{SS} [ sin(pix^2)/x^2] dy dx

- amistre64

y=1
y=0
x=1
x=cbrt(y) can be changed to make it easier to start with a dy

- anonymous

Amistre, brah, one day you're gonna have to give in and learn some LaTeX.

- amistre64

lol.....nvr ;)

- amistre64

heres our domain

##### 1 Attachment

- amistre64

we can use this to switch to dydx

- Mini

thats what i did, makes the first one extremely easy

- Mini

u just ad a y t hen its all good

- amistre64

0 < y < x^3
0 < x < 1 right?

- Mini

ya then u do the first one, and its just the regular eq with a y added to it

- amistre64

yes :)

- amistre64

i gotta work thru it to see where se go tho :)

- Mini

then u sub it all in and do the bounds and the x's cancel til u get xsin(pix^2)

- Mini

thats why i skipped though, i was pretty sure my work til there was accurate

- Mini

now ive been doing u = pix^2
du = 2pix

- amistre64

sin(pix^2)
-------- (x^3) = x sin(pix^2)
x^2
lol.... its not that i dont trust you; i just dont trust me :)

- amistre64

{S} x sin(pix^2) dx ; [0,1] right?

- Mini

so then dx = du/2pix

- amistre64

forget the u sub for the moment....

- Mini

oh yeah kinda working ahead, ive done it up until here so many times haha

- Mini

oh okay, well i tried by parts too, that pellet was NOT COOL

- amistre64

if this was:
2pi x sin(pix^2) you could solve it right?

- Mini

well yeah thats easy, because it cancels itself

- Mini

its just in terms of say udu

- amistre64

2pi
[S] --- x sin(pix^2) dx then multiply by a useful form of 1 lol
2pi

- amistre64

drag out the bottom 2pi and integrate it lol

- Mini

but i still have the 1/2pi

- Mini

no no see thats not my problem, he has the answer a 1 which is throwing me off

- amistre64

ok...let me see what I gets :) i see the issue, but not the solution :)

- Mini

follow me on my u sub i have u = pix^2
du = 2pixdx
dx = du/2pix

- Mini

when i completely substitute dx from the original problem to this du/2pix, do i have to account or add anything to that? or do i just literally plug it in the equation

- amistre64

-cos(pi x^2)
---------- at 0 = -1/(2pi)
2pi
-cos(pi x^2)
---------- at 1 = 1/(2pi) right?
2pi

- amistre64

nah, just plug it in; its good...

- Mini

no wait dont u have to change the bounds?

- amistre64

1 1
--- + --- = pi .... so thats the answer
2pi 2pi

- Mini

like say u OF 1 @ u = pix^2

- amistre64

only if you wanna work with u and not pix^2

- amistre64

same difference tho

- Mini

so after i do the integration as long as i plug u back in before i use the bounds i dont have the change them correct?

- amistre64

correct; i always mess that part up so I just re-stsute it back :)

- Mini

yea i always just confuse myself and think too much into it

- amistre64

but i get pi as an answer.... as long as it was all done right

- amistre64

err....1/pi lol

- Mini

yeah i just got that too, he still has 1 as the answer though, pisses me off

- Mini

what goods a test review if you have the wrong answers listed!

- amistre64

tell him to prove it; show him your work for verification :)

- Mini

i wont see him until my exam on monday, cant really do that haha

- amistre64

i know; right answers build confidence :)

- Mini

well i know ive done it a dozen times and come up with the same thing, so im not worried about it, hes had wrong answers before

- anonymous

Just use wolfram (or equivalent) to check answer.

- Mini

thats the ONLY thing that was confusing me is because it didnt match my answers so i thought i did my u sub wrong

- amistre64

your u sub was fine; a little more work than I would have put into it; but fine

- Mini

i used wolfram earlier and got the same general answer with an infinite bounds

- Mini

which is essentially the same thing

- amistre64

i just read up on double ints last night; they remind me of nested functions in programing

- Mini

i wont worry about it now though
however...

- Mini

if you arent busy i have a long intensive problem for u lol

- amistre64

give me a shot at it; i gotta warm up for calc1 on monday :)

- Mini

lets see

- Mini

fxy if f(x,y) = sqrt(xy) - 1/(((x^2)y)

- Mini

ill just type it as i would say it to clarify, squareroot of both x and y, minus

- Mini

one over x squared times y

- amistre64

\[\frac{\sqrt{xy}-1}{x^2y}\]

- Mini

no no

- Mini

the x^2y

- Mini

is ONLY under the 1, nothing else

- amistre64

\[\sqrt{xy} -\frac{1}{x^2y}\]

- Mini

i was working on a dry erase board so i dont have the work anymore but i ended up doing i think 4 product rules lol

- Mini

yeah thats it

- amistre64

x^(-2) y^(-1)

- amistre64

what are we to do with this function?

- amistre64

x^(1/2) y^(1/2) - x^(-2) y^(-1)

- Mini

u find der with respect to x then y

- Mini

but when you find the der with respect to y, u use what you found with respect to x

- amistre64

dF
-\[\frac{dF}{dy} = \frac{\sqrt{x}}{2 \sqrt{y}} + \frac{\ln(y)}{x^2}\]
dx

- amistre64

\[\frac{dF}{dx} = \frac{\sqrt{y}}{2\sqrt{x}} + \frac{2}{xy}\]

- Mini

no

- Mini

u do dx first, then when u get that der

- amistre64

prolly dint understand the questionthen lol

- Mini

u do d/dy to tHAT one

- amistre64

ohh... is my dx right lol

- Mini

yeah its screwy because we thought that too, but it seemed to easy and then once u get the two different der what do u do

- Mini

hold on lemme work it again

- Mini

\[1/2(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy\]

- Mini

thats the first one

- amistre64

\[D[(dF/dx)]/dy = \frac{1}{4\sqrt{xy}} - \frac{2x}{(xy)^2}\]

- Mini

change everything to chain its easier to do, cause our answers arent matching up

- amistre64

that last is messed

- amistre64

working in a dim light and no paper :)

- amistre64

the first term is good, or shoul dbe

- amistre64

but +2/x * 1/y = 2ln(y)/x

- Mini

dont even worry about ln, just makes it worse than it already is, but i dont think your first is right

- Mini

(xy)^1/2 in terms of X

- amistre64

Dx(sqrt(xy)) = sqrt(x)/2sqrt(y)

- amistre64

x and y are transposed, but that should be it

- amistre64

Csqrt(x) = C/2sqrt(x) right?

- Mini

one half goes to the front then u have (xy) then u subtract one and get (xy)^(-1/2) all divided by 2

- Mini

then chain the inside and get y

- amistre64

the y term is constant with respect to x.... right?

- Mini

yes thats why when u chain the inside u only get y, not x

- Mini

its like 9x so to speak

- amistre64

Csqrt(x) - C/2sqrt(x)

- amistre64

sqrt(y)/C -> 1/C2sqrt(y)
ends up 1/4sqrt(xy)

- Mini

dont see where u are getting this other half from, you only get a half from the chain rule because its a sqrt

- amistre64

sqrt(xy) = sqrt(x)* sqrt(y)

- Mini

yes, but i dont see why you would do that in the first place

- amistre64

to work them seperately; do dx first then dy and come to the proper solution..

- amistre64

that is what you said to do right?

- Mini

no that doesnt make any sense

- Mini

no you have to do with respect to x to the WHOLE equation first

- Mini

once you find with respect to x u hvae to apply with respect to y after that

- amistre64

oh, like implicit then?

- Mini

no lol

- amistre64

then what you want is to hold y constant and dx first right?

- Mini

yes i thought we already said that earlier, y is just y, it does nothing now

- amistre64

so i split the sqrt to work with only the x

- amistre64

sqrt(y)
------
2sqrt(x)

- Mini

no

- Mini

nevermind lol

- amistre64

you say no...but tell me why not?

- Mini

okay we have the orignal problem blah blah

- Mini

we take the derivative of that with ONLY respect to x

- Mini

once we get that function we take the derivative of that with respect to y

- amistre64

and since it is a sum we can do each term seperately...lets work the first term.

- amistre64

sqrt(xy) = sqrt(x) * sqrt(y) by definition that is appropriate

- Mini

u just chain rule so .5(xy)^(-1/2) times y

- Mini

no quit doin that!

- Mini

just look at sqrt(xy) as (xy)^(1/2)

- amistre64

ok

- Mini

now instead of splitting stuff up, u just made the whole thing a chain rule which is reaaaaaaaaally easy

- Mini

you kno what chain rule is right?

- amistre64

sqrt(Cx) ->> C/2sqrt(Cx) right?

- Mini

yes

- Mini

because its just (Cx)^(1/2)

- Mini

do you kno what the chain rule is though?

- amistre64

so that translates to:
y
------- correct?
2sqrt(xy)

- Mini

yes!

- amistre64

lol...no, i just do double integrals for a hobby ;)

- Mini

well with this problem its easier if you just put everything to a negative exponent

- Mini

its verrrrrrrry long, and that makes it faster, especially when u are trying to do it in terms of x or y

- amistre64

easier is a subjective term;

- Mini

well bringing something to the front, minus 1, then der of the inside is pretty easy, dont think thats subjective

- amistre64

so we got:
y
------ and should dy that now right?
2sqrt(xy)

- Mini

no u still have to do dx to the last part, that was only the first part lol

- amistre64

the last term can wait; its part of another section of the addition

- Mini

the eq is \[(xy)^{1/2}-1/(x ^{2}y)\]

- amistre64

right; so the first term is derivable n its own right without the second

- Mini

you can, you are going to have to end up doing the second, and for thought it seems simpler to just do all the dx first

- Mini

now you just product the 1st term get y, then product the second term

- amistre64

the length of this posting is messing up my system :) ill let you have at it alone ;)

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