Trigonometric Integration Using Recursion Formula

[Harmony]

Homework Statement

Integrate cos^n x cos nx

Homework Equations

Integration by part equations, trigonometric integrals

The Attempt at a Solution

I was given the hint that this integration involves integration by part and trigonometry integrals. I tried integration by part, by assigning v=cos^n x and du/dx=cos nx, but that failed. I thought of converting the cos nx, so that the term cos x may appear and enables me to integrate cos^n x, but so far all my attempt have failed.

 

[Gib Z]

Well, why not try to express cos (nx) in terms of of cos(x) and sin(x). To do that, use Eulers Identity: e^{ix} = \cos x + i \sin x, so that \cos (nx) + i\sin (nx) = (\cos x + i \sin x)^n, expand the RHS using the binomial theorem and separate real and imaginary coefficients.

 

[VietDao29]

Homework Statement

Integrate cos^n x cos nx

Homework Equations

Integration by part equations, trigonometric integrals

The Attempt at a Solution

I was given the hint that this integration involves integration by part and trigonometry integrals. I tried integration by part, by assigning v=cos^n x and du/dx=cos nx, but that failed. I thought of converting the cos nx, so that the term cos x may appear and enables me to integrate cos^n x, but so far all my attempt have failed.

Apart from Gib Z's approach, I have the feeling that you can also do it via Integration by Parts (twice, I think). Perhaps, you wouldn't mind showing us your work, and where you got stuck, so that we can help, or check the steps for you, would you? :)

 

[Gib Z]

Try VietDao29's solution, much simpler, should try that first =] Mine takes ages :(

 

[Harmony]

∫ cos^n x cos nx = (sin nx cos^n x)/n + ∫ sin nx sin x cos^(n-1) x dx

I got stuck here. Converting sin nx sin x into [cos (n-1)x - cos (n+1)x]/2 didn't help.

 

[VietDao29]

Nah, just continue Integrating by Parts by choosing u = sin(x) cos^{n - 1}(x), and dv = sin(nx) dx (later on, you'll find that this is a very common method to solve many Integration by Parts problems).

Just try it, and see if you get stuck any more. :)

 

[Harmony]

I integrate again as you advised, and eventually this term appear:
...(I skip the part where there is no integrals)...-1/n ∫ 2 cos^n x cos nx - cos^(n-2) x dx

How do I get rid of cos^(n-2) x ?

 

[VietDao29]

I integrate again as you advised, and eventually this term appear:
...(I skip the part where there is no integrals)...-1/n ∫ 2 cos^n x cos nx - cos^(n-2) x dx

How do I get rid of cos^(n-2) x ?

You will try to find a Recursion Formula for it.

Say, let:

I_{\alpha , \beta} = \int \cos ^ \alpha (x) \cos (\beta x) dx

You seem to have forget some brackets, and some constants in the result you provided..

You'll eventually end up with:

I_{n, n} = \mbox{something} + \mbox{something} \times I_{n, n} + \mbox{something} \times I_{n - 2, n}

After some isolation, and manipulations, you'll find the relation between I_{n, n}, and I_{n - 2, n}.

Besides, it's easy to calculate:
I_{0, n} = \int \cos(nx) dx
and:
I_{1, n} = \int \cos(x) \cos(nx) dx, right?

Now, say, you need to calculate: I_{7, 7}, you start from I_{1, 7} (which is easy to calculate, eh?), then, by using the relation between I_{n, n}, and I_{n - 2, n}, you'll be able to find I_{3, 7}, do the same, you'll get I_{5, 7}, and finally, I_{7, 7}, as desired.

If you want to calculate I_{8, 8}, you'll start from I_{0, 8}..

So, every I_{n, n}'ll eventually boil down to either I_{1, n}, or I_{0, n}. This is how Recursion Formula works.

You your final answer will look something like:

\left \{ \begin{array}{l} I_{n, n} = \mbox{something } I_{n - 2, n} \\ I_{0, n} = .. \\ I_{1, n} = ... \end{array} \right.

Is it clear? :)