Primitive of Exponential of a x by Sine of b x/Proof 4

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


Let $a, b, x \in \R$ be real numbers.

Suppose $a \ne 0 \ne b$.

Denote $\ds f_1 = \map \exp {a x} \map \cos {b x}$, $f_2 = \map \exp {a x} \map \sin {b x}$.

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

Let $\struct {\map {\CC^1} \R, +, \, \cdot \,}_\R$ be 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 $x$.

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:

$\ds \mathbf D^{-1} = \frac 1 {a^2 + b^2} \begin{bmatrix}

a & -b \\ b & a \end{bmatrix}$


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

a & -b \\ b & a \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \frac 1 {a^2 + b^2} \begin{bmatrix} -b \\ a \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:

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

Integrate with respect to $x$:

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

where $C$ is an arbitrary constant.

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