Question

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)

Answer 1

u = pi x^2

du = 2pi x dx

dx = du/2pi x

Answer 2

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

Answer 3

so i can pull out 1/2pi in front

Answer 4

yep; it aint gonna change any :)

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

(the answer isn't 1)

Answer 9

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

Answer 10

whatch....


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

Answer 11

4pix lol...srry

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

hard to type a double integral with this crappy window!

Answer 17

your u was ok but we messed up the derivative of it lol

u = 2pix^2

du = 4pi x dx

dx = du/4pi x

Answer 18

uh why is u = 2pi...?

Answer 19

u = pix^2

Answer 20

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

Answer 21

0<y<1; and cbrt(y)<x<1 is your bounds right?

Answer 22

you think changeing them would make it easier?

Answer 23

hold on phone lol

Answer 24

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

Answer 25

y=1
y=0

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

Answer 26

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

Answer 27

lol.....nvr ;)

Answer 28

heres our domain

Answer 29

we can use this to switch to dydx

Answer 30

thats what i did, makes the first one extremely easy

Answer 31

u just ad a y t hen its all good

Answer 32

0 < y < x^3

0 < x < 1 right?

Answer 33

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

Answer 34

yes :)

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

sin(pix^2)
-------- (x^3) = x sin(pix^2)
x^2

lol.... its not that i dont trust you; i just dont trust me :)

Answer 40

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

Answer 41

so then dx = du/2pix

Answer 42

forget the u sub for the moment....

Answer 43

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

Answer 44

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

Answer 45

if this was:

2pi x sin(pix^2) you could solve it right?

Answer 46

well yeah thats easy, because it cancels itself

Answer 47

its just in terms of say udu

Answer 48

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

Answer 49

drag out the bottom 2pi and integrate it lol

Answer 50

but i still have the 1/2pi

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

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

Answer 55

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

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

Answer 56

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

Answer 57

no wait dont u have to change the bounds?

Answer 58

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

Answer 59

like say u OF 1 @ u = pix^2

Answer 60

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

Answer 61

same difference tho

Answer 62

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?

Answer 63

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

Answer 64

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

Answer 65

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

Answer 66

err....1/pi lol

Answer 67

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

Answer 68

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

Answer 69

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

Answer 70

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

Answer 71

i know; right answers build confidence :)

Answer 72

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

Answer 73

Just use wolfram (or equivalent) to check answer.

Answer 74

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

Answer 75

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

Answer 76

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

Answer 77

which is essentially the same thing

Answer 78

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

Answer 79

i wont worry about it now though
however...

Answer 80

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

Answer 81

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

Answer 82

lets see

Answer 83

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

Answer 84

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

Answer 85

one over x squared times y

Answer 86

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

Answer 87

no no

Answer 88

the x^2y

Answer 89

is ONLY under the 1, nothing else

Answer 90

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

Answer 91

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

Answer 92

yeah thats it

Answer 93

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

Answer 94

what are we to do with this function?

Answer 95

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

Answer 96

u find der with respect to x then y

Answer 97

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