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

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

Differential Equations
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there's a dx there..so, do u mean to ntegrate ?
*integrate ?
yup. i tried trigonometric substitution but i would somehow end up with the answer "a" only

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x = asecu dx = asecutanu sqrt(x^2 - a^2) = atanu therefor:
|dw:1372397190010:dw|
hey thats absolutely correct! but don't forget we haven't integrated yet!
\(\huge \int a \: du=..?\)
da?
oh, and one error, \(dx = a \sec u\ tanu \: du\)
not da, here the variable us 'u' , so 'du'
oh. so it becomes au?
correct :) au +c now substitute back for 'u' x=a sec u so, u=...?
u = asec/x wait. does sec still have u or do i change it to x?
u = asec/x ......no \(\large x = a \sec u \\ \large (x/a)=\sec u \\ \large u = \sec^{-1}(x/a)\) the inverse secant function!
gotthis ?
ahhhh... so it becomes a(sec^-1 [x/a]) right?
correct! don't forget the +c part :)
wait. does that mean i have to cancel a?
you cannot cancel 'a' the other a is inside the inverse secant function
a(sec^-1 [x/a]) +c will be the final answer
ohh...got it! thank you :)
welcome ^_^ ask if anymore doubts...
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 ?
ohh...differential equations ? correct, go on... x/a = sec [(y+c)/a ] or x = a sec [(y+c)/a] ....
ahhhh....so thats how it is. thanks har!! :D
welcome ^_^

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