Chapter 3 Section 8

3.8 Definite Integrals of Sine & Cosine

Preliminary Results

• Rewriting a product of two sine terms as a difference of cosine terms.
  \begin{equation}\label{ch03-08-eq01}
  \sin A \sin B=\frac{1}{2}\left (\cos (A-B)-\cos (A+B)\right )\tag{3.14}.
  \end{equation}
  This is accomplished as follows:
  \begin{align*}
  \cos (A-B)-\cos (A+B)
  &=\cos A \cos B +\sin A \sin B – (\cos A \cos B -\sin A \sin B)=\\
  &=2 \sin A \sin B,
  \end{align*}

 

• Rewriting a product of a sine term with a cosine term as a sum of sine terms.
  \begin{equation}\label{ch03-08-eq02}
  \sin A \cos B=\frac{1}{2}\left (\sin (A+B)+\sin (A+B)\right )\tag{3.15}.
  \end{equation}
  This is accomplished as follows:
  \begin{align*}
  \sin (A+B)+\sin (A-B)
  &=\sin A \cos B +\sin B \cos a – (\sin A \cos B -\sin B \cos A)=\\
  &=2 \sin A \sin B,
  \end{align*}

 

• The continuity of sine and cosine everywhere.

 

• The limit of the difference quotient for \(\sin x\):
  \begin{equation}\label{ch03-08-eq03}
  \lim_{x\to 0}\frac{\sin x}{x} = 1\tag{3.16}.
  \end{equation}

 

• Telescoping sums (review 3.2.2).

3.8.1 Definite Integral of Sine

 
Proposition 1. Given \(a,\, b \in \mathbb{R}\) with \(a < b\), then \begin{equation}\label{ch03-08-eq04} \int_a^b\sin x \,dx = -\cos b + \cos a\tag{3.17}. \end{equation}

Proof.

Step 1. Partition of the interval.
\[\Delta x= \frac{b-a}{n}, \;\; x_i=a+i \Delta x=a+\frac{(b-a)i}{n}, \;\; I_i=[x_{i-1},x_i].\]

Step 2. Height and area of the rectangular regions.
\[h_i=\sin \left(x_i\right)=\sin \left(a+ i \Delta x\right),      A_i=\sin \left(x_i\right)\Delta x =\sin \left(a+ i\Delta x\right)\Delta x.\]

Step 3. Computation of \(R(\sin x,n)\).
\[R(\sin x,n)=\sum _{i=1}^n \sin \left(x_i\right)\Delta x=\sum _{i=1}^n \sin (a+i \Delta x)\Delta x=\Delta x\sum _{i=1}^n \sin (a+i \Delta x).\]
Multiply and divide each term in the sum by \(\displaystyle \sin \left(\frac{\Delta x}{2}\right)\). Note that it is possible to do this because for \(n\) sufficiently large we have
\[0 < \frac{\Delta x}{2} = \frac{b-a}{2n}<\pi \Longrightarrow \sin \left(\frac{\Delta x}{2}\right)\neq 0.\] Then \begin{equation}\label{ch03-08-eq05} R(\sin x,n)=\sum _{i=1}^n \sin\left(x_i\right)\Delta x =\Delta x\frac{1}{\sin \left(\frac{\Delta x}{2}\right)} \sum _{i=1}^n \sin (a+i \Delta x)\sin \left(\frac{\Delta x}{2}\right)\tag{3.18}. \end{equation} We are going to show that the sum in (\ref{ch03-08-eq02}) is telescoping. This will be accomplished by rewriting each of the products in the sum as a difference of cosine terms (see (\ref{ch03-08-eq01})). \begin{align*} \sin (a+i \Delta x)\sin \left(\frac{\Delta x}{2}\right) &\underset{?}{=}\hphantom{-}\frac{1}{2}\left(\cos \left(a+i \Delta x-\frac{\Delta x}{2}\right)-\cos \left(a+i \Delta x+\frac{\Delta x}{2}\right)\right)=\\ &\underset{?}{=}-\frac{1}{2}\left( \cos \left(a + i \Delta x + \frac{\Delta x}{2}\right) - \cos \left(a +(i-1) \Delta x + \frac{\Delta x}{2}\right) \right). \end{align*} Set \[a_i =-\frac{1}{2}\cos \left(a + i \Delta x + \frac{\Delta x}{2}\right), \, a_{i-1} =-\frac{1}{2}\cos \left(a +(i-1) \Delta x + \frac{\Delta x}{2}\right).\] It follows that the Riemann sum in (\ref{ch03-08-eq02}) is telescoping, and thus (see Proposition 2 in 3.2.2) \begin{align*} \sum _{i=1}^n \sin (a+i \Delta x)\sin \left(\frac{\Delta x}{2}\right) &=\sum _{i=1}^n \left(a_i-a_{i-1}\right) \underset{?}{=} a_n-a_0 =\\ &\underset{?}{=}-\frac{1}{2}\left( \cos \left(a + n \Delta x + \frac{\Delta x}{2}\right) - \cos \left(a +(0) \Delta x + \frac{\Delta x}{2}\right) \right) = \\ &\underset{?}{=} -\frac{1}{2}\left( \cos \left(a + (b-a) + \frac{\Delta x}{2}\right) - \cos \left(a + \frac{\Delta x}{2}\right) \right) = \\ &= -\frac{1}{2}\left( \cos \left(b + \frac{\Delta x}{2}\right) - \cos \left(a + \frac{\Delta x}{2}\right) \right). \end{align*} Therefore, (\ref{ch03-08-eq05}) can be rewritten as follows \begin{align*} \frac{\Delta x}{\sin \left(\frac{\Delta x}{2}\right)} \sum _{i=1}^n \sin (a+i \Delta x)\sin \left(\frac{\Delta x}{2}\right) &= \frac{\Delta x}{\sin \left(\frac{\Delta x}{2}\right)} \left[ -\frac{1}{2}\left( \cos \left(b + \frac{\Delta x}{2}\right) - \cos \left(a + \frac{\Delta x}{2}\right) \right) \right]=\\ &= -\frac{\frac{\Delta x}{2}}{\sin \left(\frac{\Delta x}{2}\right)} \left( \cos \left(b+\frac{\Delta x}{2}\right)-\cos \left(a+\frac{\Delta x}{2}\right)\right). \end{align*} Step 4. Evaluation of \(\displaystyle \lim_{n \to \infty} R(\sin x,n)\).

Since \(\displaystyle \Delta x =\frac{b-a}{n}\) with \(b-a\) constant, then
\(\Delta x \longrightarrow 0 \Longleftrightarrow n\longrightarrow \infty\),
that is, increasing the number of intervals in the partition is equivalent to let \(\Delta x \to 0\). Thus,
\begin{align*}
\lim_{n \to \infty}\sum _{i=1}^n \sin\left(x_i\right)\Delta x
&=\lim_{\Delta x \to 0}\frac{\frac{\Delta x}{2}}{\sin \left(\frac{\Delta x}{2}\right)}\left(\cos \left(a+\frac{\Delta x}{2}\right)-\cos \left(b+\frac{\Delta x}{2}\right)\right)=\\
&\underset{?}{=} (1)(-\cos b + \cos a)=-\cos b + \cos a.
\end{align*}

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3.8.2 Definite Integral of Cosine

 
Proposition 2. Given \(a,\, b \in \mathbb{R}\) with \(a < b\), then \begin{equation}\label{ch03-08-eq06} \int_a^b \cos x\,dx = \sin b - \sin a\tag{3.19}. \end{equation}

Proof.

Step 1. Partition of the interval.
\[\Delta x= \frac{b-a}{n}, \;\; x_i=a+i \Delta x=a+\frac{(b-a)i}{n}, \;\; I_i=[x_{i-1},x_i].\]

Step 2. Height and area of the rectangular regions.
\[\displaystyle h_i=f\left(x_i\right)=\cos \left(a+ i \Delta x\right),\;\;\;
A_i=f\left(x_i\right)\Delta x =\cos \left(a+ i\Delta x\right)\Delta x.\]

Step 3. Computation of \(R(\cos x,n)\).
\[R(\cos x,n) = \sum _{i=1}^n \cos\left(x_i\right)\Delta x=\sum _{i=1}^n \cos (a+i \Delta x)\Delta x=\Delta x\sum _{i=1}^n \cos (a+i \Delta x).\]
Multiply and divide by \(\displaystyle \sin \left(\frac{\Delta x}{2}\right)\).
We can assume that \(\displaystyle \sin \left(\frac{\Delta x}{2}\right)\neq 0\) (see Step 3 in the proof of Proposition 1).
\[\sum _{i=1}^n \cos\left(x_i\right)\Delta x=
\Delta x\frac{1}{\sin \left(\frac{\Delta x}{2}\right)}
\sum _{i=1}^n \cos (a+i \Delta x)\sin \left(\frac{\Delta x}{2}\right).\]
To compute the sum \(\displaystyle \sum _{i=1}^n \cos (a+i \Delta x)\sin \left(\frac{\Delta x}{2}\right)\), we are going to see that it is telescoping by rewriting each term as a sum of sines (see (\ref{ch03-08-eq02})) together with the fact that sine is odd:
\begin{align*}
\sin \left(\frac{\Delta x}{2}\right)\cos (a+i \Delta x)
&\underset{?}{=}\frac{1}{2}\left(\sin \left(a+i \Delta x+\frac{\Delta x}{2}\right)-\sin \left(a+i \Delta x-\frac{\Delta x}{2}\right)\right)=\\
&=\frac{1}{2}\left(
\sin \left(a+i\Delta x + \frac{\Delta x}{2}\right)-
\sin \left(a+(i-1)\Delta x + \frac{\Delta x}{2}\right)
\right).
\end{align*}
Thus (see Proposition 2 in 3.2.2),
\begin{align*}
\sum _{i=1}^n \sin \left(\frac{\Delta x}{2}\right)\cos (a+i \Delta x) & =
\sum _{i=1}^n \frac{1}{2}\left(
\sin \left(a+i\Delta x + \frac{\Delta x}{2}\right)-
\sin \left(a+(i-1)\Delta x + \frac{\Delta x}{2}\right)
\right) =\\
&=\frac{1}{2}\left(
\sin \left(a+n\Delta x + \frac{\Delta x}{2}\right)-
\sin \left(a+(1-1)\Delta x + \frac{\Delta x}{2}\right)
\right) =\\
&=\frac{1}{2}\left(
\sin \left(b + \frac{\Delta x}{2}\right)-
\sin \left(a + \frac{\Delta x}{2}\right)
\right).
\end{align*}
\begin{align*}
\Longrightarrow \sum _{i=1}^n \cos\left(x_i\right)\Delta x
&=\Delta x\frac{1}{\sin \left(\frac{\Delta x}{2}\right)}\sum _{i=1}^n \cos (a+i \Delta x)\sin \left(\frac{\Delta x}{2}\right)=\\
&=\Delta x\frac{1}{\sin \left(\frac{\Delta x}{2}\right)}\frac{1}{2}\left(\sin \left(b+\frac{\Delta x}{2}\right)-\sin \left(a+\frac{\Delta x}{2}\right)\right)=\\
&=\frac{\frac{\Delta x}{2}}{\sin \left(\frac{\Delta x}{2}\right)}\left(\sin \left(b+\frac{\Delta x}{2}\right)-\sin \left(a+\frac{\Delta x}{2}\right)\right).
\end{align*}

Step 4. Evaluation of \(\displaystyle \lim_{n \to \infty}R(\cos x,n)\)).

Since \(\Delta x \longrightarrow 0 \Longleftrightarrow n\longrightarrow \infty\)
\[
\lim_{n \to \infty}\sum _{i=1}^n \cos\left(x_i\right)\Delta x=\lim_{\Delta x \to 0}\frac{\frac{\Delta x}{2}}{\sin \left(\frac{\Delta x}{2}\right)}\left(\sin \left(b+\frac{\Delta
x}{2}\right)-\sin \left(a+\frac{\Delta x}{2}\right)\right)=\sin b -\sin a.\]

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