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

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

$\blacksquare$


Also see


Sources