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

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Theorem

$\displaystyle \int e^{a x} \sin^n b x \ \mathrm d x = \frac {e^{a x} \sin^{n - 1} b x} {a^2 + n^2 b^2} \left({a \sin b x - n b \cos b x}\right) + \frac {n \left({n - 1}\right) b^2} {a^2 + n^2 b^2} \int e^{a x} \sin^{n - 2} b x \ \mathrm d x + C$


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:

\(\displaystyle u\) \(=\) \(\displaystyle \sin^n b x\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d u} {\mathrm d x}\) \(=\) \(\displaystyle n b \sin^{n - 1} b x \cos b x\) Derivative of $\sin a x$, Derivative of Power, Chain Rule for Derivatives


and let:

\(\displaystyle \frac {\mathrm d v} {\mathrm d x}\) \(=\) \(\displaystyle e^{a x}\)
\(\displaystyle \implies \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle \frac {e^{a x} } a\) Primitive of $e^{a x}$


Then:

\(\displaystyle \int e^{a x} \sin^n b x \ \mathrm d x\) \(=\) \(\displaystyle \sin^n b x \left({\frac {e^{a x} } a}\right) - \int \left({\frac {e^{a x} } a}\right) n b \sin^{n - 1} b x \cos b x \ \mathrm d x + C\) Integration by Parts
\(\text {(1)}: \quad\) \(\displaystyle \) \(=\) \(\displaystyle \frac {e^{a x} \sin^n b x} a - \frac {n b} a \int e^{a x} \sin^{n - 1} b x \cos b x \ \mathrm d x + C\) Primitive of Constant Multiple of Function


From Primitive of $e^{a x} \sin^{n - 1} b x \cos b x$: Lemma 1:

$\displaystyle \int e^{a x} \sin^{n - 1} b x \cos b x \ \mathrm d x = \frac {e^{a x} \sin^{n - 1} b x \left({a \cos b x + b \sin b x}\right)} {a^2 + n b^2} + \frac {\left({n - 1}\right) a b} {a^2 + n b^2} \left({\int e^{a x} \sin^n b x \ \mathrm d x - \int e^{a x} \sin^{n - 2} b x \ \mathrm d x}\right) + C$


Hence:

\(\displaystyle \) \(\) \(\displaystyle \int e^{a x} \sin^n b x \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {e^{a x} \sin^n b x} a - \frac {n b} a \int e^{a x} \sin^{n - 1} b x \cos b x \ \mathrm d x + C\) from $(1)$
\(\displaystyle \) \(=\) \(\displaystyle \frac {e^{a x} \sin^n b x} a - \frac {n b} {a \left({a^2 + n b^2}\right)} e^{a x} \sin^{n - 1} b x \left({a \cos b x + b \sin b x}\right)\) Primitive of $e^{a x} \sin^n b x$: Lemma 1
\(\displaystyle \) \(\) \(\, \displaystyle - \, \) \(\displaystyle \frac {n \left({n - 1}\right) b^2} {a^2 + n b^2} \int e^{a x} \sin^n b x \ \mathrm d x + \frac {n \left({n - 1}\right) b^2} {a^2 + n b^2} \int e^{a x} \sin^{n - 2} b x \ \mathrm d x + C\)
\(\displaystyle \implies \ \ \) \(\displaystyle \) \(\) \(\displaystyle \left({1 + \frac {n \left({n - 1}\right) b^2} {a^2 + n b^2} }\right) \int e^{a x} \sin^n b x \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {e^{a x} \sin^n b x} a - \frac {n b} {a \left({a^2 + n b^2}\right)} e^{a x} \sin^{n - 1} b x \left({a \cos b x + b \sin b x}\right)\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \frac {n \left({n - 1}\right) b^2} {a^2 + n b^2} \int e^{a x} \sin^{n - 2} b x \ \mathrm d x + C\)
\(\displaystyle \implies \ \ \) \(\displaystyle \) \(\) \(\displaystyle \frac {a^2 + n b^2 + n^2 b^2 - n b^2} {a^2 + n b^2} \int e^{a x} \sin^n b x \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {e^{a x} \sin^n b x} a - \frac {n b} {a \left({a^2 + n b^2}\right)} e^{a x} \sin^{n - 1} b x \left({a \cos b x + b \sin bx}\right)\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \frac {n \left({n - 1}\right) b^2} {a^2 + n b^2} \int e^{a x} \sin^{n - 2} b x \ \mathrm d x + C\)
\(\displaystyle \implies \ \ \) \(\displaystyle \) \(\) \(\displaystyle \left({a^2 + n^2 b^2}\right) \int e^{a x} \sin^n b x \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {a^2 + n b^2} a e^{a x} \sin^n b x - \frac {n b} a e^{a x} \sin^{n - 1} b x \left({a \cos b x + b \sin b x}\right)\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \left({n \left({n - 1}\right) b^2}\right) \int e^{a x} \sin^{n - 2} b x \ \mathrm d x + C\)
\(\displaystyle \implies \ \ \) \(\displaystyle \) \(\) \(\displaystyle \left({a^2 + n^2 b^2}\right) \int e^{a x} \sin^n b x \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle e^{a x} \sin^{n - 1} b x \left({a \sin b x - n b \cos b x}\right)\) Primitive of $e^{a x} \sin^n b x$: Lemma 2
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \left({n \left({n - 1}\right) b^2}\right) \int e^{a x} \sin^{n - 2} b x \ \mathrm d x + C\)
\(\displaystyle \implies \ \ \) \(\displaystyle \) \(\) \(\displaystyle \int e^{a x} \sin^n b x \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {e^{a x} \sin^{n - 1} b x} {a^2 + n^2 b^2} \left({a \sin b x - n b \cos b x}\right) + \frac {n \left({n - 1}\right) b^2} {a^2 + n^2 b^2} \int e^{a x} \sin^{n - 2} b x \ \mathrm d x + C\)

$\blacksquare$


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