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

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

$\dfrac {a^2 + n b^2} a e^{a x} \cos^n b x + \dfrac {n b} a e^{a x} \cos^{n - 1} b x \paren {a \sin b x - b \cos b x} = e^{a x} \cos^{n - 1} b x \paren {a \cos b x + n b \sin b x}$

Proof

\(\ds \)

\(\)

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

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

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

$\blacksquare$