# Primitive of Exponential of a x by Power of Cosine of b x

## Theorem

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

## Proof

With a view to expressing the primitive in the form:

$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

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

and let:

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

Then:

 $\displaystyle \int e^{a x} \cos^n b x \rd x$ $=$ $\displaystyle \cos^n b x \paren {\frac {e^{a x} } a} - \int \paren {\frac {e^{a x} } a} \paren {-n b \cos^{n - 1} b x \sin b x} \rd x + C$ Integration by Parts $\text {(1)}: \quad$ $\displaystyle$ $=$ $\displaystyle \frac {e^{a x} \cos^n b x} a + \frac {n b} a \int e^{a x} \cos^{n - 1} b x \sin b x \rd x + C$ Primitive of Constant Multiple of Function
$\displaystyle \int e^{a x} \cos^{n - 1} b x \sin b x \rd x = \frac {e^{a x} \cos^{n - 1} b x \paren {a \sin b x - b \cos b x} } {a^2 + n b^2} - \frac {\paren {n - 1} a b} {a^2 + n b^2} \paren {\int e^{a x} \cos^n b x \rd x - \int e^{a x} \cos^{n - 2} b x \rd x} + C$

Hence:

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

$\blacksquare$