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

## Theorem

$\displaystyle \int e^{a x} \sin^n b x \ \mathrm d x = \frac {e^{a x} \sin^{n - 1} b x} {a^2 + n^2 b^2} \left({a \sin b x - n b \cos b x}\right) + \frac {n \left({n - 1}\right) b^2} {a^2 + n^2 b^2} \int e^{a x} \sin^{n - 2} b x \ \mathrm d x + C$

## Proof

With a view to expressing the primitive in the form:

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

let:

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

and let:

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

Then:

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

Hence:

 $\displaystyle$  $\displaystyle \int e^{a x} \sin^n b x \ \mathrm d x$ $\displaystyle$ $=$ $\displaystyle \frac {e^{a x} \sin^n b x} a - \frac {n b} a \int e^{a x} \sin^{n - 1} b x \cos b x \ \mathrm d x + C$ from $(1)$ $\displaystyle$ $=$ $\displaystyle \frac {e^{a x} \sin^n b x} a - \frac {n b} {a \left({a^2 + n b^2}\right)} e^{a x} \sin^{n - 1} b x \left({a \cos b x + b \sin b x}\right)$ Primitive of $e^{a x} \sin^n b x$: Lemma 1 $\displaystyle$  $\, \displaystyle - \,$ $\displaystyle \frac {n \left({n - 1}\right) b^2} {a^2 + n b^2} \int e^{a x} \sin^n b x \ \mathrm d x + \frac {n \left({n - 1}\right) b^2} {a^2 + n b^2} \int e^{a x} \sin^{n - 2} b x \ \mathrm d x + C$ $\displaystyle \implies \ \$ $\displaystyle$  $\displaystyle \left({1 + \frac {n \left({n - 1}\right) b^2} {a^2 + n b^2} }\right) \int e^{a x} \sin^n b x \ \mathrm d x$ $\displaystyle$ $=$ $\displaystyle \frac {e^{a x} \sin^n b x} a - \frac {n b} {a \left({a^2 + n b^2}\right)} e^{a x} \sin^{n - 1} b x \left({a \cos b x + b \sin b x}\right)$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \frac {n \left({n - 1}\right) b^2} {a^2 + n b^2} \int e^{a x} \sin^{n - 2} b x \ \mathrm d x + C$ $\displaystyle \implies \ \$ $\displaystyle$  $\displaystyle \frac {a^2 + n b^2 + n^2 b^2 - n b^2} {a^2 + n b^2} \int e^{a x} \sin^n b x \ \mathrm d x$ $\displaystyle$ $=$ $\displaystyle \frac {e^{a x} \sin^n b x} a - \frac {n b} {a \left({a^2 + n b^2}\right)} e^{a x} \sin^{n - 1} b x \left({a \cos b x + b \sin bx}\right)$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \frac {n \left({n - 1}\right) b^2} {a^2 + n b^2} \int e^{a x} \sin^{n - 2} b x \ \mathrm d x + C$ $\displaystyle \implies \ \$ $\displaystyle$  $\displaystyle \left({a^2 + n^2 b^2}\right) \int e^{a x} \sin^n b x \ \mathrm d x$ $\displaystyle$ $=$ $\displaystyle \frac {a^2 + n b^2} a e^{a x} \sin^n b x - \frac {n b} a e^{a x} \sin^{n - 1} b x \left({a \cos b x + b \sin b x}\right)$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \left({n \left({n - 1}\right) b^2}\right) \int e^{a x} \sin^{n - 2} b x \ \mathrm d x + C$ $\displaystyle \implies \ \$ $\displaystyle$  $\displaystyle \left({a^2 + n^2 b^2}\right) \int e^{a x} \sin^n b x \ \mathrm d x$ $\displaystyle$ $=$ $\displaystyle e^{a x} \sin^{n - 1} b x \left({a \sin b x - n b \cos b x}\right)$ Primitive of $e^{a x} \sin^n b x$: Lemma 2 $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \left({n \left({n - 1}\right) b^2}\right) \int e^{a x} \sin^{n - 2} b x \ \mathrm d x + C$ $\displaystyle \implies \ \$ $\displaystyle$  $\displaystyle \int e^{a x} \sin^n b x \ \mathrm d x$ $\displaystyle$ $=$ $\displaystyle \frac {e^{a x} \sin^{n - 1} b x} {a^2 + n^2 b^2} \left({a \sin b x - n b \cos b x}\right) + \frac {n \left({n - 1}\right) b^2} {a^2 + n^2 b^2} \int e^{a x} \sin^{n - 2} b x \ \mathrm d x + C$

$\blacksquare$