Question

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

Answer 1

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

Answer 2

*integrate ?

Answer 3

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

Answer 4

x = asecu
dx = asecutanu

sqrt(x^2 - a^2) = atanu

therefor:

Answer 5

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

Answer 6

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

Answer 7

da?

Answer 8

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

Answer 9

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

Answer 10

oh. so it becomes au?

Answer 11

correct :)
au +c
now substitute back for 'u'

x=a sec u
so, u=...?

Answer 12

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

Answer 13

u = asec/x ......no
\(\large x = a \sec u \\ \large (x/a)=\sec u \\ \large u = \sec^{-1}(x/a)\)
the inverse secant function!

Answer 14

gotthis ?

Answer 15

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

Answer 16

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

Answer 17

wait. does that mean i have to cancel a?

Answer 18

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

Answer 19

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

Answer 20

ohh...got it! thank you :)

Answer 21

welcome ^_^
ask if anymore doubts...

Answer 22

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 ?

Answer 23

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

Answer 24

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

Answer 25

welcome ^_^