Trigonometric Integration Using Recursion Formula

In summary, the Recursion Formula for integrating cos(nx) is: I_{n, n} = \mbox{something} I_{n - 2, n}
  • #1
Harmony
203
0

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.
 
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  • #2
Well, why not try to express cos (nx) in terms of of cos(x) and sin(x). To do that, use Eulers Identity: [tex]e^{ix} = \cos x + i \sin x[/tex], so that [tex]\cos (nx) + i\sin (nx) = (\cos x + i \sin x)^n[/tex], expand the RHS using the binomial theorem and separate real and imaginary coefficients.
 
  • #3
Harmony said:

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? :)
 
  • #4
Try VietDao29's solution, much simpler, should try that first =] Mine takes ages :(
 
  • #5
∫ 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.
 
  • #6
Nah, just continue Integrating by Parts by choosing u = sin(x) cosn - 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. :)
 
  • #7
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 ?
 
  • #8
Harmony said:
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:

[tex]I_{\alpha , \beta} = \int \cos ^ \alpha (x) \cos (\beta x) dx[/tex]

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

You'll eventually end up with:

[tex]I_{n, n} = \mbox{something} + \mbox{something} \times I_{n, n} + \mbox{something} \times I_{n - 2, n}[/tex]

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

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

Now, say, you need to calculate: I7, 7, you start from I1, 7 (which is easy to calculate, eh?), then, by using the relation between In, n, and In - 2, n, you'll be able to find I3, 7, do the same, you'll get I5, 7, and finally, I7, 7, as desired.

If you want to calculate I8, 8, you'll start from I0, 8..

So, every In, n'll eventually boil down to either I1, n, or I0, n. This is how Recursion Formula works.

You your final answer will look something like:

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

Is it clear? :)
 
Last edited:

Related to Trigonometric Integration Using Recursion Formula

1. What is trigonometric integration?

Trigonometric integration is a mathematical technique used to find the integral of functions involving trigonometric functions like sine, cosine, and tangent. It involves using properties of trigonometric functions and substitution to evaluate the integral.

2. How is trigonometric integration different from regular integration?

Trigonometric integration deals specifically with integrals involving trigonometric functions, while regular integration can involve any type of function. Trigonometric integration also requires the use of trigonometric identities and substitution to solve the integral.

3. What are some common trigonometric integrals?

Some common trigonometric integrals include integrals of sine, cosine, and tangent functions, as well as integrals of the reciprocal trigonometric functions (cosecant, secant, and cotangent).

4. What are some tips for solving trigonometric integrals?

Some tips for solving trigonometric integrals include using trigonometric identities to simplify the integral, using substitution to convert the integral into a simpler form, and using trigonometric substitution for integrals involving radical expressions.

5. Why is trigonometric integration important?

Trigonometric integration is important because it allows us to find the area under a curve involving trigonometric functions, which has many real-world applications in fields such as physics, engineering, and finance. It also helps us to solve differential equations and evaluate complex mathematical models.

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