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anyone know how to differentiate [(a^2)(dx)]/ [x(sqrt(x^2  a^2)]?
 9 months ago
 9 months ago
anyone know how to differentiate [(a^2)(dx)]/ [x(sqrt(x^2  a^2)]?
 9 months ago
 9 months ago

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hartnnBest ResponseYou've already chosen the best response.2
there's a dx there..so, do u mean to ntegrate ?
 9 months ago

Hidden_TwilightBest ResponseYou've already chosen the best response.1
yup. i tried trigonometric substitution but i would somehow end up with the answer "a" only
 9 months ago

Hidden_TwilightBest ResponseYou've already chosen the best response.1
x = asecu dx = asecutanu sqrt(x^2  a^2) = atanu therefor:
 9 months ago

Hidden_TwilightBest ResponseYou've already chosen the best response.1
dw:1372397190010:dw
 9 months ago

hartnnBest ResponseYou've already chosen the best response.2
hey thats absolutely correct! but don't forget we haven't integrated yet!
 9 months ago

hartnnBest ResponseYou've already chosen the best response.2
\(\huge \int a \: du=..?\)
 9 months ago

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

hartnnBest ResponseYou've already chosen the best response.2
not da, here the variable us 'u' , so 'du'
 9 months ago

Hidden_TwilightBest ResponseYou've already chosen the best response.1
oh. so it becomes au?
 9 months ago

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

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

hartnnBest 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!
 9 months ago

Hidden_TwilightBest ResponseYou've already chosen the best response.1
ahhhh... so it becomes a(sec^1 [x/a]) right?
 9 months ago

hartnnBest ResponseYou've already chosen the best response.2
correct! don't forget the +c part :)
 9 months ago

Hidden_TwilightBest ResponseYou've already chosen the best response.1
wait. does that mean i have to cancel a?
 9 months ago

hartnnBest ResponseYou've already chosen the best response.2
you cannot cancel 'a' the other a is inside the inverse secant function
 9 months ago

hartnnBest ResponseYou've already chosen the best response.2
a(sec^1 [x/a]) +c will be the final answer
 9 months ago

Hidden_TwilightBest ResponseYou've already chosen the best response.1
ohh...got it! thank you :)
 9 months ago

hartnnBest ResponseYou've already chosen the best response.2
welcome ^_^ ask if anymore doubts...
 9 months ago

Hidden_TwilightBest 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 ?
 9 months ago

hartnnBest 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] ....
 9 months ago

Hidden_TwilightBest ResponseYou've already chosen the best response.1
ahhhh....so thats how it is. thanks har!! :D
 9 months ago
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