Link between Pascal's triangle and integrals of trigonometric functions

**zass** · Nov4-09, 06:33 AM

My lecturer keeps simplifying trigonometric integrals in one line such as
$LaTeX Code: \\int^{2\\pi}_{0}sin^{4}(t)dt=\\frac{3\\pi}{4}$
and writes pascals triangle next to it. Just wondering what's the link between them? I'm sure it's obvious and easy, I'd just like to have an fast way of dealing with these

**HallsofIvy** · Nov4-09, 07:52 AM

I have no idea what your teacher is doing! However you can express powers of sin(x) in terms of things like sin(4x). The "double angle identity",

can be rewritten as

LaTeX Code: cos(2x)= cos^2(x)- (1- cos^2(x))= 2cos^2(x)- 1

or as

LaTeX Code: cos(2x)= (1- sin^2(x))- sin^2(x)= 1- 2sin^2(x)

.

Those can be "solved" as

and

.

So sin⁴(x)= (sin²(x))²= ((1/2)(1- cos(2x))²[/itex]

LaTeX Code: = (1/4)(1- 2cos(2x)+ cos^2(2x))

. And

LaTeX Code: cos^2(2x)= (1/2)(1+ cos(2(2x)))= (1/2)(1+ cos(4x)

so

LaTeX Code: sin^4(x)= 1/4- (1/2)cos(2x)+ (1/2)(1+ cos(4x))

LaTeX Code: = 3/4- (1/2)cos(2x)+ (1/2)cos(4x)

.

Since the integrals of cos(2x) and cos(4x), from 0 to $LaTeX Code: 2\\pi$ , are 0, the only thing remaining is that "1/4" and its integral from 0 to $LaTeX Code: 2\\pi$ is, of course, $LaTeX Code: (1/4)(2\\pi)= \\pi/2$ .

**arildno** · Nov4-09, 08:16 AM

Halls:
That should be:
$LaTeX Code: \\frac{1}{4}(1-2\\cos(2x)+\\frac{1}{2}(1+\\cos(4x)))$
yielding 3/8 to integrate, agreeing with the teacher's answer.

**g_edgar** · Nov4-09, 09:28 AM

Maybe your teacher is using the complex exponential version of trig functions...

$LaTeX Code: \\sin^4 t = \\left(\\frac{e^{it}-e^{-it}}{2i}\\right)^4$

and Pascal's triangle is of obvious help in expanding the numerator of the power.

**arildno** · Nov4-09, 10:16 AM

The correct relation is fairly easy to gain by repeated use of partial integration:

1. Let $LaTeX Code: I_{n}=\\int_{0}^{2\\pi}\\sin^{n}tdt=-\\cos(t)\\sin^{n-1}t\\mid_{t=0}^{t=2\\pi}+\\int_{0}^{2\\pi}(n-1)\\cos^{2}t\\sin^{n-2}t=(n-1)\\int_{0}^{2\\pi}(\\sin^{n-2}t-\\sin^{n}t)dt=(n-1)I_{n-2}-(n-1)I_{n}$

2. Thus, we have gained the recurrence relation:
$LaTeX Code: I_{n}=(n-1)I_{n-2}-(n-1)I_{n}\\to{I}_{n}=\\frac{n-1}{n}I_{n-2}, n\\geq{2}$

(The case is identical if cosine constitutes our base, rather than sine, the exponent "n" remaining the same)
3. We note that $LaTeX Code: I_{0}=2\\pi, I_{1}=0$
Thus, we easily see that for "n" odd, the integral will be 0.
Henceforth, we assume even n=2p for natural number p.

4. We may write, for any particular "n", the value as a "sew-saw"-pattern starting with "n" as the value in the leftmost denominator:
$LaTeX Code: I_{n}=(\\frac{n-1}{n}*\\frac{n-3}{n-2}*.....*\\frac{1}{2})*2\\pi$

5. This can again be twiddled into an explicit form as follows:
With n=2p, the denominator $LaTeX Code: n*(n-2)*(n-4)...2=2^{p}(p!)$ , where p! is the factorial of p.
Multiplying both the numerator and denominator with this expression yields:
$LaTeX Code: I_{n}=\\frac{n!}{2^{n}(p!)^{2}}*2\\pi=\\pi*2^{1-n}*\\frac{n!}{(n-p)!p!}=\\pi*2^{1-n}*\\binom{n}{p}$

The binomial coefficients are closely related to..Pascal's triangle.

**Gerenuk** · Nov4-09, 12:04 PM

Originally Posted by g_edgar

Maybe your teacher is using the complex exponential version of trig functions...
$LaTeX Code: \\sin^4 t = \\left(\\frac{e^{it}-e^{-it}}{2i}\\right)^4$
and Pascal's triangle is of obvious help in expanding the numerator of the power.

That looks like a good idea.

I'd prefer a slight improvement
$LaTeX Code: \\sin^4 t = \\Im\\left(e^{it}\\left(\\frac{e^{it}-e^{-it}}{2i}\\right)^3\\right)$
Then it's easier to convert back to tri-functions.