Euler's Formula/Real Domain/Proof 6

Theorem

Let $\theta \in \R$ be a real number.

Then:

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

Proof

\(\ds \int \frac {\d x} {\sqrt {x^2 + 1} }\)

\(=\)

\(\ds \map \ln {\sqrt {x^2 + 1} + x }\)

Primitive of Reciprocal of Root of x squared plus a squared/Logarithm Form and $a = 1$

\(\ds \int \frac {i\cos \theta \rd \theta } {\sqrt {1 - \sin^2 \theta} }\)

\(=\)

\(\ds \map \ln {\sqrt {1 - \sin^2 \theta} + i \sin \theta }\)

$x = i \sin \theta$; $\d x = i\cos \theta \rd \theta$

\(\ds \int \frac {i\cos \theta \rd \theta } {\sqrt {\cos^2 \theta} }\)

\(=\)

\(\ds \map \ln {\sqrt {\cos^2 \theta} + i \sin \theta }\)

Sum of Squares of Sine and Cosine

\(\ds i\int \rd \theta\)

\(=\)

\(\ds \map \ln {\cos \theta + i \sin \theta }\)

\(\ds i \theta\)

\(=\)

\(\ds \map \ln {\cos \theta + i \sin \theta }\)

\(\ds e^{i \theta}\)

\(=\)

\(\ds \cos \theta + i \sin \theta\)

$\blacksquare$