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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Hidden_Twilight
 one year ago
Best ResponseYou've already chosen the best response.1btw. 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 ?

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

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