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

### NY Lost Funds

 cyphactor And 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 planesDoes that make sense? Phylo where do I begin to solve Integral[1/(x^2 * sqr(1+x^2))]split fractions?*partial fractions8or trigonometric substitution? excelblue try trig substitutiontry what you think may work, and do itif you seem stuck, try another methodor even try the technique of letting u = the square root part (eg 1/sqrt(1+x^2)) Phylo integration by parts?nah, that was last chapter, it probably won't workI'm thinking trig. sub. but I don't see what to do... lieven try x=tan u Phylo will try: what would make the denominator a quadratic denominatoroh, non-linear?% Integrate[Cotan(x), x] mbot Phylo: (Cotan*x^2)/2 Phylo % Integrate[Cotan(x)^2, x] mbot Phylo: (Cotan*x^3)/3 |Steve| % Integrate[Cot[x],x] mbot |Steve|: Log[Sin[x]] Phylo ... |Steve| Mathematica can't read your mind. Phylo thx, I realized it when I saw the Cotan"*"x^2% Integrate[Cot[x]^2, x] mbot Phylo: -x - Cot[x] thermoplyae cyphactor: Sorry I went idle for so long. That's all correct.And on that note, goodnight Phylo I substitued tan(u) for xnow I have Integrate[1/(Tan(u)^2), u] but I have to convert back to xhmm% Integrate[1/(Tan(u)^2), u] mbot Phylo: -(1/(Tan*u)) Phylo % Integrate[1/(Tan[u]^2), u] mbot Phylo: -u - Cot[u] lieven phylo: I think you've made a mistake in working it out Phylo somewhere% Integrate[1/(x^2*SQR[1+x^2]), x] mbot Phylo: Integrate[1/(x^2*SQR[1 + x^2]), x]