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.

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

From Differentiation of Exponential of a x by Cosine of b x and Exponential of a x by Sine of b x wrt x as Invertible Matrix, $D$ is expressible as:

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

and is invertible.

By Inverse of Matrix is Scalar Product of Adjugate by Reciprocal of Determinant:

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

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:

$f_1 = \dfrac \d {\d x} \dfrac {a f_1 + b f_2} {a^2 + b^2}$

Integrating with respect to $x$:

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

where $C$ is an arbitrary constant.

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

$\blacksquare$

Sources

• 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 2.1$: Continuous and linear maps. Linear transformations