#math - Wed 16 May 2007 between 00:40 and 00:54

andrew123can anyone integrate using substitution: sin[(sqrt)x] / (sqrt)x
Safroleu = sqrt(x), so du = 1/(2sqrt(x)) dx
Copter2looks like a sin there though :o
Safrolethus we get int( sin(sqrt(x))/sqrt(x) dx ) = 2*int( sin(u) du )
= -2cos(u) = -2cos(sqrt(x))
TheBlueWizardequivalently, x = u^2, dx = 2 u du, and then substituting is pretty straightforward
[n01d]hi guys, whats the function/algorithm to calcalculate sine?
andrew123safrole: if u = sqrt(x) wudnt we get sinu/u du
Safroleif u = sqrt(x) then du = 1/(2sqrt(x))
du = 1/(2sqrt(x)) dx, sorry
so then 2 du = 1/sqrt(x) dx
so then we get int( sin(sqrt(x))/sqrt(x) dx ) = int( sin(u) du )
= 2*int( sin(u) du )
I forgot the 2
andrew123but we are integrating in regards to u.. so dont we need du/dx = 1/2sqrt(x) then dx = 2sqrt(x).du
Safroleandrew you're mixing things here that need nod be mixed
we want a substitution for 1/sqrt(x) dx
I found one
bsmntbombdoodis >, < defined on C?
Safrole< . > you mean?
An inner product?
[n01d]can someone help me whats the manual to solve sine. not using calculator
bsmntbombdoodno, greater/less than
SafroleC is *not* an ordered field
Eclipsor[n01d]: ?
Safroleyou can impose a so called lexicographical order
Eclipsorsolve sine in what way
Safrolebut that still has some nuances
But in general, no > and < are not defined for C
shinygerbil[n01d]: do you mean like the Taylor series of sin(x)?
Safrolethe Taylor series would require knowledge of cosine
shinygerbilhehe, true
Safrolethat's kind of circular.
bsmntbombdoodSafrole: no it wouldn't
shinygerbilno, wait
well, kinda
bsmntbombdoodcos(0) is easy to find
shinygerbilactually doing a Taylor expansion on sin(x) would

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