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

## Theorem

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

## Proof

With a view to expressing the primitive in the form:

$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

 $\ds u$ $=$ $\ds x$ $\ds \leadsto \ \$ $\ds \frac {\d u} {\d x}$ $=$ $\ds 1$ Derivative of Identity Function

and let:

 $\ds \frac {\d v} {\d x}$ $=$ $\ds e^{a x} \cos b x$ $\ds \leadsto \ \$ $\ds v$ $=$ $\ds \frac {e^{a x} \paren {a \cos b x + b \sin b x} } {a^2 + b^2}$ Primitive of $e^{a x} \cos b x$

Then:

 $\ds \int x e^{a x} \sin b x \rd x$ $=$ $\ds x \paren {\frac {e^{a x} \paren {a \cos b x + b \sin b x} } {a^2 + b^2} } - \int \paren {\frac {e^{a x} \paren {a \cos b x + b \sin b x} } {a^2 + b^2} } \rd x + C$ Integration by Parts $\ds$ $=$ $\ds \frac {x e^{a x} \paren {a \cos b x + b \sin b x} } {a^2 + b^2}$ Linear Combination of Primitives $\ds$  $\, \ds - \,$ $\ds \frac a {a^2 + b^2} \int e^{a x} \cos b x \rd x - \frac b {a^2 + b^2} \int e^{a x} \sin b x \rd x + C$ $\ds$ $=$ $\ds \frac {x e^{a x} \paren {a \cos b x + b \sin b x} } {a^2 + b^2}$ $\ds$  $\, \ds - \,$ $\ds \frac a {a^2 + b^2} \paren {\frac {e^{a x} \paren {a \cos b x + b \sin b x} } {a^2 + b^2} }$ Primitive of $e^{a x} \cos b x$ $\ds$  $\, \ds - \,$ $\ds \frac b {a^2 + b^2} \paren {\frac {e^{a x} \paren {a \sin b x - b \cos b x} } {a^2 + b^2} } + C$ Primitive of $e^{a x} \sin b x$ $\ds$ $=$ $\ds \frac {x e^{a x} \paren {a \cos b x + b \sin b x} } {a^2 + b^2} - \frac {e^{a x} \paren {\paren {a^2 - b^2} \cos b x - 2 a b \sin b x} } {\paren {a^2 + b^2}^2} + C$ simplifying

$\blacksquare$