# just an integral :)

• Jan 24th 2006, 01:23 PM
grozot
just an integral :)
Can anyone integrate this?
Integral ((Cos[x])^n)*Cos[(n+1)*x] dx
• Jan 24th 2006, 11:49 PM
CaptainBlack
Quote:

Originally Posted by grozot
Can anyone integrate this?
Integral ((Cos[x])^n)*Cos[(n+1)*x] dx

For what its worth QuickMath gives the result in the attached image.

This uses Mathematica as its processing engine (and I beleive that $_2F_1$ denotes the hypergeometric function).
• Jan 25th 2006, 02:35 AM
grozot
thanks, mathematica gave that result also, but i still have no idea how to do that on my exam
• Jan 25th 2006, 02:40 AM
CaptainBlack
Quote:

Originally Posted by grozot
thanks, mathematica gave that result also, but i still have no idea how to do that on my exam

Mathematica does not always give the most compact/simple form
for integrals, but the fact that there has been no substantive replys
to your post suggests that this is not a well known or simple integral

RonL
• Jan 25th 2006, 09:18 AM
CaptainBlack
Quote:

Originally Posted by grozot
thanks, mathematica gave that result also, but i still have no idea how to do that on my exam

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 $\mathbb{Z}_+$ it
might well give a simpler answer.

RonL
• Jan 25th 2006, 12:23 PM
grozot
i realized i didnt tell you that integral is from 0 to Pi/2.
and i dont know how to tell m that n is positive integer but that shoud be easy so ill figure it out.
and i have no idea why but the result is 0
• Jan 25th 2006, 12:35 PM
CaptainBlack
Quote:

Originally Posted by grozot
i realized i didnt tell you that integral is from 0 to Pi/2.
and i dont know how to tell m that n is positive integer but that shoud be easy so ill figure it out.
and i have no idea why but the result is 0

Now you tell us! ;) Look at the (anti)symmetry of the integrand about the
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
• Jan 25th 2006, 12:41 PM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
Now you tell us! ;) Look at the (anti)symmetry of the integrand about the
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

Just checked some special cases (n=2 and 3), and the integral is
non-zero over 0-pi/2, but zero over 0-pi.

RonL
• Jan 26th 2006, 06:23 AM
grozot
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]]
• Jan 26th 2006, 07:45 AM
CaptainBlack
Quote:

Originally Posted by grozot
this is what mathematica has to say
In[18]:=
n\[Element]Naturals
Integrate[(Cos[x]^n)*Cos[(n+2)x],x]

Cos[(n+2)x] ???? I thought it was Cos[(n+1)x].

RonL
• Jan 26th 2006, 09:01 AM
grozot
sorry, i was in a hurry every time i posted something here
so i made mistakes...what can i say...something happens...