       Re: can not understand the symbol or the equation for an integration result

• To: mathgroup at smc.vnet.net
• Subject: [mg122253] Re: can not understand the symbol or the equation for an integration result
• From: dimitris <dimmechan at yahoo.com>
• Date: Sat, 22 Oct 2011 06:08:20 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <j7rhjp\$hel\$1@smc.vnet.net>

```On Oct 21, 1:30 pm, Jing <jing.gu... at yahoo.com> wrote:
> Hi,
>
> I am trying to integrate a equation,
> 1/72 (8 Sqrt r^3 -
>    Sqrt r^2 Sqrt[r^2 + 2 y (y - Sqrt Sqrt[r^2 - y^2])] -
>    2 Sqrt y^2 Sqrt[r^2 + 2 y (y - Sqrt Sqrt[r^2 - y^2])] +
>    6 y Sqrt[(r^2 - y^2) (r^2 + 2 y (y - Sqrt Sqrt[r^2 - y^2]))])
>
> But the result is kind of confusing.First, the variable y is changed to another variable y -> -r Sin[t].
> exp1 = Simplify[1/72 (8 Sqrt r^3 -
>    Sqrt r^2 Sqrt[r^2 + 2 y (y - Sqrt Sqrt[r^2 - y^2])] -
>    2 Sqrt y^2 Sqrt[r^2 + 2 y (y - Sqrt Sqrt[r^2 - y^2])] +
>    6 y Sqrt[(r^2 - y^2) (r^2 + 2 y (y - Sqrt Sqrt[r^2 - y^2]))])
>  /. {y -> -r Sin[t]}, r > 0 && 0 < t < Pi/6]
>
> result:-(1/72) r^3 (3 Sin[2 t] Sqrt[2 - Cos[2 t] + Sqrt Sin[2 t]] +
>    Sqrt (-8 + Sqrt[1 + 2 Sin[t]^2 + Sqrt Sin[2 t]] +
>       2 Sin[t]^2 Sqrt[1 + 2 Sin[t]^2 + Sqrt Sin[2 t]]))
>
> Then,do integration :
> m19 = Integrate[-r Cos[t] exp1, t, Assumptions -> {r > 0}]
> final result:
> 1/72 r^4 (-8 Sqrt Sin[t] + (3 t (Sqrt Cos[t] + 3 Sin[t]))/(
>    2 Sqrt[2 - Cos[2 t] + Sqrt Sin[2 t]]) - (
>    Sqrt[2 - Cos[2 t] +
>      Sqrt Sin[2 t]] (6 (I + Sqrt) Cos[t] +
>       3 (I + Sqrt) Cos[
>         3 t] + (3 + I Sqrt) (-8 Sin[t] + Sin[3 t])))/(
>    8 (I + Sqrt)))
>
> I don't know what the I is. Is it the imaginary number? Why it can appear?
>
> Thanks

Yes is the imaginary unit.
First of all Mathematica does not evaluate integrals like a human.
It uses advanced algorithms specific for symbolic integration.

But I don't understand what did you expect to appear in the result?

Dimitris

```

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