anonymous
  • anonymous
anyone know how to differentiate [(a^2)(dx)]/ [x(sqrt(x^2 - a^2)]?
Differential Equations
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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hartnn
  • hartnn
there's a dx there..so, do u mean to ntegrate ?
hartnn
  • hartnn
*integrate ?
anonymous
  • anonymous
yup. i tried trigonometric substitution but i would somehow end up with the answer "a" only

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More answers

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

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