Editing Contour integration (section)

===Example 3a – trigonometric integrals, the general procedure===
The above method may be applied to all integrals of the type
<math display=block> \int_0^{2\pi} \frac{P\big(\sin(t),\sin(2t),\ldots,\cos(t),\cos(2t),\ldots\big)}{Q\big(\sin(t),\sin(2t),\ldots,\cos(t),\cos(2t),\ldots\big)}\, dt</math>

where {{mvar|P}} and {{mvar|Q}} are polynomials, i.e. a rational function in trigonometric terms is being integrated. Note that the bounds of integration may as well be {{pi}} and −{{pi}}, as in the previous example, or any other pair of endpoints 2{{pi}} apart.

The trick is to use the substitution {{math|1=''z'' = ''e<sup>it</sup>''}} where {{math|1=''dz'' = ''ie<sup>it</sup> dt''}} and hence
<math display=block> \frac{1}{iz} \,dz = dt.</math>

This substitution maps the interval {{closed-closed|0, 2π}} to the unit circle. Furthermore,
<math display=block> \sin(k t) = \frac{e^{i k t} - e^{- i k t}}{2 i} = \frac{z^k - z^{-k}}{2i}</math>
and
<math display=block> \cos(k t) = \frac{e^{i k t} + e^{- i k t}}{2} = \frac{z^k + z^{-k}}{2}</math>
so that a rational function {{math|''f''(''z'')}} in {{mvar|z}} results from the substitution, and the integral becomes
<math display=block> \oint_{|z|=1} f(z) \frac{1}{iz}\, dz </math>
which is in turn computed by summing the residues of {{math|''f''(''z''){{sfrac|1|''iz''}}}} inside the unit circle.

[[Image:TrigonometricToComplex.png|right]]
The image at right illustrates this for
<math display=block> I = \int_0^\frac{\pi}{2} \frac{1}{1 + (\sin t)^2}\, dt,</math>
which we now compute. The first step is to recognize that
<math display=block> I = \frac14 \int_0^{2\pi} \frac{1}{1 + (\sin t)^2} \,dt.</math>

The substitution yields
<math display=block> \frac{1}{4} \oint_{|z|=1} \frac{4 i z}{z^4 - 6z^2 + 1}\, dz = \oint_{|z|=1} \frac{i z}{z^4 - 6z^2 + 1}\, dz.</math>

The poles of this function are at {{math|1 ± {{sqrt|2}}}} and {{math|−1 ± {{sqrt|2}}}}. Of these, {{math|1 + {{sqrt|2}}}} and {{math|−1 − {{sqrt|2}}}} are outside the unit circle (shown in red, not to scale), whereas {{math|1 − {{sqrt|2}}}} and {{math|−1 + {{sqrt|2}}}} are inside the unit circle (shown in blue). The corresponding residues are both equal to {{math|−{{sfrac|''i''{{sqrt|2}}|16}}}}, so that the value of the integral is
<math display=block> I = 2 \pi i \; 2 \left( - \frac{\sqrt{2}}{16} i \right) = \pi \frac{\sqrt{2}}{4}.</math>