Primitive of Exponential of a x by Power of Sine of b x/Lemma 2

Lemma for Primitive of $e^{a x} \sin^n b x$

$\dfrac {a^2 + n b^2} a e^{a x} \sin^n b x - \dfrac {n b} a e^{a x} \sin^{n - 1} b x \left({a \cos b x + b \sin b x}\right) = e^{a x} \sin^{n - 1} b x \left({a \sin b x - n b \cos b x}\right)$

Proof

\(\ds \)

\(=\)

\(\ds \frac {a^2 + n b^2} a e^{a x} \sin^n b x - \frac {n b} a e^{a x} \sin^{n - 1} b x \left({a \cos b x + b \sin b x}\right)\)

\(\ds \)

\(=\)

\(\ds \frac {a^2 + n b^2} a e^{a x} \sin^{n - 1} b x \sin b x - \frac {n b} a e^{a x} \sin^{n - 1} b x a \cos b x - \frac {n b} a e^{a x} \sin^{n - 1} b x b \sin b x\)

\(\ds \)

\(=\)

\(\ds e^{a x} \sin^{n - 1} b x \left({\frac {a^2 + n b^2} a \sin b x - \frac {n b} a a \cos b x - \frac {n b} a b \sin b x}\right)\)

\(\ds \)

\(=\)

\(\ds e^{a x} \sin^{n - 1} b x \left({a \sin b x + \frac {n b^2} a \sin b x - n b \cos b x - \frac {n b^2} a \sin b x}\right)\)

\(\ds \)

\(=\)

\(\ds e^{a x} \sin^{n - 1} b x \left({a \sin b x - n b \cos b x}\right)\)

$\blacksquare$