next up previous index
Next: 9.4 Indefinite Integrals Up: 9. Trigonometric Functions Previous: 9.2 Calculation of   Index

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

\begin{displaymath}\int_a^b \cos = \sin(b) - \sin(a). \end{displaymath}

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

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

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

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

\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).

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

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

to get

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


\begin{displaymath}\sum(\cos,P_n,S_n)={{\Big({{\Delta_n}\over 2}\Big)}\over

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

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

9.44   Exercise. A Let $[a,b]$ be an interval in $\mbox{{\bf R}}$. Show that
\int_a^b\sin =
\cos(a)-\cos(b) .
\end{displaymath} (9.45)

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

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

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

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

\begin{displaymath}\int_a^b \cos = \sin(b) - \sin(a). \end{displaymath}


\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

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

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

9.48   Exercise. A Find the area of the set

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

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 $c>0$.

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

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

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

\begin{displaymath}(BB' +CC' + \ldots +LM):AM = A'B:BA,\end{displaymath}

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})}.
\end{displaymath} (9.52)

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

\begin{displaymath}\int_0^\phi \sin = 1 - \cos(\phi).\end{displaymath}

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

\phi &=& \mbox{length of arc} (AL)\\
{\phi\over n} &=& \mbox{length of arc} (AB).


\begin{displaymath}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),\end{displaymath}


\begin{displaymath}AM = 1-\cos(\phi).


\begin{displaymath}P_n = \{ 0,{2\phi\over 2n+1},{4\phi\over 2n+1} ,\cdots,
{2n\phi \over 2n+1}, \phi\},\end{displaymath}


\begin{displaymath}S_n = \{0, {\phi\over n}, {2\phi\over n},\cdots,{n\phi\over n}\}.\end{displaymath}

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

\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).

By Archimedes' formula, we conclude that
\sum(\sin,P_n,S_n) = (1-\cos(\phi)){\phi\over 2n+1} \cdot{A'B\over BA}.
\end{displaymath} (9.53)

We have

\mbox{length arc}(BA) &=& {\phi\over n},\\
\mbox{length arc}(BA') &=& \pi -{\phi\over n}.

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})}
\end{displaymath} (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,
\end{displaymath} (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

\begin{displaymath}\int_0^\phi \sin = 1 - \cos(\phi).\end{displaymath}

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