Trigonometric Integrals

Section 8.2 Trigonometric Integrals

As indicated at the beginning of this chapter, we can evaluate certain definite real integrals with the aid of the residue theorem. One way to do this is by interpreting the definite integral as the parametric form of an integral of an analytic function along a simple closed contour. Suppose that we want to evaluate an integral of the form

\begin{equation} \int_0^{2\pi}F(\cos \theta ,\sin \theta)\,d\theta\text{.}\tag{8.2.1} \end{equation}

where \(F(u,v)\) is a function of the two real variables \(u\) and \(v\text{.}\) Consider the unit circle \(C_1(0)\) with parametrization

\begin{equation*} C_1^{+}(0) :z=\cos \theta +i\sin \theta =e^{i\theta}, \text{ for } 0 \le \theta \le 2\pi\text{,} \end{equation*}

which gives the symbolic differentials

\begin{align} dz \amp = (-\sin \theta +i\cos \theta) d\theta =ie^{i\theta}\,d\theta, \text{ and }\notag\\ d\theta \amp = \frac{dz}{ie^{i\theta}}=\frac{dz}{iz}\text{.}\tag{8.2.2} \end{align}

Combining \(z=\cos \theta +i\sin \theta\) with \(\frac{1}{z}=\cos \theta -i\sin \theta\text{,}\) we obtain

\begin{equation} \cos \theta =\frac{1}{2}\left(z+\frac{1}{z}\right), \text{ and } \sin \theta =\frac{1}{2i}\left(z-\frac{1}{z}\right)\text{.}\tag{8.2.3} \end{equation}

Using the substitutions for \(\cos \theta\text{,}\) \(\sin \theta\text{,}\) and \(d\theta\) in Expression (8.2.1) transforms the definite integral into the contour integral

\begin{equation*} \int_0^{2\pi}F(\cos \theta, \sin \theta)\,d\theta = \int\limits_{C_1^{+}(0)}f(z)\,dz\text{.} \end{equation*}

where the new integrand is \(\displaystyle f(z) = \frac{F\Big(\frac{1}{2}(z+\frac{1}{z}), \; \frac{1}{2i}(z-\frac{1}{z})\Big)}{iz}\text{.}\)

Suppose that \(f\) is analytic inside and on the unit circle \(C_1(0)\text{,}\) except at the points \(z_1, \, z_2, \, \ldots , \, z_n\) that lie interior to \(C_1(0)\text{.}\) Then the residue theorem gives

\begin{equation} \int_0^{2\pi}F(\cos \theta ,\sin \theta)\,d\theta = 2\pi i\sum_{k=1}^n\mathrm{Res}[f,z_k]\text{.}\tag{8.2.4} \end{equation}

The situation is illustrated in Figure 8.2.1.

Figure 8.2.1. Changing variables: a definite integral on \([0,2\pi]\) to a contour integral around \(C\)

Example 8.2.2.

Evaluate \(\int_0^{2\pi}\frac{1}{1+3\cos^2\theta}\,d\theta\) by using complex analysis.

Solution.

Using Substitutions (8.2.2) and (8.2.3), we transform the integral to

\begin{equation*} \int\limits_{C_1^{+}(0)}\frac{1}{1+3(\frac{z+z^{-1}}{2})^2}\left(\frac{1}{iz}\right)dz = \int\limits_{C_1^{+}(0)}\frac{-i4z}{3z^4+10z^2+3}\,dz = \int\limits_{C_1^{+}(0)}f(z)\,dz\text{.} \end{equation*}

where \(f(z) =\frac{-i4z}{3z^4+10z^2+3}\text{.}\) The singularities of \(f\) are poles located at the points where \(3(z^2)^2+10(z^2) +3=0\text{.}\) Using the quadratic formula, we see that the singular points satisfy the relation \(z^2=\frac{-10\pm \sqrt{100-36}}{6 }=\frac{-5\pm 4}{3}\text{.}\) Hence the only singularities that lie inside the unit circle are simple poles corresponding to the solutions of \(z^2=-\frac{1}{3}\text{,}\) which are the two points \(z_1=\frac{i}{\sqrt{3}}\) and \(z_2=-\frac{i}{\sqrt{3}}\text{.}\) We use Theorem 8.1.7 and L’Hôpital’s rule to get the residues at \(z_k\text{,}\) for \(k=1, \, 2\text{:}\)

\begin{align*} \mathrm{Res}[f,z_k] \amp = \lim_{z \to z_k}\frac{-i4z(z-z_k)}{3z^4+10z^2+3}\\ \amp = \lim_{z \to z_k}\frac{-i4(2z-z_k)}{12z^3+20z}\\ \amp = \frac{-i4z_k}{12z_k^3+20z_k}\\ \amp = \frac{-i}{3z_k^2+5}\text{.} \end{align*}

Since \(z_k=\frac{\pm i}{\sqrt{3}}\) and \(z_k^2=-\frac{1}{3}\text{,}\) the residues for each \(z_k\) are given by \(\mathrm{Res}[f,z_k] =-\frac{i}{3(-\frac{1}{3})+5} = -\frac{i}{4}\text{.}\) Now use Equation (8.2.4) to compute the value of the integral:

\begin{equation*} \int_0^{2\pi}\frac{1}{1+3\cos^2\theta}\,d\theta = 2\pi i\left(\frac{-i}{4}+\frac{-i}{4}\right)= \pi \end{equation*}

Example 8.2.3.

Evaluate \(\int_0^{2\pi}\frac{1}{1+3\cos^2t}\,dt\) by using a computer algebra system.

Solution.

We can obtain the antiderivative of \(\frac{1}{1+3\cos^2t}\) by using software such as Mathematica or MAPLE. It is \(\int \frac{1}{1+3\cos^2t}\,dt = \frac{-\mathrm{Arctan}(2\cot t)}{2} = g(t)\text{.}\) Since \(\cot 0\) and \(\cot 2\pi\) are not defined, the computations for both \(g(0)\) and \(g(2\pi)\) are indeterminate. The graph \(s=g(t)\) shown in Figure 8.2.4 reveals another problem: The integrand \(\frac{1}{1+3\cos^2t}\) is a continuous function for all \(t\text{,}\) but the function \(g\) has a discontinuity at \(\pi\text{.}\) This condition appears to be a violation of the fundamental theorem of calculus, which asserts that the integral of a continuous function must be differentiable and hence continuous. The problem is that \(g(t)\) is not an antiderivative of \(\frac{1}{1+3\cos^2t}\) for all \(t\) in the interval \([0,2\pi ]\text{.}\) Oddly, it is the antiderivative at all points except \(0\text{,}\) \(\pi\text{,}\) and \(2\pi\text{,}\) which you can verify by computing \(g\,'(t)\) and showing that it equals \(\frac{1}{1+3\cos^2t}\) whenever \(g(t)\) is defined.

Figure 8.2.4. Graph of \(g(t) = \int \frac{1}{1+3\cos^2t}\ dt = \frac{-\mathrm{Arctan}(2\cot t)}{2}\)

The integration algorithm used by computer algebra systems here (the Risch-Norman algorithm) gives the antiderivative \(g(t) = \frac{-\mathrm{Arctan}(2\cot t)}{2}\text{,}\) and we must take great care in using this information.

We get the proper value of the integral by using \(g(t)\) on the open subintervals \((0,\pi )\) and \((\pi ,2\pi )\) where it is continuous, and taking appropriate limits:

\begin{align*} {3} \int_0^{2\pi}\frac{1}{1+3\cos^2t}\,dt \amp = \left[\int_0^{\pi} \frac{1}{1+3\cos^2t}\,dt\right] \amp \amp + \left[\int_{\pi}^{2\pi}\frac{1}{1+3\cos^2t}\,dt\right]\\ \amp = \left[\lim_{{t \to \pi^{-},}{\,s \to 0^{+}}}\int_{s}^t\frac{1}{1+3\cos^2t}\,dt\right] \amp \amp + \left[\lim_{{t \to 2\pi^{-},}{\,s \to \pi^{+}}}\int_{s}^t\frac{1}{1+3\cos^2t}\,dt\right]\\ \amp = \left[\lim_{t \to \pi^-}g(t) -\lim_{s \to 0^+}g(s)\right] \amp \amp + \left[\lim_{t \to 2\pi^{-}}g(t) -\lim_{s \to \pi^+}g(s)\right]\\ \amp = \left[\frac{\pi}{4}-\frac{-\pi}{4}\right] \amp \amp + \left[\frac{\pi}{4}-\frac{-\pi}{4}\right]\\ \amp = \pi\text{.} \end{align*}

Example 8.2.5.

Evaluate \(\int_0^{2\pi}\frac{\cos 2\theta}{5-4\cos \theta}\,d\theta\text{.}\)

Solution.

For values of \(z\) that lie on the unit circle \(C_1(0)\text{,}\) we have

\begin{equation*} z^2 = \cos 2\theta +i\sin 2\theta, \text{ and } z^{-2}=\cos 2\theta -i\sin 2\theta\text{.} \end{equation*}

We solve for \(\cos 2\theta\) and \(\sin 2\theta\) to obtain the substitutions

\begin{equation*} \cos 2\theta = \frac{1}{2}(z^2+z^{-2}), \text{ and } \sin 2\theta = \frac{1}{2i}(z^2-z^{-2})\text{.} \end{equation*}

Using the identity for \(\cos 2\theta\) along with Substitutions (8.2.2) and (8.2.3), we rewrite the integral as

\begin{equation*} \int_{C_1^{+}(0)}\frac{\frac{1}{2}(z^2+z^{-2})}{5-4(\frac{z+z^{-1}}{2})}\left(\frac{1}{iz}\right)dz = \int_{C_1^{+}(0)}\frac{i(z^4+1)}{2z^2(z-2) (2z-1)}\,dz = \int\limits_{C_1^{+}(0)}f(z)\,dz\text{,} \end{equation*}

where \(f(z) =\frac{i(z^4+1)}{2z^2(z-2) (2z-1)}\text{.}\) The singularities of \(f\) lying inside \(C\) are poles located at the points \(0\) and \(\frac{1}{2}\text{.}\) We use Theorem 8.1.7 to get the residues:

\begin{align*} \mathrm{Res}[f,0] \amp = \lim\limits_{z \to 0}\frac{d}{dz}\left[z^2f(z)\right] = \lim\limits_{z \to 0}\frac{d}{dz}\left[i\frac{(z^4+1)}{2(2z^2-5z+2)}\right]\\ \amp = \lim\limits_{z \to 0}\left[i\frac{4z^3(2z^2-5z+2) - (4z-5)(z^4+1)}{2(2z^2-5z+2)^2}\right]\\ \amp = \frac{5i}{8}\text{,} \end{align*}

and

\begin{equation*} \mathrm{Res}\left[f,\frac{1}{2}\right] =\lim_{z \to \frac{1}{2}}\left(z-\frac{1}{2}\right)f(z) = \lim_{z \to \frac{1}{2}}\frac{i(z^4+1)}{4z^2(z-2)} = -\frac{ 17i}{24}\text{.} \end{equation*}

Therefore we conclude that

\begin{equation*} \int_0^{2\pi}\frac{\cos 2\theta}{5-4\cos \theta}\,d\theta = 2\pi i\left(\frac{5i}{8}-\frac{17i}{24}\right) = \frac{\pi}{6} \end{equation*}

Exercises Exercises

Use residues to evaluate the following integrals:

1.

\(\int_0^{2\pi}\frac{1}{3\cos \theta +5}d\theta\text{.}\)

Solution.

\(\frac{\pi}{2}\text{.}\)

2.

\(\int_0^{2\pi}\frac{1}{4\sin \theta +5}d\theta\text{.}\) \label {8.2.2}

3.

\(\int_0^{2\pi}\frac{1}{15\sin^2\theta +1}d\theta\text{.}\)

Solution.

\(\frac{\pi}{2}\text{.}\)

4.

\(\int_0^{2\pi}\frac{1}{5\cos^2\theta +4}d\theta\text{.}\)

5.

\(\int_0^{2\pi}\frac{\sin^2\theta}{5+4\cos \theta}\,d\theta\text{.}\)

Solution.

\(\frac{\pi}{4}\text{.}\)

6.

\(\int_0^{2\pi}\frac{\sin^2\theta}{5-3\cos \theta}\,d\theta\text{.}\)

7.

\(\int_0^{2\pi}\frac{1}{(5+3\cos \theta)^2}\,d\theta\text{.}\)

Solution.

\(\frac{5\pi}{32}\text{.}\)

8.

\(\int_0^{2\pi}\frac{1}{(5+4\cos \theta)^2}\,d\theta\text{.}\)

9.

\(\int_0^{2\pi}\frac{\cos 2\theta}{5+3\cos \theta}d\theta\text{.}\)

Solution.

\(\frac{\pi}{18}\text{.}\)

10.

\(\int_0^{2\pi}\frac{\cos 2\theta}{13-12\cos \theta}\,d\theta\text{.}\)

11.

\(\int_0^{2\pi}\frac{1}{(1+3\cos^2\theta)^2}\,d\theta\text{.}\)

Solution.

\(\frac{5\pi}{8}\text{.}\)

12.

\(\int_0^{2\pi}\frac{1}{(1+8\cos^2\theta)^2}\,d\theta\text{.}\)

13.

\(\int_0^{2\pi}\frac{\cos^23\theta}{5-4\cos 2\theta}\,d\theta\text{.}\)

Solution.

\(\frac{3\pi}{8}\text{.}\)

14.

\(\int_0^{2\pi}\frac{\cos^23\theta}{5-3\cos 2\theta}\,d\theta\text{.}\)

15.

\(\int_0^{2\pi}\frac{1}{a\cos \theta +b\sin \theta +d}\,d\theta\text{,}\) where \(a, \, b\text{,}\) and \(d\) are real, and \(a^2+b^2\lt d^2\text{.}\)

Solution.

\(\frac{2\pi}{\sqrt{d^2-a^2-b^2}}\text{.}\)

16.

\(\int_0^{2\pi}\frac{1}{a\cos^2\theta +b\sin^2\theta+d}\,d\theta\text{,}\)where \(a, \, b\text{,}\) and \(d\) are real, \(a>d\text{,}\) and \(b>d\text{.}\)

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