9.3 Integrals of the Trigonometric Functions

Proof: Let be any interval in
. Then is piecewise
monotonic on and hence is integrable. Let
be
the regular partition of into equal subintervals, and let

be the sample for consisting of the midpoints of the intervals of .

Let so that and for . Then

Multiply both sides of this equation by and use the identity

to get

Thus

(By taking large enough we can guarantee that , and then , so we haven't divided by .) Thus by theorem 9.37

The proof is similar to the proof of (9.43). The magic factor is the same as in that proof.

This is a natural generalization of the convention for in definition 5.67.

Proof: We will prove the first formula. The proof of the second is similar. If then the conclusion follows from theorem 9.43.

If then

so the conclusion follows in all cases.

**Statement A:**

If a polygon be inscribed in a segment of a circle so that all its sides excluding the base are equal and their number even, as , being the middle point of segment, and if the lines , , parallel to the base and joining pairs of angular points be drawn, then

where is the middle point of and is the diameter through .[2, page 29]We will now show that this result can be reformulated in modern notation as follows.

**Statement B:**
Let be a number in , and let be a positive integer.
Then there exists a partitition-sample sequence
for ,
such that

In exercise (9.56) you are asked to show that
(9.52) implies that

Proof that statement A implies statement B: Assume that statement A is true. Take the circle to have radius equal to , and let

Then

and

Let

and

Then is a partition of with mesh equal to , and is a sample for , so is a partition-sample sequence for , and we have

By Archimedes' formula, we conclude that

We have

By the formula for the length of a chord (9.30) we have

Equation (9.52) follows from (9.53) and (9.54).

Prove statement A above. Note that (see the figure
below statement A)