# Primitive of Exponential of a x by Cosine of b x/Proof 3

## Theorem

$\ds \int e^{a x} \cos b x \rd x = \frac {e^{a x} \paren {a \cos b x + b \sin b x} } {a^2 + b^2} + C$

## Proof

Let $a, b \in \R_{>0}$ be real constants.

Let $f_1$ and $f_2$ be the real functions defined as:

 $\ds \forall x \in \R: \,$ $\ds \map {f_1} x$ $=$ $\ds \map \exp {a x} \map \cos {b x}$ $\ds \map {f_2} x$ $=$ $\ds \map \exp {a x} \map \sin {b x}$

Let $\map \CC \R$ denote the space of continuous real-valued functions.

Let $\struct {\map {\CC^1} \R, +, \, \cdot \,}_\R$ denote the vector space of continuously differentiable real-valued functions.

Let $S = \span \set {f_1, f_2} \subset \map {\CC^1} \R$ be a vector space.

Let $D : S \to S$ be the derivative with respect to $x$.

$\mathbf D = \begin{bmatrix} a & b \\ -b & a \end{bmatrix}$

and is invertible.

$\mathbf D^{-1} = \dfrac 1 {a^2 + b^2} \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$

Then:

 $\ds \mathbf D^{-1} \begin{bmatrix} 1 \\ 0 \end {bmatrix}$ $=$ $\ds \dfrac 1 {a^2 + b^2} \begin{bmatrix} a & -b \\ b & a \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ $\ds$ $=$ $\ds \dfrac 1 {a^2 + b^2} \begin{bmatrix} a \\ b \end{bmatrix}$

Application of $\mathbf D$ on both sides on the left and writing out explicitly in terms of $f_1$ and $f_2$ yields:

$f_1 = \dfrac \d {\d x} \dfrac {a f_1 + b f_2} {a^2 + b^2}$
$\ds \int f_1 \rd x = \frac {a f_1 + b f_2} {a^2 + b^2} + C$

where $C$ is an arbitrary constant.

Substitute definitions of $f_1$ and $f_2$ to get the desired result.

$\blacksquare$