Calculus/Integration techniques/Integration by Complexifying

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Integration techniques/Integration by Complexifying

This technique requires an understanding and recognition of complex numbers. Specifically Euler's formula:

\cos \theta + i\cdot \sin \theta = e^{i \cdot \theta}

Recognize, for example, that the real portion:

\mathrm{Re}\{ e^{i \cdot \theta} \} = \cos \theta

Given an integral of the general form:

\int e^{x} \cos {2x} \; dx

We can complexify it:

\int \mathrm{Re}\{ e^{x} (\cos {2x} + i\cdot \sin {2x}) \} \; dx

\int \mathrm{Re}\{ e^{x} (e^{i 2x}) \} \; dx

With basic rules of exponents:

\int \mathrm{Re}\{ e^{x + i2x} \} \; dx

It can be proven that the "real portion" operator can be moved outside the integral:

\mathrm{Re}\{ \int e^{x(1 + 2i)} \; dx \}

The integral easily evaluates:

\mathrm{Re}\{ \frac{e^{x(1 + 2i)}}{1 + 2i} \}

Multiplying and dividing by (1-2i):

\mathrm{Re} \{ \frac{1 - 2i}{5} e^{x(1 + 2i)} \}

Which can be rewritten as:

\mathrm{Re} \{ \frac{1 - 2i}{5} e^{x} e^{i2x} \}

Applying Euler's forumula:

\mathrm{Re} \{ \frac{1 - 2i}{5} e^{x} (\cos 2x + i\cdot \sin 2x) \}


\mathrm{Re} \{ \frac{e^{x}}{5} (\cos 2x +2 \sin 2x) + i\cdot \frac{e^{x}}{5} (\sin 2x -2 \cos 2x) \}

Taking the Real part of this expression:

 \frac{e^{x}}{5} (\cos 2x +2 \sin 2x)


\int e^{x} \cos {2x} \; dx = \frac{e^{x}}{5} (\cos 2x +2 \sin 2x)+C