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      <text>13.6.4.1 Harder trigonometric integrals</text>
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      <text><![CDATA[<h3><span>Harder trigonometric integrals</span><br></h3>
<p>The following seemingly innocent integrals are examples, important in engineering, of trigonometric integrals that <strong>cannot</strong> be evaluated as <strong>indefinite</strong> integrals:</p> 
<ul>
<li>\({\displaystyle\int\sin (x^2)\,\mathrm{d}x}\phantom{--}\) and \(\phantom{--}{\displaystyle\int\cos (x^2)\,\mathrm{d}x}\phantom{---}\)These are called Fresnel integrals.</li> 
<li>\({\displaystyle\int\frac{\sin x}{x}\,\mathrm{d}x}\phantom{---}\)This is called the Sine integral.</li>
</ul> 
<p><strong>Definite</strong> integrals of this type, which are what normally arise in applications, have to be evaluated by <strong>approximate numerical methods</strong>.</p>
<p>Fresnel integrals with limits arise in wave and antenna theory and the Sine integral with limits in filter theory. It is useful sometimes to be able to visualize the definite integral. For example consider</p> 
<p>\(F(t) = {\displaystyle\int_0^t\dfrac{\sin x}{x}\,\mathrm{d}x}, t&gt;0\)</p>
<p>Clearly, \(F(0)={\displaystyle\int_0^0\dfrac{\sin x}{x}\,\mathrm{d}x} = 0\).</p>
<p> Recall the graph of \(\dfrac{\sin x}{x}\) against \(x\)</p><p><img src="@@PLUGINFILE@@/sine%20x%20over%20x.svg" alt="The sine of x over x as x increases" width="336" height="156" class="img-responsive atto_image_button_text-bottom"><br></p>

<p>For any positive value of \(t\), \(F(t)\) is the shaded area shown (the area interpretation of a definite integral was covered earlier in this Workbook). As \(t\) increases from \(0\) to \(\pi\), it follows that \(F(t)\) increases from \(0\) to a maximum value \(F(\pi)={\displaystyle\int_0^{\pi}\frac{\sin x}{x}\,\mathrm{d}x}\) whose value could be determined numerically (it is actually about 1.85). As \(t\) further increases from \(\pi\) to \(2\pi\) the value of \(F(t)\) will decrease to a local minimum at \(2\pi\) because the \(\dfrac{\sin x}{x}\) curve is below the \(x\)-axis between \(\pi\) and \(2\pi\). Note that the area below the curve <strong>is</strong> considered to be negative in this application. Continuing to argue in this way we can obtain the shape of the \(F(t)\) graph in the following Figure: (can you see why the oscillations decrease in amplitude?)</p><p><img src="@@PLUGINFILE@@/Translated%20sine%20x%20over%20x.svg" alt="" width="372" height="205" role="presentation" class="img-responsive atto_image_button_text-bottom"><br></p>

<p>The result \({\displaystyle\int_0^{\infty}\frac{\sin x}{x}\,\mathrm{d}x} = \dfrac{\pi}{2}\) is clearly illustrated in the graph (you are not expected to know how this result is obtained). Methods for solving such problems are dealt with in a later Section.</p>]]></text>
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