Posted by T.Gracken on September 04, 2002 at 17:41:18:
In Reply to: integration help posted by colin on September 04, 2002 at 16:14:33:
: How do I integrate this?
: Can someone explain the technique please?
: I don't need the actual answer, just help with the approach.
: I can do dx/Sqrt(x) fine (2Sqrt(x)) but this seems to be a beast of another nature...
for notation purposes, i will use int[A]dx to mean the integral of A with respect to x.
for lack of anything better to do (and not wanting to think too hard on this), here’s the approach I took...
First, I rationalized the denominator.
so int[ 1/(a + sqrt(x))]dx = int[ (a - sqrt(x)) / (a2 - x) ]dx
then I separated the integrand to get
int[ a/(a2 - x) ]dx - int[ sqrt(x) / (a2 - x) ]dx
The left integral should be straightforward, but the right is not so straightforward.
on the right integral ... int[ sqrt(x) / (a2 - x) ]dx, I used the following substitution:
sqrt(x) = a*sin(u) ........................note: x=a2*sin2(u) ..................another note: don’t forget to calculate dx for substitution purposes.
This (along with the left integral) will eventually lead to
int[ 1/(a + sqrt(x))]dx
= -a*ln |x - a2| - 2a*ln |(a+sqrt(x)) / sqrt(a2-x)| + 2sqrt(x).
You can verify the result using differentiation.
If the trig substitution throws you off, you might want to try integrating the right integral with a two step method... that is, first, let u = sqrt(x), then u2 = x, so du = 2u(du), yadda, yadda, yadda... Then use a trig substitution [that would be u = a*sin(t)]. this will eventually end in the same result, just more back substitution to rewrite the expression in terms of x.
I had about a page (hand-written) of work from the trig substitution to the end result, so don’t be alarmed if the work gets a bit messy.
good luck, and hope that helps.
moreover (where else but math do we use such a word?) I hope I didn’t screw up the “text and tags”!
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