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

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Theorem

$\ds \int e^{a x} \sin^n b x \rd x = \frac {e^{a x} \sin^{n - 1} b x} {a^2 + n^2 b^2} \paren {a \sin b x - n b \cos b x} + \frac {n \paren {n - 1} b^2} {a^2 + n^2 b^2} \int e^{a x} \sin^{n - 2} b x \rd x + C$


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:

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


and let:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds e^{a x}\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds \frac {e^{a x} } a\) Primitive of $e^{a x}$


Then:

\(\ds \int e^{a x} \sin^n b x \rd x\) \(=\) \(\ds \sin^n b x \paren {\frac {e^{a x} } a} - \int \paren {\frac {e^{a x} } a} n b \sin^{n - 1} b x \cos b x \rd x + C\) Integration by Parts
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \frac {e^{a x} \sin^n b x} a - \frac {n b} a \int e^{a x} \sin^{n - 1} b x \cos b x \rd x + C\) Primitive of Constant Multiple of Function


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

$\ds \int e^{a x} \sin^{n - 1} b x \cos b x \rd x = \frac {e^{a x} \sin^{n - 1} b x \paren {a \cos b x + b \sin b x} } {a^2 + n b^2} + \frac {\paren {n - 1} a b} {a^2 + n b^2} \paren {\int e^{a x} \sin^n b x \rd x - \int e^{a x} \sin^{n - 2} b x \rd x} + C$


Hence:

\(\ds \) \(\) \(\ds \int e^{a x} \sin^n b x \rd x\)
\(\ds \) \(=\) \(\ds \frac {e^{a x} \sin^n b x} a - \frac {n b} a \int e^{a x} \sin^{n - 1} b x \cos b x \rd x + C\) from $(1)$
\(\ds \) \(=\) \(\ds \frac {e^{a x} \sin^n b x} a - \frac {n b} {a \paren {a^2 + n b^2} } e^{a x} \sin^{n - 1} b x \paren {a \cos b x + b \sin b x}\) Primitive of $e^{a x} \sin^n b x$: Lemma 1
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \frac {n \paren {n - 1} b^2} {a^2 + n b^2} \int e^{a x} \sin^n b x \rd x + \frac {n \paren {n - 1} b^2} {a^2 + n b^2} \int e^{a x} \sin^{n - 2} b x \rd x + C\)
\(\ds \leadsto \ \ \) \(\ds \) \(\) \(\ds \paren {1 + \frac {n \paren {n - 1} b^2} {a^2 + n b^2} } \int e^{a x} \sin^n b x \rd x\)
\(\ds \) \(=\) \(\ds \frac {e^{a x} \sin^n b x} a - \frac {n b} {a \paren {a^2 + n b^2} } e^{a x} \sin^{n - 1} b x \paren {a \cos b x + b \sin b x}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \frac {n \paren {n - 1} b^2} {a^2 + n b^2} \int e^{a x} \sin^{n - 2} b x \rd x + C\)
\(\ds \leadsto \ \ \) \(\ds \) \(\) \(\ds \frac {a^2 + n b^2 + n^2 b^2 - n b^2} {a^2 + n b^2} \int e^{a x} \sin^n b x \rd x\)
\(\ds \) \(=\) \(\ds \frac {e^{a x} \sin^n b x} a - \frac {n b} {a \paren {a^2 + n b^2} } e^{a x} \sin^{n - 1} b x \paren {a \cos b x + b \sin b x}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \frac {n \paren {n - 1} b^2} {a^2 + n b^2} \int e^{a x} \sin^{n - 2} b x \rd x + C\)
\(\ds \leadsto \ \ \) \(\ds \) \(\) \(\ds \paren {a^2 + n^2 b^2} \int e^{a x} \sin^n b x \rd x\)
\(\ds \) \(=\) \(\ds \frac {a^2 + n b^2} a e^{a x} \sin^n b x - \frac {n b} a e^{a x} \sin^{n - 1} b x \paren {a \cos b x + b \sin b x}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {n \paren {n - 1} b^2} \int e^{a x} \sin^{n - 2} b x \rd x + C\)
\(\ds \leadsto \ \ \) \(\ds \) \(\) \(\ds \paren {a^2 + n^2 b^2} \int e^{a x} \sin^n b x \rd x\)
\(\ds \) \(=\) \(\ds e^{a x} \sin^{n - 1} b x \paren {a \sin b x - n b \cos b x}\) Primitive of $e^{a x} \sin^n b x$: Lemma 2
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {n \paren {n - 1} b^2} \int e^{a x} \sin^{n - 2} b x \rd x + C\)
\(\ds \leadsto \ \ \) \(\ds \) \(\) \(\ds \int e^{a x} \sin^n b x \rd x\)
\(\ds \) \(=\) \(\ds \frac {e^{a x} \sin^{n - 1} b x} {a^2 + n^2 b^2} \paren {a \sin b x - n b \cos b x} + \frac {n \paren {n - 1} b^2} {a^2 + n^2 b^2} \int e^{a x} \sin^{n - 2} b x \rd x + C\)

$\blacksquare$


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