#math - Tue 20 Feb 2007 between 04:27 and 04:55

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cyphactorAnd if I am understanding this correctly it is a vector which is perpendicular to the two planes because I am taking the normal vectors of the two planes and cross-producting them which gives me another vector which is perpendicular (normal) to the original two planes normal vectors. Hence, when I use that resulting normal from the cross-product I get a plane which is perpendicular to that normal. Hence, perpendicular two the two original planes
Does that make sense?
Phylowhere do I begin to solve Integral[1/(x^2 * sqr(1+x^2))]
split fractions?
*partial fractions8
or trigonometric substitution?
excelbluetry trig substitution
try what you think may work, and do it
if you seem stuck, try another method
or even try the technique of letting u = the square root part (eg 1/sqrt(1+x^2))
Phylointegration by parts?
nah, that was last chapter, it probably won't work
I'm thinking trig. sub. but I don't see what to do...
lieventry x=tan u
Phylowill try: what would make the denominator a quadratic denominator
oh, non-linear?
% Integrate[Cotan(x), x]
mbotPhylo: (Cotan*x^2)/2
Phylo% Integrate[Cotan(x)^2, x]
mbotPhylo: (Cotan*x^3)/3
|Steve|% Integrate[Cot[x],x]
mbot|Steve|: Log[Sin[x]]
Phylo...
|Steve|Mathematica can't read your mind.
Phylothx, I realized it when I saw the Cotan"*"x^2
% Integrate[Cot[x]^2, x]
mbotPhylo: -x - Cot[x]
thermoplyaecyphactor: Sorry I went idle for so long. That's all correct.
And on that note, goodnight
PhyloI substitued tan(u) for x
now I have Integrate[1/(Tan(u)^2), u] but I have to convert back to x
hmm
% Integrate[1/(Tan(u)^2), u]
mbotPhylo: -(1/(Tan*u))
Phylo% Integrate[1/(Tan[u]^2), u]
mbotPhylo: -u - Cot[u]
lievenphylo: I think you've made a mistake in working it out
Phylosomewhere
% Integrate[1/(x^2*SQR[1+x^2]), x]
mbotPhylo: Integrate[1/(x^2*SQR[1 + x^2]), x]

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