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MathGroup Archive 2007

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Re: Indefinite Integral

  • To: mathgroup at smc.vnet.net
  • Subject: [mg77558] Re: Indefinite Integral
  • From: dimitris <dimmechan at yahoo.com>
  • Date: Wed, 13 Jun 2007 07:21:53 -0400 (EDT)
  • References: <f4j0ij$j7n$1@smc.vnet.net>

Hello Joel.

I use Mathematica 5.2 in the following.

Here is the integrand.

In[1]:=
int[a_, b_, p_, x_] := Exp[(-p)*x]*(Sin[x]/(a^2*Sin[x]^2 +
b^2*Cos[x]^2)^(3/2))

As you may notice (otherwise you would not have posted the query!)
Mathematica returns the indefinite integral unevaluated.

Here are two subcases of the parameters that Mathematica
returns an antiderivative.

In[17]:=
Integrate[int[a, a, p, x], x]
Out[17]=
-((Cos[x] + p*Sin[x])/(E^(p*x)*((a^2)^(3/2)*(1 + p^2))))

In[18]:=
Integrate[int[a, b, 0, x], x]
Out[18]=
-((Sqrt[2]*Cos[x])/(a^2*Sqrt[a^2 + b^2 + (-a^2 + b^2)*Cos[2*x]]))

But I am sure you have also checked this.

When Mathematica does not succeed, we should not give up
but trying to help it!
Here (I assume that you have consult the well known collection
of tables in literature; I didn't have time to search them for your
integral) I
tried various approaches as substitutions, integration by parts etc
but with
unsatisfactory results.

For the end I left one approach which you should always have in mind
for parametric integrals like yours!

This approach consists of the following steps:

Take the indefinite integral with respect to a and b.

In[27]:=
Int=Integrate[int[a, b, p, x], a, b]
Out[27]=
-((Sqrt[a^2 + b^2 + (-a^2 + b^2)*Cos[2*x]]*Csc[x]*Sec[x]^2)/
(E^(p*x)*(Sqrt[2]*a*b)))

(*Check*)

In[34]:=
Simplify[D[%, a, b] == int[a, b, p, x]]
Out[34]=
True

Now take the antiderivative of Int wrt x.

In[95]:=
Integrate[Int, x]

Out[95]=
-((1/(Sqrt[2]*a*b))*((Sqrt[a^2 + b^2 + (-a^2 + b^2)*Cos[2*x]]*Sec[x])/
E^(p*x) +
    (3*a^2 + b^2)*p*Integrate`TableMatchTrig1[((-(1/2))*I*Csc[2*x])/
E^((-I + p)*x), x, 0] +
    4*b^2*Integrate`TableMatchTrig1[((1/2)*Csc[2*x])/E^((-I + p)*x),
x, 0] +
    3*a^2*p*Integrate`TableMatchTrig1[((1/2)*I*Csc[2*x])/E^((I +
p)*x), x, 0] +
    b^2*p*Integrate`TableMatchTrig1[((1/2)*I*Csc[2*x])/E^((I + p)*x),
x, 0] +
    4*b^2*Integrate`TableMatchTrig1[((1/2)*Csc[2*x])/E^((I + p)*x), x,
0] -
    a^2*p*Integrate`TableMatchTrig1[((-(1/2))*I*Csc[2*x])/E^((-3*I +
p)*x), x, 0] +
    b^2*p*Integrate`TableMatchTrig1[((-(1/2))*I*Csc[2*x])/E^((-3*I +
p)*x), x, 0] -
    a^2*p*Integrate`TableMatchTrig1[((1/2)*I*Csc[2*x])/E^((3*I +
p)*x), x, 0] +
    b^2*p*Integrate`TableMatchTrig1[((1/2)*I*Csc[2*x])/E^((3*I +
p)*x), x, 0]))

Not exactly a readable result; but still a result!
I am wondering what version 6 returns here!

Take now the derivatives wrt to a,b of the last expression and you
get

In[135]:=
res=FullSimplify[D[%, a, b]] (*wait a couple of minutes for the
simplification!*)

Out[135]=
(1/Sqrt[2])*((4*Cos[x]*Sin[x]^2)/(E^(p*x)*(a^2 + b^2 + (-a^2 +
b^2)*Cos[2*x])^(3/2)) +
   (1/a^2 + 3/b^2)*p*Integrate`TableMatchTrig1[((-(1/2))*I*Csc[2*x])/
E^((-I + p)*x), x, 0] +
   (4*Integrate`TableMatchTrig1[((1/2)*Csc[2*x])/E^((-I + p)*x), x,
0])/a^2 +
   (1/(a^2*b^2))*((3*a^2 +
b^2)*p*Integrate`TableMatchTrig1[((1/2)*I*Csc[2*x])/E^((I + p)*x), x,
0] +
     4*b^2*Integrate`TableMatchTrig1[((1/2)*Csc[2*x])/E^((I + p)*x),
x, 0] -
     (a - b)*(a + b)*p*(Integrate`TableMatchTrig1[((-
(1/2))*I*Csc[2*x])/E^((-3*I + p)*x), x, 0] +
       Integrate`TableMatchTrig1[((1/2)*I*Csc[2*x])/E^((3*I + p)*x),
x, 0])))

And this is the desired result! A RESULT? This mesh?
Now if I could see your face I am sure that you would look quite
curious
about how someone could use this result!
You asked for a closed form solution in Mathematica!
That is, what Mathematica (in collaboration with mind...) returns!

I sent my reply to Daniel Lichtblau of WRI for more details.

Best Regards
Dimitris




 /  Joel Storch       :
> I would like a closed-form expression for
>
> Integrate[Exp[-p x] Sin[x]/(a^2 Sin[x]^2+b^2 Cos[x]^2)^(3/2),x]
>
> where a,b & p are positive constants



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