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Hidden_Twilight
Group Title
anyone know how to differentiate [(a^2)(dx)]/ [x(sqrt(x^2  a^2)]?
 one year ago
 one year ago
Hidden_Twilight Group Title
anyone know how to differentiate [(a^2)(dx)]/ [x(sqrt(x^2  a^2)]?
 one year ago
 one year ago

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hartnn Group TitleBest ResponseYou've already chosen the best response.2
there's a dx there..so, do u mean to ntegrate ?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
*integrate ?
 one year ago

Hidden_Twilight Group TitleBest ResponseYou've already chosen the best response.1
yup. i tried trigonometric substitution but i would somehow end up with the answer "a" only
 one year ago

Hidden_Twilight Group TitleBest ResponseYou've already chosen the best response.1
x = asecu dx = asecutanu sqrt(x^2  a^2) = atanu therefor:
 one year ago

Hidden_Twilight Group TitleBest ResponseYou've already chosen the best response.1
dw:1372397190010:dw
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
hey thats absolutely correct! but don't forget we haven't integrated yet!
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
\(\huge \int a \: du=..?\)
 one year ago

Hidden_Twilight Group TitleBest ResponseYou've already chosen the best response.1
da?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
oh, and one error, \(dx = a \sec u\ tanu \: du\)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
not da, here the variable us 'u' , so 'du'
 one year ago

Hidden_Twilight Group TitleBest ResponseYou've already chosen the best response.1
oh. so it becomes au?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
correct :) au +c now substitute back for 'u' x=a sec u so, u=...?
 one year ago

Hidden_Twilight Group TitleBest ResponseYou've already chosen the best response.1
u = asec/x wait. does sec still have u or do i change it to x?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
u = asec/x ......no \(\large x = a \sec u \\ \large (x/a)=\sec u \\ \large u = \sec^{1}(x/a)\) the inverse secant function!
 one year ago

Hidden_Twilight Group TitleBest ResponseYou've already chosen the best response.1
ahhhh... so it becomes a(sec^1 [x/a]) right?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
correct! don't forget the +c part :)
 one year ago

Hidden_Twilight Group TitleBest ResponseYou've already chosen the best response.1
wait. does that mean i have to cancel a?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
you cannot cancel 'a' the other a is inside the inverse secant function
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
a(sec^1 [x/a]) +c will be the final answer
 one year ago

Hidden_Twilight Group TitleBest ResponseYou've already chosen the best response.1
ohh...got it! thank you :)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
welcome ^_^ ask if anymore doubts...
 one year ago

Hidden_Twilight Group TitleBest ResponseYou've already chosen the best response.1
btw. what if the other sie has y? a(sec^1 [x/a]) = y (y was already integrated) a(sec^1 [x/a]) = y + C (sec^1 [x/a]) =( y + C)/a ?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
ohh...differential equations ? correct, go on... x/a = sec [(y+c)/a ] or x = a sec [(y+c)/a] ....
 one year ago

Hidden_Twilight Group TitleBest ResponseYou've already chosen the best response.1
ahhhh....so thats how it is. thanks har!! :D
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
welcome ^_^
 one year ago
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