Posted by Soroban on August 15, 2002 at 23:21:06:
In Reply to: Have you learnt about trigonometric substitutions - yet? (n/t) posted by Subhotosh Khan on August 15, 2002 at 08:46:27:
: : Can anyone help me with both an answer AND an explanation.
: : INTEGRATE: 1 / x((x^2 - 4)^1/2)
: : and
: : INTEGRATE: 1 / (x^1/2)(1 - 2(x^1/2))
: : Thank you for taking the time.
: *********************************
: These problems - GENERALLY - become easier to deal with after trigonometric substitution:
: INT[1 / x((x^2 - 4)^1/2)] dx
: substitute:
: x = 2*sec T ...................(1)
: dx = 2*sec T * tan T dT
: INT[1 / x((x^2 - 4)^1/2)] dx
: = INT[1/(2*sec T) * ((4 sec^2 T - 4)^1/2)]*2*sec T * tan T dT
: = INT[1/(2*sec T) * ( 2*tan T)]*2*sec T * tan T dT
: = INT[ 2*tan^2 T] dT
Sorry, Subhotosh... your own fractions confused you.
It cancels down to: (1/2)INT dT = (1/2)T + C
Since x = 2 sec T, then T = arcsec(T/2).
So the answer is (1/2)arcsec(T/2) + C
By the way, there IS a recognizable formula for the arcsecant form, with which it could
have been solved in one step.
: The second one - does not need trigonometric sustitution. You can reduce it algebraically ---
: 1 / (x^1/2) (1 - 2(x^1/2))
: = x^(-1/2) - 2
: Now this should be easy....
Once again, a thousand pardons!
The denominator is: x^(1/2) - 2x, which requires some work.
I saw it as du/u: u = 1 - 2x^(1/2)
then du = -x^(-1/2) dx [which we have!]
The problem becomes: -INT du/u = -ln|u| + C
I apologize again for pointing out your errors, Subhotosh. I'm sure you saw them at the
very instant you clicked on SUBMIT and (like me) wished we could un-post our writings.
I've read your responses, and I like and respect your tips, hints, and explanations. (I am
definitely NOT a Doctor You-know-who.) And thank you for your kind words at the other site.