intergrate this! (1 Viewer)

[blackfriday]

int: sin^2(1/x)

this is actually a question from uni which i cant figure out so im wondering if any of you kids can figure it out.

cheers

 

[zeek]

cos2(1/x) = cos^2(1/x) - sin^2(1/x)
= 1 - 2sin^2(@)
.'. sin^2(1/x) = (1 - cos 2(1/x))/2

.'. int [sin^2 (1/x) dx] = int [1/2 dx] - int [(cos 2(1/x))2 dx]
= x/2 - [(1/2).1/(-2/x^2).sin 2(1/x)] + C
= x/2 + ((x^2)/4).sin 2(1/x) + C

sorry about the formating


EDIT: WOOPS my answer is wrong... hmmmm gonna try something else and i'll get back

 

[Riviet]

Interesting question.

I used by parts with u=sin²(1/x) v'=x, then I got

xsin²(1/x) + 2 int. 1/x.sin(1/x) dx

Used by parts again with u=sin(1/x) and v'=1/x, and I got

xsin²(1/x) + 2[ lnx.sin(1/x) + int. (lnx)/x² . cos(1/x) dx]

At this point I tried a substitution of u=1/x, which simplified the blue integral into int. ln(1/u).cosu du

You should be able to finish it off now with by parts, hope that helps.

Edit: looks like something's wrong.

 

[zeek]

"I used by parts with u=sin2(1/x) v'=x, then I got

xsin2(1/x) + 2 int. 1/x.sin(1/x) dx"

How were you able to get the int [1/x sin(1/x) dx] using the above variables?
I'm assuming you made a mistake with labelling v=x as v'=x.

If you let:

u = sin^2(1/x) du/dx = -(2/x^2).cos(1/x).sin(1/x) = (-1/x^2).sin(2/x) [by the function of a function rule and chain rule]
dv/dx = 1 v = x

int [sin^2 (1/x) dx] = xsin^2 (1/x) + int [ (1/x).sin(2/x) dx ]

Now seperately....

int [(1/x).sin(2/x) dx]

db/dx = sin(2/x) c = 1/x
b = (x^2/2).cos(2/x) dc/dx = -1/x^2

.'. int [(1/x).sin(2/x) dx] = (x/2).cos(2/x) + int [(1/2).cos(2/x) dx]
= (x/2).cos(2/x) - (x^2/4)sin(2/x)

Now after substituting this integration into the previous one....

.'. int [sin^2(1/x) dx] = xsin^2(1/x) + (x/2).cos(2/x) - (x^2/4)sin(2/x) + C

Im not sure if this is completely correct (im trying to differentiate it now LOL) but the process looks ok so i don't see any problems with it... YET... LOL

 

[blackfriday]

.'. int [(1/x).sin(2/x) dx] = (x/2).cos(2/x) + int [(1/2).cos(2/x) dx]
= (x/2).cos(2/x) - (x^2/4)sin(2/x)

seems like you can get the u.v part but integrating to u.dv part seems to be the problem. the answer is actually: 1/2*x-1/2*cos(2/x)*x-Si(2/x) according to maple but i have no idea what the Si is.

 

[Riviet]

Wow, the answer's so... compact.
I tried it again and got a very long expression, differentiated it and got nowhere.

 

[robbo_145]

seems like you can get the u.v part but integrating to u.dv part seems to be the problem. the answer is actually: 1/2*x-1/2*cos(2/x)*x-Si(2/x) according to maple but i have no idea what the Si is.

Yeah the wolfram integrator gives the following
Integrate[Sin[1/x]^2, x] ==

-(x*Cos[2/x])/2 + (x - 2*SinIntegral[2/x])/2

for information on the sinintegral http://mathworld.wolfram.com/SineIntegral.html

i think we can deduce this is beyond Yr12 level integration (and we haven't covered it in 1st year uni maths yet either)

 

[SeDaTeD]

Did the question actually ask you to find the primitive or a definite integral?

 

[blackfriday]

definite integral between zero and infinity and asked whether or not the integral coverged. i know the answer is pi/2 but i cant show it properly.

 

[Slidey]

If you get a question, provide ALL the information; including limits of integration. It can change how one approaches the question.

 

[blackfriday]

i actually wanted to know what the integral was out of pure interest.

 

[who_loves_maths]

If this appeared in your Calculus course under Improper Integrals this semester I'm sure you would have been taught the integral (rather than having to derive it empirically yourself):

si(0) = Int{infinity to 0}[Sinc(x).dx] = -pi/2 ; where the Sinc function is defined as the continuous function:

_______ [ 1 , x = 0
Sinc(x) = |
_______ [ (Sin(x))/x , x elsewhere


Knowing this, it isn't very hard to go on to show I = Int{0 to inf}[(Sin(1/x))^2 dx] = pi/2 :

I = [x/2 - (x/2)(Cos(2/x)]{0 to inf} - Int{inf to 0}[Sinc(v).dv]

the function: F(x) = x/2 - (x/2)(Cos(2/x) has limits: lim{x -> inf} F = lim(x -> 0) F

thus [x/2 - (x/2)(Cos(2/x)]{0 to inf} = 0

Hence, I = -Int{inf to 0}[Sinc(v).dv]

= pi/2

as required.