Primitive of Exponential of a x by Power of Cosine of b x/Lemma 1

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

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

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:

\(\ds u\)

\(=\)

\(\ds \cos^{n - 1} b x\)

\(\ds \leadsto \ \ \)

\(\ds \frac {\d u} {\d x}\)

\(=\)

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

Derivative of $\cos a x$, Derivative of Power, Chain Rule for Derivatives

and let:

\(\ds \frac {\d v} {\d x}\)

\(=\)

\(\ds e^{a x} \sin b x\)

\(\ds \leadsto \ \ \)

\(\ds v\)

\(=\)

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

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

Then:

\(\ds \)

\(\)

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

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

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

Integration by Parts

\(\ds \)

\(\)

\(\, \ds - \, \)

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

\(\ds \)

\(=\)

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

\(\ds \)

\(\)

\(\, \ds + \, \)

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

Linear Combination of Primitives

\(\ds \)

\(\)

\(\, \ds - \, \)

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

\(\ds \leadsto \ \ \)

\(\ds \)

\(\)

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

gathering terms

\(\ds \)

\(=\)

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

\(\ds \)

\(\)

\(\, \ds + \, \)

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

\(\ds \leadsto \ \ \)

\(\ds \)

\(\)

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

\(\ds \)

\(=\)

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

simplifying

\(\ds \)

\(\)

\(\, \ds + \, \)

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

\(\ds \)

\(=\)

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

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds \paren {n - 1} a b \int e^{a x} \cos^{n - 2} b x \paren {1 - \cos^2 b x} \rd x + C\)

Sum of Squares of Sine and Cosine

\(\ds \)

\(=\)

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

simplifying

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds \paren {n - 1} a b \int e^{a x} \cos^{n - 2} b x \rd x\)

\(\ds \)

\(\)

\(\, \ds - \, \)

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

\(\ds \leadsto \ \ \)

\(\ds \)

\(\)

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

\(\ds \)

\(=\)

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

simplifying

\(\ds \)

\(\)

\(\, \ds + \, \)

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

\(\ds \)

\(\)

\(\, \ds - \, \)

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

and so rearranging:

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

$\blacksquare$