## anonymous 5 years ago 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)

1. amistre64

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

2. amistre64

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

3. anonymous

so i can pull out 1/2pi in front

4. amistre64

yep; it aint gonna change any :)

5. anonymous

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

6. amistre64

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

7. anonymous

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

8. anonymous

(the answer isn't 1)

9. anonymous

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

10. amistre64

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

11. amistre64

4pix lol...srry

12. anonymous

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

13. amistre64

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

14. amistre64

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

15. anonymous

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

16. anonymous

hard to type a double integral with this crappy window!

17. 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

18. anonymous

uh why is u = 2pi...?

19. anonymous

u = pix^2

20. anonymous

$\iint$ \.[\iint \]

21. amistre64

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

22. amistre64

you think changeing them would make it easier?

23. anonymous

hold on phone lol

24. amistre64

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

25. amistre64

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

26. anonymous

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

27. amistre64

lol.....nvr ;)

28. amistre64

heres our domain

29. amistre64

we can use this to switch to dydx

30. anonymous

thats what i did, makes the first one extremely easy

31. anonymous

u just ad a y t hen its all good

32. amistre64

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

33. anonymous

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

34. amistre64

yes :)

35. amistre64

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

36. anonymous

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

37. anonymous

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

38. anonymous

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

39. 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 :)

40. amistre64

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

41. anonymous

so then dx = du/2pix

42. amistre64

forget the u sub for the moment....

43. anonymous

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

44. anonymous

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

45. amistre64

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

46. anonymous

well yeah thats easy, because it cancels itself

47. anonymous

its just in terms of say udu

48. amistre64

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

49. amistre64

drag out the bottom 2pi and integrate it lol

50. anonymous

but i still have the 1/2pi

51. anonymous

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

52. amistre64

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

53. anonymous

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

54. anonymous

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

55. amistre64

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

56. amistre64

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

57. anonymous

no wait dont u have to change the bounds?

58. amistre64

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

59. anonymous

like say u OF 1 @ u = pix^2

60. amistre64

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

61. amistre64

same difference tho

62. anonymous

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?

63. amistre64

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

64. anonymous

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

65. amistre64

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

66. amistre64

err....1/pi lol

67. anonymous

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

68. anonymous

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

69. amistre64

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

70. anonymous

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

71. amistre64

i know; right answers build confidence :)

72. anonymous

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

73. anonymous

Just use wolfram (or equivalent) to check answer.

74. anonymous

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

75. amistre64

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

76. anonymous

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

77. anonymous

which is essentially the same thing

78. amistre64

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

79. anonymous

i wont worry about it now though however...

80. anonymous

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

81. amistre64

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

82. anonymous

lets see

83. anonymous

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

84. anonymous

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

85. anonymous

one over x squared times y

86. amistre64

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

87. anonymous

no no

88. anonymous

the x^2y

89. anonymous

is ONLY under the 1, nothing else

90. amistre64

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

91. anonymous

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

92. anonymous

yeah thats it

93. amistre64

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

94. amistre64

what are we to do with this function?

95. amistre64

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

96. anonymous

u find der with respect to x then y

97. anonymous

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

98. amistre64

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

99. amistre64

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

100. anonymous

no

101. anonymous

u do dx first, then when u get that der

102. amistre64

prolly dint understand the questionthen lol

103. anonymous

u do d/dy to tHAT one

104. amistre64

ohh... is my dx right lol

105. anonymous

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

106. anonymous

hold on lemme work it again

107. anonymous

$1/2(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy$

108. anonymous

thats the first one

109. amistre64

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

110. anonymous

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

111. amistre64

that last is messed

112. amistre64

working in a dim light and no paper :)

113. amistre64

the first term is good, or shoul dbe

114. amistre64

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

115. anonymous

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

116. anonymous

(xy)^1/2 in terms of X

117. amistre64

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

118. amistre64

x and y are transposed, but that should be it

119. amistre64

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

120. anonymous

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

121. anonymous

then chain the inside and get y

122. amistre64

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

123. anonymous

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

124. anonymous

its like 9x so to speak

125. amistre64

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

126. amistre64

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

127. anonymous

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

128. amistre64

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

129. anonymous

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

130. amistre64

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

131. amistre64

that is what you said to do right?

132. anonymous

no that doesnt make any sense

133. anonymous

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

134. anonymous

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

135. amistre64

oh, like implicit then?

136. anonymous

no lol

137. amistre64

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

138. anonymous

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

139. amistre64

so i split the sqrt to work with only the x

140. amistre64

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

141. anonymous

no

142. anonymous

nevermind lol

143. amistre64

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

144. anonymous

okay we have the orignal problem blah blah

145. anonymous

we take the derivative of that with ONLY respect to x

146. anonymous

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

147. amistre64

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

148. amistre64

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

149. anonymous

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

150. anonymous

no quit doin that!

151. anonymous

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

152. amistre64

ok

153. anonymous

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

154. anonymous

you kno what chain rule is right?

155. amistre64

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

156. anonymous

yes

157. anonymous

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

158. anonymous

do you kno what the chain rule is though?

159. amistre64

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

160. anonymous

yes!

161. amistre64

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

162. anonymous

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

163. anonymous

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

164. amistre64

easier is a subjective term;

165. anonymous

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

166. amistre64

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

167. anonymous

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

168. amistre64

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

169. anonymous

the eq is $(xy)^{1/2}-1/(x ^{2}y)$

170. amistre64

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

171. anonymous

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

172. anonymous

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

173. amistre64

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