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

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1. hartnn

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

2. hartnn

*integrate ?

3. Hidden_Twilight

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

4. Hidden_Twilight

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

5. Hidden_Twilight

|dw:1372397190010:dw|

6. hartnn

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

7. hartnn

$$\huge \int a \: du=..?$$

8. Hidden_Twilight

da?

9. hartnn

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

10. hartnn

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

11. Hidden_Twilight

oh. so it becomes au?

12. hartnn

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

13. Hidden_Twilight

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

14. 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!

15. hartnn

gotthis ?

16. Hidden_Twilight

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

17. hartnn

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

18. Hidden_Twilight

wait. does that mean i have to cancel a?

19. hartnn

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

20. hartnn

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

21. Hidden_Twilight

ohh...got it! thank you :)

22. hartnn

welcome ^_^ ask if anymore doubts...

23. Hidden_Twilight

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 ?

24. hartnn

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

25. Hidden_Twilight

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

26. hartnn

welcome ^_^