9.3 Integrals of the Trigonometric Functions

9.43   Theorem (Integral of $\cos$) Let $[a,b]$ be an interval in $\mbox{{\bf R}}$. Then the cosine function is integrable on $[a,b]$, and

$$\int_a^b \cos = \sin(b) - \sin(a).$$

Proof: Let $[a,b]$ be any interval in $\mbox{{\bf R}}$. Then $\cos$ is piecewise monotonic on $[a,b]$ and hence is integrable. Let $P_n=\{x_0,x_1,\cdots ,x_n\}$ be the regular partition of $[a,b]$ into $n$ equal subintervals, and let

$$S_n=\Big\{ {{x_0+x_1}\over 2},{{x_1+x_2}\over 2},\cdots ,{{x_{n-1}+x_n}\over 2}\Big\}$$

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

Let $${\Delta _n={{b-a}\over n}}$$ so that $x_i-x_{i-1}=\Delta_n$ and $${ {{x_{i-1}+x_i}\over 2}=x_{i-1}+{{\Delta_n}\over 2}}$$ for $1\leq i\leq n$. Then

$$\begin{eqnarray*} \sum (\cos ,P_n,S_n) &=& \sum_{i=1}^n\cos\Big( x_{i-1}+{{\Delt... ... \Delta_n\sum_{i=1}^n\cos\Big( x_{i-1}+{{\Delta_n}\over 2}\Big). \end{eqnarray*}$$

Multiply both sides of this equation by $${\sin\Big({{\Delta_n}\over 2}\Big)}$$ and use the identity

$$\sin(t)\cos(s)={1\over 2}[\sin(s+t)-\sin(s-t)]$$

to get

$$\begin{eqnarray*} \sin\Big({{\Delta_n}\over 2}\Big)\sum(\cos,P_n,S_n) &=& \Delta... ...-\sin(x_0)]\\ &=& {{\Delta_n}\over 2}\Big(\sin(b)-\sin(a)\Big). \end{eqnarray*}$$

Thus

$$\sum(\cos,P_n,S_n)={{\Big({{\Delta_n}\over 2}\Big)}\over {\sin\Big({{\Delta_n}\over 2}\Big)}}\Big(\sin(b)-\sin(a)\Big).$$

(By taking $n$ large enough we can guarantee that $${ {{\Delta_n}\over 2}<\pi}$$, and then $${ \sin \Big( {{\Delta_n}\over 2}\Big)\neq 0}$$, so we haven't divided by $0$.) Thus by theorem 9.37

$$\begin{eqnarray*} \int_a^b\cos &=&\lim\{\sum(\cos,P_n,S_n)\}\\ &=& \lim \left\{... ...sin(a) \Big) \cdot 1 = \sin(b) - \sin(a). \mbox{ $\diamondsuit$} \end{eqnarray*}$$

9.44   Exercise. A Let $[a,b]$ be an interval in $\mbox{{\bf R}}$. Show that

$$\int_a^b\sin = \cos(a)-\cos(b) .$$ (9.45)

The proof is similar to the proof of (9.43). The magic factor $${\sin\Big({\Delta_n \over 2}\Big)}$$ is the same as in that proof.

9.46   Notation ( $${\int_b^af}$$.) If $f$ is integrable on the interval $[a,b]$, we define

$$\int_b^a f=-\int_a^b f \mbox{ or } \int_b^a f(t)dt=-\int_a^b f(t)dt.$$

This is a natural generalization of the convention for $A_b^a f$ in definition 5.67.

9.47   Theorem (Integrals of $\sin$ and $\cos$.) Let $a$ and $b$ be any real numbers. Then

$$\int_a^b \cos = \sin(b) - \sin(a).$$

and

$$\int_a^b\sin = \cos(a)-\cos(b) .$$

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

If $b<a$ then

$$\displaystyle {\int_a^b\cos=-\int_b^a\cos=-[\sin (a)-\sin (b)]=\sin (b)-\sin (a)},$$

so the conclusion follows in all cases. $\diamondsuit$

9.48   Exercise. A Find the area of the set

$$S_0^\pi (\sin)=\{(x,y)\colon 0\leq x\leq\pi \mbox{ and } 0\leq y\leq\sin x\}.$$

Draw a picture of $S_0^\pi (\sin)$.

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 $a<b$ and $c>0$.

$$\begin{eqnarray*} \int_a^b\sin (cx)dx &=& {1\over c}\int_{ca}^{cb}\sin x\; dx \\ &=& {{-\cos (cb)+\cos (ca)}\over c} \end{eqnarray*}$$

$$\begin{eqnarray*} \int_a^b\cos (cx)dx &=& {1\over c} \int_{ca}^{cb}\cos x\; dx \\ &=& {{\sin (cb)-\sin (ca)}\over c} \end{eqnarray*}$$

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 $LAL'$ so that all its sides excluding the base are equal and their number even, as $LK\ldots A \ldots K'L'$, $A$ being the middle point of segment, and if the lines $BB'$, $CC'$,$\ldots$ parallel to the base $LL'$ and joining pairs of angular points be drawn, then

$$(BB' +CC' + \ldots +LM):AM = A'B:BA,$$

where $M$ is the middle point of $LL'$ and $AA'$ is the diameter through $M$.[2, page 29]

We will now show that this result can be reformulated in modern notation as follows.

Statement B: Let $\phi$ be a number in $[0,\pi]$, and let $n$ be a positive integer. Then there exists a partitition-sample sequence $(\{P_n\},\{S_n\})$ for $[0,\phi]$, such that

$$\sum(\sin,P_n,S_n) = (1-\cos(\phi)){\phi\over {2n+1}}\ {\cos({\phi\over 2n})\over \sin({\phi\over 2n})}.$$ (9.52)

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

$$\int_0^\phi \sin = 1 - \cos(\phi).$$

Proof that statement A implies statement B: Assume that statement A is true. Take the circle to have radius equal to $1$, and let

$$\begin{eqnarray*} \phi &=& \mbox{length of arc} (AL)\\ {\phi\over n} &=& \mbox{length of arc} (AB). \end{eqnarray*}$$

Then

$$BB'+CC'+\ldots +LM = 2\sin({\phi\over n}) +2\sin({2\phi\over n}) +\cdots+2\sin({(n-1)\phi\over n}) + \sin(\phi),$$

and

$$AM = 1-\cos(\phi).$$

Let

$$P_n = \{ 0,{2\phi\over 2n+1},{4\phi\over 2n+1} ,\cdots, {2n\phi \over 2n+1}, \phi\},$$

and

$$S_n = \{0, {\phi\over n}, {2\phi\over n},\cdots,{n\phi\over n}\}.$$

Then $P_n$ is a partition of $[0,\phi]$ with mesh equal to ${2\phi\over 2n+1}$, and $S_n$ is a sample for $P_n$, so $(\{P_n\},\{S_n\})$ is a partition-sample sequence for $[0,\phi]$, and we have

$$\begin{eqnarray*} \sum(\sin,P_n,S_n)&=& {2\phi\over 2n+1}\Big(\sin({\phi\over n}... ...) +\cdots +\sin({(n-1)\phi\over n}) +{1\over 2}\sin(\phi)\Big). \end{eqnarray*}$$

By Archimedes' formula, we conclude that

$$\sum(\sin,P_n,S_n) = (1-\cos(\phi)){\phi\over 2n+1} \cdot{A'B\over BA}.$$ (9.53)

We have

$$\begin{eqnarray*} \mbox{length arc}(BA) &=& {\phi\over n},\\ \mbox{length arc}(BA') &=& \pi -{\phi\over n}. \end{eqnarray*}$$

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

$${A'B\over BA} = {\mbox{chord}(AB')\over \mbox{chord}(BA)} ... ...n})\over 2})} ={\cos({\phi\over 2n})\over\sin({\phi\over 2n})}$$ (9.54)

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

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

$$AM = AF+FP+PG+GQ +\cdots + HR+RM,$$ (9.55)

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

9.56   Exercise. Assuming equation (9.52), show that

$$\int_0^\phi \sin = 1 - \cos(\phi).$$