Re: trig substitution
- To: mathgroup at smc.vnet.net
- Subject: [mg45009] Re: trig substitution
- From: drbob at bigfoot.com (Bobby R. Treat)
- Date: Fri, 12 Dec 2003 04:42:49 -0500 (EST)
- References: <br41ju$iuf$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Here's a way to reason toward that integral.
We know (or should know) the following integral:
Integrate[1/(1 + x^2), x]
ArcTan[x]
Your integrand is the square of that one, of course. But if we
subtract your integrand minus a constant c times the other one, we get
this:
Together[1/(1 + x^2)^2 - c/(1 + x^2)]
(1 - c - c*x^2)/(1 + x^2)^2
The constant part of the numerator gets us back to the original
problem, but the other part gives a term like
(c*x^2)/(1 + x^2)^2
If we can integrate this, we can integrate the original problem. But
how?
We expect (or should expect) the integral to be something like
(a*x)/(1 + x^2)
(...from thinking about what happens when we differentiate a fraction.
The denominator gets squared, and the numerator is 2x times the
numerator, so it's no problem to get a squared term up top if we have
a x up there already. There's another term, of course, so the
derivative is actually a polynomial like the one above.)
So let's just solve for the constants!
s = SolveAlways[
1/(1 + x^2)^2 ==
D[(a*x + b)/(1 + x^2),
x] + c/(1 + x^2), x]
{{b -> 0, a -> 1/2, c -> 1/2}}
It follows that the original integrand is equal to
D[(a*x)/(1 + x^2), x] +
c/(1 + x^2) /. First[s]
Simplify[%]
-(x^2/(1 + x^2)^2) + 1/(1 + x^2)
1/(1 + x^2)^2
and hence the integral is
(a*x)/(1 + x^2) + Integrate[
c/(1 + x^2), x] /.
First[s]
x/(2*(1 + x^2)) + ArcTan[x]/2
And that's Mathematica's answer, of course.
Integrate[1/(1 + x^2)^2, x]
(1/2)*(x/(1 + x^2) + ArcTan[x])
Bobby
"Tom Radomski" <gtg852q at mail.gatech.edu> wrote in message news:<br41ju$iuf$1 at smc.vnet.net>...
> I have a problem that I'm having a lot of problems trying to solve. I found
> the answer to it on the online integrator but I don't understand the work
> behind it. I someone could explain the math behind it that would be greatly
> appreciated.
> Question: 1/(1+x^2)^2
> ans: [x/(2(1+x^2))]+[(arctan x)/2]
> Thanks,
> Tom