For what its worth QuickMath gives the result in the attached image.Originally Posted by grozot
This uses Mathematica as its processing engine (and I beleive that denotes the hypergeometric function).
For what its worth QuickMath gives the result in the attached image.Originally Posted by grozot
This uses Mathematica as its processing engine (and I beleive that denotes the hypergeometric function).
Having thought about this integral and Mathematica, I think that M. mightOriginally Posted by grozot
just be a bit to clever for its own good. It does not know that we want
n to be a positive integer.
If we make M. assume the domain of n is it
might well give a simpler answer.
RonL
Now you tell us! Look at the (anti)symmetry of the integrand about theOriginally Posted by grozot
mid point of the interval.
I was looking at this earlier today, but I don't have my notes with me now,
though I thought it was over an interval of length pi or 2pi that the integral
was zero. However memory can be misleading.
RonL
this is what mathematica has to say
In[18]:=
n\[Element]Naturals
Integrate[(Cos[x]^n)*Cos[(n+2)x],x]
Out[18]=
n\[Element]Naturals
Out[19]=
\!\(\(Cos[x]\^\(1 + n\)\ Sin[\((1 + n)\)\ x]\)\/\(1 + n\)\)
Out[19] is 0 over 0=pi/2
this is just trying to understand how all that is true:
In[20]:=
\!\(D[\(Cos[x]\^\(1 + n\)\ Sin[\((1 + n)\)\ x]\)\/\(1 + n\), x]\)
Out[20]=
\!\(Cos[x]\^\(1 + n\)\ Cos[\((1 + n)\)\ x] -
Cos[x]\^n\ Sin[x]\ Sin[\((1 + n)\)\ x]\)
In[21]:=
\!\(Simplify[
Cos[x]\^\(1 + n\)\ Cos[\((1 + n)\)\ x] -
Cos[x]\^n\ Sin[x]\ Sin[\((1 + n)\)\ x]]\)
Out[21]=
\!\(Cos[x]\^n\ Cos[\((2 + n)\)\ x]\)
so i guess i should somehow integrate this
Cos[x]\^\(1 + n\)\ Cos[\((1 + n)\)\ x] -
Cos[x]\^n\ Sin[x]\ Sin[\((1 + n)\)\ x]]