Question

double integral of x^3cos(xy)dydx from
o to pi/4 and sinx to cosx

Answer 1

\[\pi/4\] ?

Answer 2

\[\int\limits_{0}^{\Pi/4}\int\limits_{sinx}^{cosx} x^3 \cos(xy) dydx\]

Answer 3

lol yeah, that's why i'm asking for help :o

Answer 4

i have no idea how to integrate that partially lol

Answer 5

firstly integrate with respect to x, keeping y constant

Answer 6

will be done by by parts

Answer 7

why am i integrating with respect to x first, if its dydx? o.O

Answer 8

thats the way partial integrals are done, first we need to do wrt to dx, and put in limits
then whatever the answer we get we have to integrate wrt to y, and put in the limiits , that will be final answer

Answer 9

but isn't that when its dxdy? when it's dydx you integrate wrt y first, and then x? o.O

Answer 10

http://tutorial.math.lamar.edu/Classes/CalcIII/IteratedIntegrals.aspx

Answer 11

ohhh i see, yeh we need to do that way, and yeh we have to interchange the limit dude

Answer 12

if i integrate with respect to x first, then do i need to change the limits?

Answer 13

wait i will do simplified form for you

Answer 14

so if its y = sinx then ... x = sin*-1y?!

Answer 15

\[\int\limits_{?}^{y/\sqrt{2}}\int\limits_{0}^{1/\sqrt{2}}x^3\cos(xy)dxdy\]

Answer 16

now do it..........

Answer 17

wait.. how did you get that? o.O

Answer 18

btw in which year in college you are?

Answer 19

senior, why?

Answer 20

well its called interchanging the limit,

Answer 21

because we need to have dx first, for simplification

Answer 22

wait, why is having dx first simplified? and would you mind explaining to me how you interchanged the limit, if you don't mind?

Answer 23

because , according to question, we have to integrate wrt to y first, but you can see , for the first integration, the limits are given in terms of x, so wee need to interchange

Answer 24

wait lets try this with a much simplier integration, cuz i'm getting really confused with what you mean.. so lets say the limits are the same as the problem given above, but we integrate 1 okay?

\[\int\limits_{0}^{\Pi/4}\int\limits_{sinx}^{cosx}1 dydx = y |(from sinx \to cosx)\]
= cosx - sinx
\[=\int\limits_{0}^{\Pi/4} cosx - sinx dx\] = sinx + cosx | (from 0 to Pi/4)

see what i mean? the limits, although in terms of x, are actually functions of y
y = sinx, and y = cosx, that's why it's dydx

i don't think we should be changing the limits =\

do you know how to partially integrate the equation in terms of y first? :o the cos(xy) is killing me lol

Answer 25

Is this the question no one has answered?

Answer 26

it is in pending

Answer 27

yes :(

Answer 28

@KingGeorge i'm so happy you're online!!

Answer 29

alright, integrating dydx would be easier So we have the following integral \[\large \int\limits\limits_{0}^{\pi/4}\int\limits\limits_{\sin x}^{\cos x} x^3 \cos(xy)\;\; dydx\]If we integrate with respect to y first, we get \[\large \int\limits_{0}^{\pi/4} \left(x^2 \sin(xy) |_{\sin x}^{\cos x}\right) dx\]

Answer 30

wait.. seriously? you don't need to do anything with the (xy)? you just integrate the cos? .. that's it? thats what i've been struggling so much witH? lol..

Answer 31

since you can consider x a constant, you're just integrating \(\cos(xy)\) with respect to y.

Answer 32

when i'm plugging it in.. is it just x^3sin(cosxx? lol

Answer 33

cosx^2?

Answer 34

Not quite. I'm getting \[x^2\sin(x \cos(x))-x^2\sin(x \sin(x))=x^2(\sin(x \cos(x))-\sin(x \sin(x))\]

Answer 35

okay so.. how the heck.. do i integrate this monstrosity? lol

Answer 36

lool

Answer 37

http://www.wolframalpha.com/input/?i=integrate+x^3+%28sin%28xcosx%29++-+sin%28xsinx%29%29+from+0+to+pi%2F4

Looks like numerical method!!

http://www.wolframalpha.com/input/?i=integrate+x^3+%28sin%28xcosx%29++-+sin%28xsinx%29%29

Answer 38

I have no idea of how to integrate that monstrosity by hand.

Answer 39

ytfreak is it in our book.. lol

Answer 40

okay what about integrating x^3sin(x^3)?

Answer 41

another montrosity
http://www.wolframalpha.com/input/?i=integrate+x^3+sin%28x^3%29

Answer 42

wtfreak is this. what is my teacher thinking...

Answer 43

i?!? there's an i int here!? you can get an i when you integrate?! what is going on.

Answer 44

Perhaps your teacher meant \(x^2 \sin(x^3)\)?

Answer 45

but isn't that really hard to do as well? did you see what @experimentX linked? lol

Answer 46

\(x^2 \sin(x^3)\) is actually pretty easy.

Answer 47

\[\int\limits x^2 \sin(x^3)\;\;dx =-{1 \over 3} \cos(x^3)+C\]

Answer 48

what... how did you get that? :o

Answer 49

I've seen enough integrals to recognize that pattern, but analytically, a u-sub of \(u=x^3\) would do the trick.

Answer 50

oooohhhhhhhh lol duh... -_-"