Re: automatic integral output simplification

*To*: mathgroup at smc.vnet.net*Subject*: [mg115192] Re: automatic integral output simplification*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Tue, 4 Jan 2011 04:23:19 -0500 (EST)

The Sin[t]^2 form appears because the coefficient of t in Cos[2 t] is EXACTLY 2, so the following results are to be expected: Integrate[-8 Sin[2 t], {t, 0, t}] Integrate[-8 Sin[2. t], {t, 0, t}] Integrate[-8 Sin[w t], {t, 0, t}] Integrate[-8 Sin[w t], {t, 0, t}] /. w -> 2 Integrate[-8 Sin[w t], {t, 0, t}] /. w -> 2 // Expand -8 Sin[t]^2 -4. + 4. Cos[2. t] (8 (-1 + Cos[t w]))/w 4 (-1 + Cos[2 t]) -4 + 4 Cos[2 t] Bobby On Mon, 03 Jan 2011 02:57:03 -0600, sean k <seaninsocal at gmail.com> wrote: > Hello group, > > I want to get around the way Mathematica simplifies the output of an > integral. > > For instance, Let's say you want to integrate -8 sin(2t) from t = {0, > t} > > int1 = \[Integral]-8 Sin[2 t] \[DifferentialD]t > ul1 = int1; > ll1 = int1 /. t -> 0; > ul1 - ll1 > > yields the correct output. > > -4 + 4 Cos[2 t] > > But if i pose that integral with limits included in it, Mathematica > outputs > > Integrate[-8 Sin[2 t], {t, 0, t}] > > -8 Sin[t]^2 > > Now both these are correct, except when the definite integral is > evaluated, "mathematicamagic" kicks in. > > My guess is that the coefficient in the argument for the sin function > is the culprit. What should I do here to make the Mathematica retain > the "2 t" inside the Sin? > > Thanks in advance for any insights. > > Sean > > > > > -- DrMajorBob at yahoo.com