# Primitive of Exponential of a x by Sine of b x

## Theorem

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

## Proof 1

 $\ds \int e^{a x} \sin b x \rd x$ $=$ $\ds \frac {e^{a x} \sin b x} a - \frac b a \int e^{a x} \cos b x \rd x$ Primitive of $e^{a x} \sin b x$: Lemma $\ds$ $=$ $\ds \frac {e^{a x} \sin b x} a - \frac b a \paren {\frac {e^{a x} \cos b x} a + \frac b a \int e^{a x} \sin b x \rd x}$ Primitive of $e^{a x} \cos b x$: Lemma $\ds$ $=$ $\ds \frac {e^{a x} a \sin b x - e^{a x} b \cos b x} {a^2} - \frac {b^2} {a^2} \int e^{a x} \sin b x \rd x$ simplifying $\ds \leadsto \ \$ $\ds \paren {1 + \frac {b^2} {a^2} } \int e^{a x} \sin b x \rd x$ $=$ $\ds \frac {e^{a x} \left({a \sin b x - b \cos b x}\right)} {a^2}$ simplifying $\ds \leadsto \ \$ $\ds \frac {a^2 + b^2} {a^2} \int e^{a x} \sin b x \rd x$ $=$ $\ds \frac {e^{a x} \paren {a \sin b x - b \cos b x} } {a^2}$ common denominator $\ds \leadsto \ \$ $\ds \int e^{a x} \sin b x \rd x$ $=$ $\ds \frac {e^{a x} \paren {a \sin b x - b \cos b x} } {a^2 + b^2}$ multiplying by $\dfrac {a^2} {a^2 + b^2}$

$\blacksquare$

## Proof 2

 $\ds \cos b x + i \sin b x$ $=$ $\ds e^{i b x}$ Euler's Formula $\ds \leadsto \ \$ $\ds e^{a x} \cos b x + i e^{a x} \sin b x$ $=$ $\ds e^{a x} e^{i b x}$ multiplying both sides by $e^{a x}$ $\ds$ $=$ $\ds e^{\paren {a + i b} x}$ Exponent Combination Laws $\ds \leadsto \ \$ $\ds \int e^{a x} \cos b x \, \d x + i \int e^{a x} \sin b x \d x$ $=$ $\ds \int e^{\paren {a + i b} x} \d x$ Linear Combination of Complex Integrals $\ds$ $=$ $\ds \frac 1 {a + i b} e^{\paren {a + i b} x} + C$ Primitive of $e^{a x}$ $\ds$ $=$ $\ds \frac {a - i b} {a^2 + b^2} e^{\paren {a + i b} x} + C$ multiplying top and bottom by $a - i b$ $\ds$ $=$ $\ds \frac {a - i b} {a^2 + b^2} e^{a x} e^{i b x} + C$ Exponent Combination Laws $\ds$ $=$ $\ds \frac {a - i b} {a^2 + b^2} e^{a x} \paren {\cos b x + i \sin b x} + C$ Euler's Formula $\ds$ $=$ $\ds \frac a {a^2 + b^2} e^{a x} \cos b x - \frac {i b} {a^2 + b^2} e^{a x} \cos b x$ $\ds$  $\, \ds + \,$ $\ds \frac {i a} {a^2 + b^2} e^{a x} \sin b x + \frac b {a^2 + b^2} e^{a x} \sin b x + C$

The result follows from equating imaginary parts.

$\blacksquare$

## Proof 3

 $\ds \int e^{a x} \sin b x \rd x$ $=$ $\ds \int e^{a x} \paren {\frac {e^{i b x} - e^{-i b x} } {2 i} } \rd x$ Sine Exponential Formulation $\ds$ $=$ $\ds \frac 1 {2 i} \int e^{a x} \paren {e^{i b x} - e^{-i b x} } \rd x$ Primitive of Constant Multiple of Function $\ds$ $=$ $\ds \frac 1 {2 i} \int \paren {e^{a x} e^{i b x} - e^{a x} e^{-i b x} } \rd x$ $\ds$ $=$ $\ds \frac 1 {2 i} \int \paren {e^{a x + i b x} - e^{a x - i b x} } \rd x$ Product of Powers $\ds$ $=$ $\ds \frac 1 {2 i} \int e^{a x + i b x} \rd x - \frac 1 {2 i} \int e^{a x - i b x} \rd x$ Linear Combination of Integrals $\ds$ $=$ $\ds \frac 1 {2 i} \int e^{\paren {a + i b} x} \rd x - \frac 1 {2 i} \int e^{\paren {a - i b} x} \rd x$ $\ds$ $=$ $\ds \frac 1 {2 i} \frac {e^{\paren {a + i b} x} } {a + i b} - \frac 1 {2 i} \frac {e^{\paren {a - i b} x} } {a - i b} + C$ Primitive of $e^{a x}$ $\ds$ $=$ $\ds \frac 1 {2 i} \frac {e^{a x + i b x} } {a + i b} - \frac 1 {2 i} \frac {e^{a x - i b x} } {a - i b} + C$ $\ds$ $=$ $\ds \frac 1 {2 i} \frac {e^{a x} e^{i b x} } {a + i b} - \frac 1 {2 i} \frac {e^{a x} e^{-i b x} } {a - i b} + C$ Product of Powers $\ds$ $=$ $\ds \frac 1 {2 i} \frac {e^{a x} e^{i b x} \paren {a - i b} } {\paren {a + i b} \paren {a - i b} } - \frac 1 {2 i} \frac {e^{a x} e^{-i b x} \paren {a + i b} } {\paren {a - i b} \paren {a + i b} } + C$ $\ds$ $=$ $\ds \frac {e^{a x} e^{i b x} \paren {a - i b} - e^{a x} e^{-i b x} \paren {a + i b} } {2 i \paren {a + i b} \paren {a - i b} } + C$ $\ds$ $=$ $\ds \frac {e^{a x} e^{i b x} \paren {a - i b} - e^{a x} e^{-i b x} \paren {a + i b} } {2 i \paren {a^2 + b^2} } + C$ Product of Complex Number with Conjugate $\ds$ $=$ $\ds \frac {a e^{a x} e^{i b x} - i b e^{a x} e^{i b x} - a e^{a x} e^{-i b x} - i b e^{a x} e^{-i b x} } {2 i \paren {a^2 + b^2} } + C$ $\ds$ $=$ $\ds \frac {e^{a x} } {\paren {a^2 + b^2} } \paren {\frac {a e^{i b x} - i b e^{i b x} - a e^{-i b x} - i b e^{-i b x} } {2 i} } + C$ $\ds$ $=$ $\ds \frac {e^{a x} } {\paren {a^2 + b^2} } \paren {a \frac {e^{i b x} - e^{-i b x} } {2 i} - b \frac {e^{i b x} + e^{-i b x} } 2} + C$ $\ds$ $=$ $\ds \frac {e^{a x} } {\paren {a^2 + b^2} } \paren {a \frac {e^{i b x} - e^{-i b x} } {2 i} - b \cos b x} + C$ Cosine Exponential Formulation $\ds$ $=$ $\ds \frac {e^{a x} } {\paren {a^2 + b^2} } \paren {a \sin b x - b \cos b x} + C$ Sine Exponential Formulation $\ds$ $=$ $\ds \frac {e^{a x} \paren {a \sin b x - b \cos b x} } {a^2 + b^2} + C$

$\blacksquare$

## Proof 4

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 $x$.

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

and is invertible.

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

Then:

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

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

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

where $C$ is an arbitrary constant.

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

$\blacksquare$