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:

and let:

Then:

From Primitive of $e^{a x} \sin^{n - 1} b x \cos b x$: Lemma 1:


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

Hence: