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The Yin-Yang Theorem:Integrating squares of sines and cosines

A class handout for Kent Merryfields Calculus II class.

1. The Theorem Stated and Proved

Heres the statement: Suppose that k > 0 and that both ka and kb are integer multiples of

2.

Then b

a

sin2(kx) dx =

b

a

cos2(kx) dx =b a

2

Paraphrasing: if youre integrating either sin2 or cos2 of a linear expression over an interval thatstarts and stops at the ends of quadrants, then the integral is half the box. Alternatively, you

can phrase the result as either sin2(kx) or cos2(kx) having an average value of1

2over any whole

number of quadrants.

We can prove this in many different ways, including the ordinary calculus-book method that

uses the half-angle identities sin2 x = 1 cos2x2

and cos2 x = 1 + cos 2x2

. But theres another,

more pictorial and memorable proof. Start by drawing a box: 0 x 2

and 0 y y. Draw thegraph of y = sin2 x within that box. We get the following picture:

0 2

0

1

y = sin2 x

The area inside the box and under the curve is

2

0

sin2 xdx. But what is the area inside the

box and above the curve? The distance from the curve to the top of the box is 1 sin2 x, and thatis cos2 x. Hence, the area inside the box and above the curve is

2

0

cos2 xdx.

Now, how are those two areas related? It appears that the picture is symmetric (and the

appearance is correct). Rotate the box by 180 about its central point that is, the point

4,

1

2

and the area below the curve becomes the area above the curve and the area above the curvebecomes the area below the curve.

For another way to see that, make the substitution u = 2x, which we can write as x =

2u.

Note that sin

2 u

= cos u. (Complementary angles.)

2

0

sin2 x dx =

0

2

sin2

2 u

(du) =

0

2

cos2 u(du)

=

2

0

cos2 u du =

2

0

cos2 x dx

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So, the area under the curve is the same as the area over the curve (and still inside the box).Those areas add together to be the area of the whole box, so each area is half the box. That is,

2

0

sin2 x dx =

2

0

cos2 x dx =1

2

2 1 =

4.

If we change x to kx, that stretches or shrinks the picture horizontally, but the same argumentstill works and we still get that the integral is half the box. If the integral covers several (a wholenumber of) quadrants, we may simply quote this theorem on each quadrant and add them together.

If this theorem has an official name, Im not aware of it. But as a handy descriptive shorthand,I like to refer to the picture above and call this the Yin-Yang Theorem.

2. Examples of Using this Theorem

2

0

cos2 d =1

2 2 1 = .

6

4

5sin2(2x) dx =1

2 (6

4)

5 = 10.

As part of this example, we note that when x = 4, 2x = 8 and when x = 6, 2x = 12, andeach of those points is the end of a quadrant.

0

cos2x

3

dx =?

In this case, we note that when x = ,x

3=

3, which is not the end of a quadrant, so the

Yin-Yang Theorem does not apply. We must compute this integral by other means. If we do so,

we get

2+

3

3

8.

In an alternating current electrical circuit, the voltage, V(t), depends on time in a cyclical usually sinusoidal fashion. For most purposes, the effective voltage is the root mean squarevoltage. To define what this means, let [a, b] be an interval that consists of a whole number ofperiods, and define:

Vrms =

1

b a

b

a

(V(t))2 dt

12

.

That is, the root mean square voltage is the square root of the average of the square of the voltage.Now, suppose the voltage is sinusoidal: for instance, V(t) = V0 sin(120t) for a 60 Hz (60 cycle persecond) current with voltage amplitude (peak voltage) equal to V0. Take some length of time thatcontains whole periods one second will do and compute the root mean square voltage

Vrms = 1

0

V20 sin2(120t) dt

1

2

=

V20

1

2

12

by half the box

=V0

2

As it turns out, for your 110 volt household current, that 110 volts is (approximately) theroot mean square voltage. The peak voltage must be equal to V0 = 110

2 157 volts, so

V(t) = 157 sin(120t).

2

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3. Integrating All the Way Around

It is quite common to need to compute the integral from 0 to 2 (one whole period) of somefunction of sines and cosines. In particular, this happens quite often in Calculus III when comput-ing various double or triple integrals in polar, cylindrical, or spherical coordinates. Not all suchfunctions are polynomials, but enough are to make that an interesting case. Consider

2

0

cosm sinn d

for nonnegative integers m and n. Note that cos and sin each take on both positive and negativevalues, and are negative exactly as much as they are positive. If either is raised to an odd power,it stands to reason that the average value would be zero.

Claim: If either m or n is odd, then

2

0

cosm sinn d = 0.

One way to prove that is to use the technique used to find the antiderivative. Suppose m isodd. Then make the substitution u = sin , du = cos d, replacing the remaining powers of cos2 by (1 sin2 ). The result is a polynomial integral in u, which gives us a polynomial in u, hence apolynomial in sin u. But that polynomial in sin u is periodic of period 2, so evaluating it at both0 and 2 gives the same value, and those values subtract to zero. The case of n odd is similar.

If both m and n are even, then cosm x sinn x is positive, and the integral is positive. If bothm and n are zero, then were integrating a constant, and the value of the definite integral in thewhole box. If one exponent is 2 and the other zero, we have the situation covered by the Yin-Yang Theorem: half the box. Other powers arent so simple, but we can still take advantage of ourknowledge. Here are two examples:

2

0

(3

2cos )3 d = 2

0

27

54cos + 36 cos2

8cos3 d

= 27 2 0 + 12 36 2 0

= 54 + 36 = 90

The second example is

2

0

sin2 cos2 d. We note that sin 2 = 2sin cos , so that

sin cos =1

2sin2. Use this:

2

0

sin2 cos2 d =1

4

2

0

sin2 2 d

=1

4 1

2 2 1 =

4

3