plz: integrate sin(lnx) ??? (1 Viewer)

[wori]

I have no idea how to do this~

 

[FinalFantasy]

I=int. sin(lnx) dx
let u=sin (lnx) and dv\dx=1 and v=x
du\dx=(1\x)cos(lnx)
I=xsin(lnx)-int. cos(lnx) dx
consider int. cos(lnx) dx
let u=cos(lnx) and dv\dx=1 and v=x
du\dx=(1\x)(-sin(lnx))
int. cos(lnx) dx=xcos(lnx)+int. sin(lnx)
.: I=xsin(lnx)-[xcos(lnx)+int. sin(lnx)]=xsin(lnx)-xcos(lnx)-I
2I=xsin(lnx)-xcos(lnx)
I=x\2(sin(lnx)-cos(lnx))

 

[who_loves_maths]

hi wori (again),

integrate sin(lnx):

let u = lnx ---> du = dx/x ---> x = e^u

so, Int[sin(lnx)dx] = Int[(e^u)sin(u)du]

now, integrate by parts:

let w = sin(u) ---> dw = cos(u)du; and, let dz = (e^u)du ---> z = e^u

therefore, Int[(e^u)sin(u)du] = sin(u)(e^u) - Int[(e^u)cos(u)du]

consider Int[(e^u)cos(u)du], and apply integration by parts again:

let m = cos(u) ---> dm = -sin(u)du; and, let dn = (e^u)du ---> n = e^u

therefore, Int[(e^u)cos(u)du] = cos(u)(e^u) + Int[(e^u)sin(u)du]

and now substitute this back:

Int[(e^u)sin(u)du] = sin(u)(e^u) - [cos(u)(e^u) + Int[(e^u)sin(u)du]]

---> Int[(e^u)sin(u)du] = (1/2)(e^u)(sinu - cosu); where u = lnx originally.

Hence, Int[sin(lnx)dx] = (1/2)[xsin(lnx) - xcos(lnx)] + c


hope this helps
{i suspect, however, that there is an easier way than this... so you might have to wait for FinalFantasy, who's usually the expert on integration around here.}

Edit: lol, FinalFantasy already got there before me... should have guessed ; hehe... next time, i'll type faster Final! (and no more double spacing !!!)

 

[FinalFantasy]

hey man. plz don't say dat.. i am no expert on integration and there's many things i can't do :\ i dun deserve to be an expert in any maths topic

 

[~ ReNcH ~]

Well, by the looks of it FinalFantasy's method is slightly shorter, but only by a few lines. Either way you have to apply integration by parts twice, from what I can see.

 

[who_loves_maths]

Originally Posted by FinalFantasy
hey man. plz don't say dat.. i am no expert on integration and there's many things i can't do :\ i dun deserve to be an expert in any maths topic

aww... no need to be modest Final. we all know how good you are at 4u integration. i consider you an expert at it :uhhuh:

 

[who_loves_maths]

Originally Posted by ~ ReNcH ~
Well, by the looks of it FinalFantasy's method is slightly shorter, but only by a few lines. Either way you have to apply integration by parts twice, from what I can see.

no not just 'slightly', FinalFantasy's method is obviously much the simpler and more expedient method here.

 

[wori]

thank you very much for ur methods !!!

I should have waited this morning (around 0:30)