# Thread: just an integral :)

1. ## just an integral :)

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

2. 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).

3. thanks, mathematica gave that result also, but i still have no idea how to do that on my exam

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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]]

10. 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

11. sorry, i was in a hurry every time i posted something here
so i made mistakes...what can i say...something happens...