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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

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?

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...

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 ^_^