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9.3 Integrals of the Trigonometric Functions

9.43   Theorem (Integral of ) Let be an interval in . Then the cosine function is integrable on , and

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

9.44   Exercise. A Let be an interval in . Show that
 (9.45)

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

9.46   Notation ( .) If is integrable on the interval , we define

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

9.47   Theorem (Integrals of and .) Let and be any real numbers. Then

and

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.

9.48   Exercise. A Find the area of the set

Draw a picture of .

9.49   Exercise. A Find the area of the shaded figure, which is bounded by the graphs of the sine and cosine functions.

9.50   Example. By the change of scale theorem we have for and .

9.51   Entertainment (Archimedes sine integral) In On the Sphere and Cylinder 1., Archimedes states the following proposition: (see the figure below statement A).

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

 (9.52)

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
 (9.53)

We have

By the formula for the length of a chord (9.30) we have
 (9.54)

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

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

 (9.55)

and each summand on the right side of (9.55) is a side of a right triangle similar to triangle .

9.56   Exercise. Assuming equation (9.52), show that

Next: 9.4 Indefinite Integrals Up: 9. Trigonometric Functions Previous: 9.2 Calculation of   Index
Ray Mayer 2007-09-07