anonymous
  • anonymous
what is the intergral of 1/1 +x^2 dy
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I think it is( x^3)/(3)+c
amistre64
  • amistre64
1/1 = 1
amistre64
  • amistre64
but that would be to obvious to be right :)

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amistre64
  • amistre64
and dy? instead of dx? is that a typo....
amistre64
  • amistre64
(S) 1/(1+x^2) dx perhaps...
amistre64
  • amistre64
1/(2x) (S) 2x/(1+x^2) dx ....
amistre64
  • amistre64
ln(1+x^2) --------- (2x) I wish I knew if I am doing this right :)
anonymous
  • anonymous
Yes, it's ∫1/(1+x^2)dx. Turns out to be arctan(x), I believe...it's on the tables too.
amistre64
  • amistre64
DOH!!.... yeah.....
amistre64
  • amistre64
and I messed p taking that "x" out of it.... should left it in if I was going to do what I was thinking of doing :)
amistre64
  • amistre64
arctan(x)... 1/sqrt(x^2+1)?? great, i just confused myself lol
anonymous
  • anonymous
let consider u = 1+x^2 then du = 2x dx and dx = 1/2x du substitute to the integral S u/1/2x du =1/2x.1/2(1+x^2)^2 =1/4x ( 1+ X^2 )^2 would be the answer....
anonymous
  • anonymous
quantum that answer is correct but how did you work it? bysetting it up 1/2 tan^-1 +1 = 2x^2 + 3/( x^2 + 1)^2 dx
amistre64
  • amistre64
(S) 1/(1+tan^2(x)) dx ??
amistre64
  • amistre64
wanna take lazy calc, real calc, and stats over the summer :)
anonymous
  • anonymous
Okay, here it is:\[\arctan(x)=y\]\[\tan(y)=x\]\[dx/dy = \sec^2(x) = tan^2(x)+1\] We want dy/dx, not dx/dy, so invert it:\[dy/dx = \frac{1}{\tan^2(y)+1}\] and because \[\tan^2(y) = x^2\] then\[dy/dx = \frac{1}{1+x^2.}\] Now we're just integrating it, instead of differentiating the opposite.
anonymous
  • anonymous
why do(S) 1/(1+tan^2(x)) dx would be???
amistre64
  • amistre64
yeah :)
anonymous
  • anonymous
if u work like that...thats brilliant idea quantum,,,but we still need to diferentiate 1 and 1+x^2 to solve the problem...
amistre64
  • amistre64
suppose I shoulda changed the variable :)
anonymous
  • anonymous
and i think the answer is 2x/(1+x^2)^2 based on Ur way...
anonymous
  • anonymous
Sorry, in the third line I should have said sec^2(y) = tan^2(y)+1, sorry if that confused anyone.
amistre64
  • amistre64
no more confused than usual :)
anonymous
  • anonymous
We established with that, that the derivative of arctan(x) = 1/(1+x^2). When we integrate the derivative, we're going to get back to arctan(x).
anonymous
  • anonymous
then tell me the idea that u're supposing tan(y)=x?
amistre64
  • amistre64
the books call it "trig" substitution... if i recall correctly
anonymous
  • anonymous
It is trig substitution. :P
anonymous
  • anonymous
the book of Calculus?...
amistre64
  • amistre64
yep; all the ones from Newton on down have had it :)
anonymous
  • anonymous
@tianpradana: If tan^-1(x) = y, then x = tan(y).
amistre64
  • amistre64
Leibnez even considered it..maybe :)
anonymous
  • anonymous
oh cheh thanks :-)
amistre64
  • amistre64
tends to be around, integration by parts, u substitutions..trig subs.... inthat general area :)
anonymous
  • anonymous
brilliant idea guys...
amistre64
  • amistre64
its the area I have been avoinding like the plague....

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