Mini
  • Mini
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)
Mathematics
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

amistre64
  • amistre64
u = pi x^2 du = 2pi x dx dx = du/2pi x
amistre64
  • amistre64
x sin(u) sin(u) ----- = ----- du 2pi x 2pi
Mini
  • Mini
so i can pull out 1/2pi in front

Looking for something else?

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

More answers

amistre64
  • amistre64
yep; it aint gonna change any :)
Mini
  • 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
  • amistre64
you need to have a 2pi up top to counter your integration
Mini
  • Mini
So when I set something dx = blah blah blah, i have to account for what im plugging in
anonymous
  • anonymous
(the answer isn't 1)
Mini
  • Mini
for instance dx = 2xdu, i need to replace dx with the 2xdu but also put in a 1/2x?
amistre64
  • amistre64
whatch.... D(-cos(2pix^2)) = 2pix sin(2pix^2) right?
amistre64
  • amistre64
4pix lol...srry
Mini
  • Mini
lol newton his answers are rarely ever what they seem to be, its also a double integration though
amistre64
  • amistre64
post the entire questionthen; its rather difficult to step in half way and try to determine where something messed up at
amistre64
  • amistre64
4pi x sin(2pix^2) -------------- dx is doable right? 4pi
Mini
  • Mini
|0to1|cube root y to 1 (sin(pix^2))/x^2 dxdy
Mini
  • Mini
hard to type a double integral with this crappy window!
amistre64
  • 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
  • Mini
uh why is u = 2pi...?
Mini
  • Mini
u = pix^2
anonymous
  • anonymous
\[\iint \] \.[\iint \]
amistre64
  • amistre64
0
amistre64
  • amistre64
you think changeing them would make it easier?
Mini
  • Mini
hold on phone lol
amistre64
  • amistre64
{SS} [ sin(pix^2)/x^2] dy dx
amistre64
  • amistre64
y=1 y=0 x=1 x=cbrt(y) can be changed to make it easier to start with a dy
anonymous
  • anonymous
Amistre, brah, one day you're gonna have to give in and learn some LaTeX.
amistre64
  • amistre64
lol.....nvr ;)
amistre64
  • amistre64
heres our domain
1 Attachment
amistre64
  • amistre64
we can use this to switch to dydx
Mini
  • Mini
thats what i did, makes the first one extremely easy
Mini
  • Mini
u just ad a y t hen its all good
amistre64
  • amistre64
0 < y < x^3 0 < x < 1 right?
Mini
  • Mini
ya then u do the first one, and its just the regular eq with a y added to it
amistre64
  • amistre64
yes :)
amistre64
  • amistre64
i gotta work thru it to see where se go tho :)
Mini
  • Mini
then u sub it all in and do the bounds and the x's cancel til u get xsin(pix^2)
Mini
  • Mini
thats why i skipped though, i was pretty sure my work til there was accurate
Mini
  • Mini
now ive been doing u = pix^2 du = 2pix
amistre64
  • 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
  • amistre64
{S} x sin(pix^2) dx ; [0,1] right?
Mini
  • Mini
so then dx = du/2pix
amistre64
  • amistre64
forget the u sub for the moment....
Mini
  • Mini
oh yeah kinda working ahead, ive done it up until here so many times haha
Mini
  • Mini
oh okay, well i tried by parts too, that pellet was NOT COOL
amistre64
  • amistre64
if this was: 2pi x sin(pix^2) you could solve it right?
Mini
  • Mini
well yeah thats easy, because it cancels itself
Mini
  • Mini
its just in terms of say udu
amistre64
  • amistre64
2pi [S] --- x sin(pix^2) dx then multiply by a useful form of 1 lol 2pi
amistre64
  • amistre64
drag out the bottom 2pi and integrate it lol
Mini
  • Mini
but i still have the 1/2pi
Mini
  • Mini
no no see thats not my problem, he has the answer a 1 which is throwing me off
amistre64
  • amistre64
ok...let me see what I gets :) i see the issue, but not the solution :)
Mini
  • Mini
follow me on my u sub i have u = pix^2 du = 2pixdx dx = du/2pix
Mini
  • 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
  • amistre64
-cos(pi x^2) ---------- at 0 = -1/(2pi) 2pi -cos(pi x^2) ---------- at 1 = 1/(2pi) right? 2pi
amistre64
  • amistre64
nah, just plug it in; its good...
Mini
  • Mini
no wait dont u have to change the bounds?
amistre64
  • amistre64
1 1 --- + --- = pi .... so thats the answer 2pi 2pi
Mini
  • Mini
like say u OF 1 @ u = pix^2
amistre64
  • amistre64
only if you wanna work with u and not pix^2
amistre64
  • amistre64
same difference tho
Mini
  • 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
  • amistre64
correct; i always mess that part up so I just re-stsute it back :)
Mini
  • Mini
yea i always just confuse myself and think too much into it
amistre64
  • amistre64
but i get pi as an answer.... as long as it was all done right
amistre64
  • amistre64
err....1/pi lol
Mini
  • Mini
yeah i just got that too, he still has 1 as the answer though, pisses me off
Mini
  • Mini
what goods a test review if you have the wrong answers listed!
amistre64
  • amistre64
tell him to prove it; show him your work for verification :)
Mini
  • Mini
i wont see him until my exam on monday, cant really do that haha
amistre64
  • amistre64
i know; right answers build confidence :)
Mini
  • 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
  • anonymous
Just use wolfram (or equivalent) to check answer.
Mini
  • 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
  • amistre64
your u sub was fine; a little more work than I would have put into it; but fine
Mini
  • Mini
i used wolfram earlier and got the same general answer with an infinite bounds
Mini
  • Mini
which is essentially the same thing
amistre64
  • amistre64
i just read up on double ints last night; they remind me of nested functions in programing
Mini
  • Mini
i wont worry about it now though however...
Mini
  • Mini
if you arent busy i have a long intensive problem for u lol
amistre64
  • amistre64
give me a shot at it; i gotta warm up for calc1 on monday :)
Mini
  • Mini
lets see
Mini
  • Mini
fxy if f(x,y) = sqrt(xy) - 1/(((x^2)y)
Mini
  • Mini
ill just type it as i would say it to clarify, squareroot of both x and y, minus
Mini
  • Mini
one over x squared times y
amistre64
  • amistre64
\[\frac{\sqrt{xy}-1}{x^2y}\]
Mini
  • Mini
no no
Mini
  • Mini
the x^2y
Mini
  • Mini
is ONLY under the 1, nothing else
amistre64
  • amistre64
\[\sqrt{xy} -\frac{1}{x^2y}\]
Mini
  • 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
  • Mini
yeah thats it
amistre64
  • amistre64
x^(-2) y^(-1)
amistre64
  • amistre64
what are we to do with this function?
amistre64
  • amistre64
x^(1/2) y^(1/2) - x^(-2) y^(-1)
Mini
  • Mini
u find der with respect to x then y
Mini
  • Mini
but when you find the der with respect to y, u use what you found with respect to x
amistre64
  • amistre64
dF -\[\frac{dF}{dy} = \frac{\sqrt{x}}{2 \sqrt{y}} + \frac{\ln(y)}{x^2}\] dx
amistre64
  • amistre64
\[\frac{dF}{dx} = \frac{\sqrt{y}}{2\sqrt{x}} + \frac{2}{xy}\]
Mini
  • Mini
no
Mini
  • Mini
u do dx first, then when u get that der
amistre64
  • amistre64
prolly dint understand the questionthen lol
Mini
  • Mini
u do d/dy to tHAT one
amistre64
  • amistre64
ohh... is my dx right lol
Mini
  • 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
  • Mini
hold on lemme work it again
Mini
  • Mini
\[1/2(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy\]
Mini
  • Mini
thats the first one
amistre64
  • amistre64
\[D[(dF/dx)]/dy = \frac{1}{4\sqrt{xy}} - \frac{2x}{(xy)^2}\]
Mini
  • Mini
change everything to chain its easier to do, cause our answers arent matching up
amistre64
  • amistre64
that last is messed
amistre64
  • amistre64
working in a dim light and no paper :)
amistre64
  • amistre64
the first term is good, or shoul dbe
amistre64
  • amistre64
but +2/x * 1/y = 2ln(y)/x
Mini
  • Mini
dont even worry about ln, just makes it worse than it already is, but i dont think your first is right
Mini
  • Mini
(xy)^1/2 in terms of X
amistre64
  • amistre64
Dx(sqrt(xy)) = sqrt(x)/2sqrt(y)
amistre64
  • amistre64
x and y are transposed, but that should be it
amistre64
  • amistre64
Csqrt(x) = C/2sqrt(x) right?
Mini
  • 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
  • Mini
then chain the inside and get y
amistre64
  • amistre64
the y term is constant with respect to x.... right?
Mini
  • Mini
yes thats why when u chain the inside u only get y, not x
Mini
  • Mini
its like 9x so to speak
amistre64
  • amistre64
Csqrt(x) - C/2sqrt(x)
amistre64
  • amistre64
sqrt(y)/C -> 1/C2sqrt(y) ends up 1/4sqrt(xy)
Mini
  • 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
  • amistre64
sqrt(xy) = sqrt(x)* sqrt(y)
Mini
  • Mini
yes, but i dont see why you would do that in the first place
amistre64
  • amistre64
to work them seperately; do dx first then dy and come to the proper solution..
amistre64
  • amistre64
that is what you said to do right?
Mini
  • Mini
no that doesnt make any sense
Mini
  • Mini
no you have to do with respect to x to the WHOLE equation first
Mini
  • Mini
once you find with respect to x u hvae to apply with respect to y after that
amistre64
  • amistre64
oh, like implicit then?
Mini
  • Mini
no lol
amistre64
  • amistre64
then what you want is to hold y constant and dx first right?
Mini
  • Mini
yes i thought we already said that earlier, y is just y, it does nothing now
amistre64
  • amistre64
so i split the sqrt to work with only the x
amistre64
  • amistre64
sqrt(y) ------ 2sqrt(x)
Mini
  • Mini
no
Mini
  • Mini
nevermind lol
amistre64
  • amistre64
you say no...but tell me why not?
Mini
  • Mini
okay we have the orignal problem blah blah
Mini
  • Mini
we take the derivative of that with ONLY respect to x
Mini
  • Mini
once we get that function we take the derivative of that with respect to y
amistre64
  • amistre64
and since it is a sum we can do each term seperately...lets work the first term.
amistre64
  • amistre64
sqrt(xy) = sqrt(x) * sqrt(y) by definition that is appropriate
Mini
  • Mini
u just chain rule so .5(xy)^(-1/2) times y
Mini
  • Mini
no quit doin that!
Mini
  • Mini
just look at sqrt(xy) as (xy)^(1/2)
amistre64
  • amistre64
ok
Mini
  • Mini
now instead of splitting stuff up, u just made the whole thing a chain rule which is reaaaaaaaaally easy
Mini
  • Mini
you kno what chain rule is right?
amistre64
  • amistre64
sqrt(Cx) ->> C/2sqrt(Cx) right?
Mini
  • Mini
yes
Mini
  • Mini
because its just (Cx)^(1/2)
Mini
  • Mini
do you kno what the chain rule is though?
amistre64
  • amistre64
so that translates to: y ------- correct? 2sqrt(xy)
Mini
  • Mini
yes!
amistre64
  • amistre64
lol...no, i just do double integrals for a hobby ;)
Mini
  • Mini
well with this problem its easier if you just put everything to a negative exponent
Mini
  • Mini
its verrrrrrrry long, and that makes it faster, especially when u are trying to do it in terms of x or y
amistre64
  • amistre64
easier is a subjective term;
Mini
  • Mini
well bringing something to the front, minus 1, then der of the inside is pretty easy, dont think thats subjective
amistre64
  • amistre64
so we got: y ------ and should dy that now right? 2sqrt(xy)
Mini
  • Mini
no u still have to do dx to the last part, that was only the first part lol
amistre64
  • amistre64
the last term can wait; its part of another section of the addition
Mini
  • Mini
the eq is \[(xy)^{1/2}-1/(x ^{2}y)\]
amistre64
  • amistre64
right; so the first term is derivable n its own right without the second
Mini
  • 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
  • Mini
now you just product the 1st term get y, then product the second term
amistre64
  • amistre64
the length of this posting is messing up my system :) ill let you have at it alone ;)

Looking for something else?

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