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



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hartnn
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there's a dx there..so, do u mean to ntegrate ?

hartnn
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*integrate ?

Hidden_Twilight
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yup. i tried trigonometric substitution but i would somehow end up with the answer "a" only

Hidden_Twilight
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x = asecu
dx = asecutanu
sqrt(x^2  a^2) = atanu
therefor:

Hidden_Twilight
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dw:1372397190010:dw

hartnn
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hey thats absolutely correct! but don't forget we haven't integrated yet!

hartnn
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\(\huge \int a \: du=..?\)

Hidden_Twilight
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da?

hartnn
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oh, and one error,
\(dx = a \sec u\ tanu \: du\)

hartnn
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not da, here the variable us 'u' , so 'du'

Hidden_Twilight
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oh. so it becomes au?

hartnn
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correct :)
au +c
now substitute back for 'u'
x=a sec u
so, u=...?

Hidden_Twilight
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u = asec/x wait. does sec still have u or do i change it to x?

hartnn
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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
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gotthis ?

Hidden_Twilight
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ahhhh... so it becomes
a(sec^1 [x/a]) right?

hartnn
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correct!
don't forget the +c part :)

Hidden_Twilight
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wait. does that mean i have to cancel a?

hartnn
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you cannot cancel 'a'
the other a is inside the inverse secant function

hartnn
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a(sec^1 [x/a]) +c will be the final answer

Hidden_Twilight
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ohh...got it! thank you :)

hartnn
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welcome ^_^
ask if anymore doubts...

Hidden_Twilight
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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
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ohh...differential equations ?
correct, go on...
x/a = sec [(y+c)/a ]
or x = a sec [(y+c)/a] ....

Hidden_Twilight
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ahhhh....so thats how it is. thanks har!! :D

hartnn
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welcome ^_^