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du/dx=iexp(ix) thus (-i/u)du=dx |
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:<math>\frac{1}{2}\int \frac{6 + e^{2ix} + e^{-2ix} }{e^{ix} + e^{-ix} + e^{3ix} + e^{-3ix}} \, dx.</math>
If we now make the [[integration by substitution|substitution]] ''u'' = ''e''<sup>''ix''</sup>, the result is the integral of a [[rational function]]:
:<math>\frac{-i}{2}\int \frac{1+6u^2 + u^4}{1 + u^2 + u^4 + u^6}\,du.</math>
Any [[rational function]] is integrable (using, for example, [[partial fractions in integration|partial fractions]]), and therefore any fraction involving trigonometric functions may be integrated as well.
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