Calculus/Integration techniques/Integration by Complexifying

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

$\cos (\theta )+i\sin (\theta )=e^{\theta i}$

Recognize, for example, that the real portion:

$\text{Re}\left\{e^{\theta i}\right\}=\cos (\theta )$

Given an integral of the general form:

$\int e^x\cos (2x)dx$

We can complexify it:

$\int \text{Re}\left\{e^x\right(\cos (2x)+i\sin (2x)\left)\right\}dx$

$\int \text{Re}\{e^x(e^{2xi})\}dx$

With basic rules of exponents:

$\int \text{Re}\{e^{x+2ix}\}dx$

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

$\text{Re}\{\int e^{x(1+2i)}dx\}$

The integral easily evaluates:

$\text{Re}\left\{\frac{e^{x(1+2i)}}{1+2i}\right\}$

Multiplying and dividing by $1-2i$ :

$\text{Re}\left\{\frac{1-2i}{5}{e}^{x(1+2i)}\right\}$

Which can be rewritten as:

$\text{Re}\left\{\frac{1-2i}{5}{e}^x{e}^{2ix}\right\}$

Applying Euler's forumula:

$\text{Re}\left\{\frac{1-2i}{5}{e}^x\right(\cos (2x)+i\sin (2x)\left)\right\}$

Expanding:

$\text{Re}\left\{\frac{e^x}{5}\right(\cos (2x)+2\sin (2x))+i\cdot \frac{e^x}{5}(\sin (2x)-2\cos (2x)\left)\right\}$

Taking the Real part of this expression:

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

So:

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