Editing Contour integration (section)

===Example 3 – trigonometric integrals===
Certain substitutions can be made to integrals involving [[trigonometric functions]], so the integral is transformed into a rational function of a complex variable and then the above methods can be used in order to evaluate the integral.

As an example, consider
<math display=block>\int_{-\pi}^\pi \frac{1 }{ 1 + 3 (\cos t)^2} \,dt.</math>

We seek to make a substitution of {{math|1=''z'' = ''e<sup>it</sup>''}}. Now, recall
<math display=block>\cos t = \frac12 \left(e^{it}+e^{-it}\right) = \frac12 \left(z+\frac{1}{z}\right)</math>
and
<math display=block>\frac{dz}{dt} = iz,\ dt = \frac{dz}{iz}.</math>

Taking {{mvar|C}} to be the unit circle, we substitute to get:

<math>\begin{align}
\oint_C \frac{1}{ 1 + 3 \left(\frac12 \left(z+\frac{1}{z}\right)\right)^2} \,\frac{dz}{iz} &= \oint_C \frac{1 }{ 1 + \frac34 \left(z+\frac{1}{z}\right)^2}\frac{1}{iz} \,dz \\
&= \oint_C \frac{-i}{ z+\frac34 z\left(z+\frac{1}{z}\right)^2}\,dz \\
&= -i \oint_C \frac{dz}{ z+\frac34 z\left(z^2+2+\frac{1}{z^2}\right)} \\
&= -i \oint_C \frac{dz}{ z+\frac34 \left(z^3+2z+\frac{1}{z}\right)} \\
&= -i \oint_C \frac{dz}{ \frac34 z^3+\frac52 z+\frac{3}{4z}} \\
&= -i \oint_C \frac{4}{ 3z^3+10z+\frac{3}{z}}\,dz \\
&= -4i \oint_C \frac{dz}{ 3z^3+10z+\frac{3}{z}} \\
&= -4i \oint_C \frac{z}{ 3z^4+10z^2+3 } \,dz \\
&= -4i \oint_C \frac{z}{ 3\left(z+\sqrt{3}i\right)\left(z-\sqrt{3}i\right)\left(z+\frac{i}{\sqrt 3}\right)\left(z-\frac{i}{\sqrt 3}\right)}\,dz \\
&= -\frac{4i}{3} \oint_C \frac{z}{\left(z+\sqrt{3}i\right)\left(z-\sqrt{3}i\right)\left(z+\frac{i}{\sqrt 3}\right)\left(z-\frac{i}{\sqrt 3}\right)}\,dz.
\end{align}</math>

The singularities to be considered are at <math>\tfrac{\pm i}{\sqrt{3}}.</math> Let {{math|''C''<sub>1</sub>}} be a small circle about <math>\tfrac{i}{\sqrt{3}},</math> and {{math|''C''<sub>2</sub>}} be a small circle about <math>\tfrac{-i}{\sqrt{3}}.</math> Then we arrive at the following:
<math display=block>\begin{align}
& -\frac{4i}{3} \left [\oint_{C_1} \frac{\frac{z}{\left(z+\sqrt{3}i\right)\left(z-\sqrt{3}i\right)\left(z+\frac{i}{\sqrt 3} \right)}}{z-\frac{i}{\sqrt 3}}\,dz +\oint_{C_2} \frac{\frac{z}{\left(z+\sqrt{3}i\right)\left(z-\sqrt{3}i\right)\left(z-\frac{i}{\sqrt 3}\right)}}{z+\frac{i}{\sqrt 3}} \, dz \right ] \\
= {} & -\frac{4i}{3} \left[ 2\pi i \left[\frac{z}{\left(z+\sqrt{3}i\right)\left(z-\sqrt{3}i\right)\left(z+\frac{i}{\sqrt 3}\right)}\right]_{z=\frac{i}{\sqrt 3}} + 2\pi i \left[\frac{z}{\left(z+\sqrt{3}i\right)\left(z-\sqrt{3}i\right)\left(z-\frac{i}{\sqrt 3}\right)} \right]_{z=-\frac{i}{\sqrt 3}}\right] \\
= {} & \frac{8\pi}{3} \left[\frac{\frac{i}{\sqrt 3}}{\left(\frac{i}{\sqrt 3}+\sqrt{3}i\right)\left(\frac{i}{\sqrt 3}-\sqrt{3}i\right)\left(\frac{i}{\sqrt 3}+\frac{i}{\sqrt 3}\right)} + \frac{-\frac{i}{\sqrt 3}}{\left(-\frac{i}{\sqrt 3}+\sqrt{3}i\right)\left(-\frac{i}{\sqrt 3}-\sqrt{3}i\right)\left(-\frac{i}{\sqrt 3}-\frac{i}{\sqrt 3}\right)} \right] \\
= {} & \frac{8\pi}{3} \left[\frac{\frac{i}{\sqrt 3}}{\left(\frac{4}{\sqrt 3}i\right)\left(-\frac{2}{i\sqrt{3}}\right)\left(\frac{2}{\sqrt{3}i}\right)}+\frac{-\frac{i}{\sqrt 3}}{\left(\frac{2}{\sqrt 3}i\right)\left(-\frac{4}{\sqrt 3}i\right)\left(-\frac{2}{\sqrt 3}i\right)}\right] \\
= {} & \frac{8\pi}{3}\left[\frac{\frac{i}{\sqrt 3}}{i\left(\frac{4}{\sqrt 3}\right)\left(\frac{2}{\sqrt 3}\right)\left(\frac{2}{\sqrt 3}\right)}+\frac{-\frac{i}{\sqrt 3}}{-i\left(\frac{2}{\sqrt 3}\right)\left(\frac{4}{\sqrt 3}\right)\left(\frac{2}{\sqrt 3}\right)}\right] \\
= {} & \frac{8\pi}{3}\left[\frac{\frac{1}{\sqrt 3}}{\left(\frac{4}{\sqrt 3}\right)\left(\frac{2}{\sqrt 3}\right)\left(\frac{2}{\sqrt 3}\right)}+\frac{\frac{1}{\sqrt 3}}{\left(\frac{2}{\sqrt 3}\right)\left(\frac{4}{\sqrt 3}\right)\left(\frac{2}{\sqrt 3}\right)}\right] \\
= {} & \frac{8\pi}{3}\left[\frac{\frac{1}{\sqrt 3}}{\frac{16}{3\sqrt{3}}}+\frac{\frac{1}{\sqrt 3}}{\frac{16}{3\sqrt{3}}} \right] \\
= {} & \frac{8\pi}{3}\left[\frac{3}{16} + \frac{3}{16} \right] \\
= {} & \pi.
\end{align}</math>