Primitive of Exponential of a x by Cosine of b x

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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 1

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

$\blacksquare$


Proof 2

\(\ds \int e^{a x} e^{i b x} \rd x\) \(=\) \(\ds i \int e^{a x} \sin b x \rd x + \int e^{a x} \cos b x \rd x\) Euler's Formula
\(\ds \leadsto \ \ \) \(\ds \int e^{a x} \cos b x \rd x\) \(=\) \(\ds \map \Re {\int e^{\paren {a + i b} x} \rd x}\)
\(\ds \) \(=\) \(\ds \map \Re {\frac {e^{\paren {a + i b} x} } {a + i b} } + C\) Primitive of Exponential of a x
\(\ds \) \(=\) \(\ds \map \Re {\frac {\paren {a - i b} e^{\paren {a + i b} x} } {a^2 + b^2} } + C\) multiplying through by $\dfrac {a - i b} {a - i b}$
\(\ds \) \(=\) \(\ds \map \Re {\frac {i a e^{a x} \sin b x + a e^{a x} \cos b x - i b \paren {i e^{a x} \sin b x + e^{a x} \cos b x} } {a^2 + b^2} } + C\) Euler's Formula
\(\ds \) \(=\) \(\ds \map \Re {\frac {i \paren {a e^{a x} \sin b x - b e^{a x} \cos b x} + \paren {a e^{a x} \cos b x + b e^{a x} \sin b x} } { a^2 + b^2} } + C\)
\(\ds \) \(=\) \(\ds \frac {e^{a x} \paren {a \cos b x + b \sin b x} } {a^2 + b^2} + C\) isolating real part

$\blacksquare$


Proof 3

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.

$\blacksquare$


Also see


Sources

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