Re: Two little integrals
- From: Lynn Kurtz <kurtzDELETE-THIS@xxxxxxx>
- Date: Sun, 10 Jul 2005 23:10:27 GMT
On 10 Jul 2005 15:52:11 -0700, "novelo" <novelo@xxxxxxxxx> wrote:
>Hello everybody. I'm stucked with two integrals.
>I was trying to solve an integral by trigonometrical substitution:
>
> 1
>Integral of: ------------
> (36-x^2)^(4/2)
>
>This is the integral of one over the square root of ((36 - x^2)^4)
>
Which is just the integral of 1 / (36 - x^2)^2. Why not rewrite it as:
1 / ((6 + x)^2*(6 - x)^2 ) and use partial fractions?
>Well... so I did all the trigonometric substitution and I get the
>following integral:
>
> 1
>--- Integral of: (secant of theta)^3 dtheta
>216
>
>My problem is that I cannot find an easy way to do the integral of (sec
>x)^3. I've tryed every trigonometric property I know, integration by
>parts, etc... and I get cyclic results that bring me nowhere.
>
I haven't checked your calculations, but Int(sec^3(x)) is usually done
by parts twice which gives the integral in terms of itself, from which
you can solve for it. The first time let u = sec(x) and dv =
sec^2(x)dx and the second time let u = tan(x) and dv = sec(x)tan(x)dx
>An the other one that I've tryed for days and I don't get even near to
>the response (by the way I've tryed this one in 4 calculators and none
>was able to do it)
Well, that settles it then eh?
>
>Square root of: [sin(square root of x)] --> This one is not urgent, it
>came outfor curiosity only, but the other one must be solvable, but,
>how!?
>
That one doesn't have a elementary antiderivative.
--Lynn
.
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- Two little integrals
- From: novelo
- Two little integrals
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