Primitive of Exponential of a x by Sine of b x

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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 Primitives
\(\ds \) \(=\) \(\ds \frac 1 {2 i} \int e^{\paren {a + i b} x} \rd x - \frac 1 {2 i} \int e^{\paren {a - i b} x} \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 {2 i} \frac {e^{\paren {a + i b} x} } {a + i b} - \frac 1 {2 i} \frac {e^{\paren {a - i b} x} } {a - i b} + C\) 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 $x$.

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:

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

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:

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

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.

$\blacksquare$


Also see


Sources

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