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

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Theorem

$\displaystyle \int x e^{a x} \sin b x \ \mathrm d x = \frac {x e^{a x} \left({a \sin b x - b \cos bx}\right)} {a^2 + b^2} - \frac {e^{a x} \left({\left({a^2 - b^2}\right) \sin b x - 2 a b \cos bx}\right)} {\left({a^2 + b^2}\right)^2} + C$


Proof

With a view to expressing the primitive in the form:

$\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \ \mathrm d x$

let:

\(\displaystyle u\) \(=\) \(\displaystyle x\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d u} {\mathrm d x}\) \(=\) \(\displaystyle 1\) Derivative of Identity Function


and let:

\(\displaystyle \frac {\mathrm d v} {\mathrm d x}\) \(=\) \(\displaystyle e^{a x} \sin b x\)
\(\displaystyle \implies \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle \frac {e^{a x} \left({a \sin b x - b \cos bx}\right)} {a^2 + b^2}\) Primitive of $e^{a x} \sin b x$


Then:

\(\displaystyle \int x e^{a x} \sin bx \ \mathrm d x\) \(=\) \(\displaystyle x \left({\frac {e^{a x} \left({a \sin b x - b \cos bx}\right)} {a^2 + b^2} }\right) - \int \left({\frac {e^{a x} \left({a \sin b x - b \cos bx}\right)} {a^2 + b^2} }\right) \ \mathrm d x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle \frac {x e^{a x} \left({a \sin b x - b \cos bx}\right)} {a^2 + b^2}\) Linear Combination of Integrals
\(\displaystyle \) \(\) \(\, \displaystyle - \, \) \(\displaystyle \frac a {a^2 + b^2} \int e^{a x} \sin b x \ \mathrm d x + \frac b {a^2 + b^2} \int e^{a x} \cos b x \ \mathrm d x + C\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {x e^{a x} \left({a \sin b x - b \cos bx}\right)} {a^2 + b^2}\)
\(\displaystyle \) \(\) \(\, \displaystyle - \, \) \(\displaystyle \frac a {a^2 + b^2} \left({\frac {e^{a x} \left({a \sin b x - b \cos bx}\right)} {a^2 + b^2} }\right)\) Primitive of $e^{a x} \sin b x$
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \frac b {a^2 + b^2} \left({\frac {e^{a x} \left({a \cos b x + b \sin bx}\right)} {a^2 + b^2} }\right) + C\) Primitive of $e^{a x} \cos b x$
\(\displaystyle \) \(=\) \(\displaystyle \frac {x e^{a x} \left({a \sin b x - b \cos bx}\right)} {a^2 + b^2} - \frac {e^{a x} \left({\left({a^2 - b^2}\right) \sin b x - 2 a b \cos bx}\right)} {\left({a^2 + b^2}\right)^2} + C\) simplifying

$\blacksquare$

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Sources