Can anyone integrate this?

Integral ((Cos[x])^n)*Cos[(n+1)*x] dx

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- Jan 24th 2006, 01:23 PMgrozotjust an integral :)
Can anyone integrate this?

Integral ((Cos[x])^n)*Cos[(n+1)*x] dx - Jan 24th 2006, 11:49 PMCaptainBlackQuote:

Originally Posted by**grozot**

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

- Jan 25th 2006, 02:40 AMCaptainBlackQuote:

Originally Posted by**grozot**

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 AMCaptainBlackQuote:

Originally 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 - Jan 25th 2006, 12:23 PMgrozot
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 PMCaptainBlackQuote:

Originally 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 - Jan 25th 2006, 12:41 PMCaptainBlackQuote:

Originally Posted by**CaptainBlack**

non-zero over 0-pi/2, but zero over 0-pi.

RonL - Jan 26th 2006, 06:23 AMgrozot
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 AMCaptainBlackQuote:

Originally Posted by**grozot**

RonL - Jan 26th 2006, 09:01 AMgrozot
sorry, i was in a hurry every time i posted something here

so i made mistakes...what can i say...something happens...