5 Harder trigonometric integrals

The following seemingly innocent integrals are examples, important in engineering, of trigonometric integrals that cannot be evaluated as indefinite integrals:

1. $\int \; <!--nolimits\; -->sin\left(x^2\right)dx$ and $\int \; <!--nolimits\; -->cos\left(x^2\right)dx$ These are called Fresnel integrals.
2. $\int \; <!--nolimits\; -->\frac{sinx}{x}dx$ This is called the Sine integral.

Definite integrals of this type, which are what normally arise in applications, have to be evaluated by approximate numerical methods .

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

$F\left(t\right)={\int \; <!--nolimits\; -->}_0^t\frac{sinx}{x}dxt>0$

Clearly, $F\left(0\right)={\int \; <!--nolimits\; -->}_0^0\frac{sinx}{x}dx=0$ . Recall the graph of $\frac{sinx}{x}$ against $x$ , $x>0$ :

Figure 14

For any positive value of $t$ , $F\left(t\right)$ 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 $F\left(t\right)$ increases from $0$ to a maximum value

$F\left(\pi \right)={\int \; <!--nolimits\; -->}_0^{\pi }\frac{sinx}{x}dx$

whose value could be determined numerically (it is actually about 1.85). As $t$ further increases from $\pi$ to $2\pi$ the value of $F\left(t\right)$ will decrease to a local minimum at $2\pi$ because the $\frac{sinx}{x}$ curve is below the $x$ -axis between $\pi$ and $2\pi$ . Note that the area below the curve is considered to be negative in this application.

Continuing to argue in this way we can obtain the shape of the $F\left(t\right)$ graph in Figure 15: (can you see why the oscillations decrease in amplitude?)

Figure 15

The result ${\int \; <!--nolimits\; -->}_0^{\infty }\frac{sinx}{x}dx=\frac{\pi }{2}$ 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 HELM booklet  31.

Exercises

You will need to refer to a Table of Trigonometric Identities to answer these questions.

1. Find
   1. $\int \; <!--nolimits\; -->{cos}^{2}xdx$
   2. ${\int \; <!--nolimits\; -->}_0^{\pi ∕2}{cos}^{2}tdt$
   3. $\int \; <!--nolimits\; -->\left({cos}^2\theta +{sin}^2\theta \right)d\theta$
2. Use the identity $sin\left(A+B\right)+sin\left(A-B\right)\equiv 2sinAcosB$ to find $\int \; <!--nolimits\; -->sin3xcos2xdx$
3. Find $\int \; <!--nolimits\; -->\left(1+{tan}^{2}x\right)dx$ .
4. The mean square value of a function $f\left(t\right)$ over the interval $t=a$ to $t=b$ is defined to be

   $\frac{1}{b-a}{\int \; <!--nolimits\; -->}_a^b{\left(f\left(t\right)\right)}^{2}dt$

   Find the mean square value of $f\left(t\right)=sint$ over the interval $t=0$ to $t=2\pi$ .

5. 1. Show that the reduction formula for $J_n=\int \; <!--nolimits\; -->{cos}^{n}xdx$ is

      $J_n=\frac{1}{n}{cos}^{n-1}\left(x\right)sinx+\frac{\left(n-1\right)}{n}{J}_{n-2}$

   2. Using the reduction formula in (a) show that

      $\int \; <!--nolimits\; -->{cos}^{5}xdx=\frac{1}{5}{cos}^{4}xsinx+\frac{4}{15}{cos}^{2}xsinx+\frac{8}{15}sinx$

   3. Show that if $J_n={\int \; <!--nolimits\; -->}_0^{\pi ∕2}{cos}^{n}xdx$ , then $J_n=\left(\frac{n-1}{n}\right)J_{n-2}$ (Wallis’ formula).
   4. Using Wallis’ formula show that ${\int \; <!--nolimits\; -->}_0^{\pi ∕2}{cos}^{6}xdx=\frac{5}{32}\pi$ .

Answer