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Q: Integration of $e^{ax}\cos^n bx$ and $e^{ax}\sin^n bx$

I know how to integration $e^{ax}\cos bx$ using $\cos bx=\frac{e^{ibx}+e^{-ibx}}{2}$.Using the same trick here I got $$\int e^{ax}\cos^n bx ~dx=\int e^{ax}\left(\frac{e^{ibx}+e^{-ibx}}{2}\right)^n~ dx$$But it doesn't look easy to solve.I google it and find the result only which is:

$$\int e^{ax}\cos^n bx ~dx=\frac{bn\sin (bx)+a\cos (bx)}{a^2+b^2n^2}e^{ax}\cos^{n-1}bx+\frac{b^2(n-1)n\int e^{ax}\cos^{n-2}bxdx}{a^2+b^2n^2}.$$

Can Anyone help me to figure out this. Any hints or solution will be appreciated.
Thanks in advance.

José Carlos Santos
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emonHR
  • 2,650

1 Answers1

1

(Differentiation is easier than integration)

Let $f_{m,a, b}(x)=f_m=e^{ax}\cos^m bx$. Take the derivative of $f_{m}$, \begin{align*} f'_{m}&=af_{m}-bmf_{m-1}\sin bx.\tag{1} \end{align*} So, if we take $g_{m}=af_{m}\color{magenta}+bmf_{m-1}\sin bx$, the derivative of $g_{m}$ will cancel $\color{blue}{\sin bx}$ out, i.e. \begin{align*} g'_{m}&=af'_{m}\hspace{8.2em}+bm\big(f'_{m-1}\sin bx\hspace{10.8em}+b\underbrace{f_{m-1}\cos bx}_{=f_{m}}\big)\\ g'_{m}&=a^2f_{m}-\color{blue}{abmf_{m-1}\sin bx}+\color{blue}{bm}\big(\color{blue}{af_{m-1}\sin bx}-b(m-1) f_{m-2}\color{orange}{\sin^2 bx}+bf_m\big)\\ g'_{m}&=(a^2+b^2m)f_m-b^2m(m-1)f_{m-2}\color{orange}{(1-\cos^2 bx)}\\ g'_{m}&=(a^2+b^2m^2)f_m-b^2m(m-1)f_{m-2}.\tag{2} \end{align*} Now, we assume $$I_{m}=\int f_m~\mathrm dx=\int e^{ax}\cos^m bx~\mathrm dx,$$ integrate $(2)$, we get the desired result

$$g_m=(a^2+b^2m^2)I_{m}-b^2m(m-1)I_{m-2}.$$

Tianlalu
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  • Thanks @Tianlalu Really appreciated.Can i use Differentiation to get Integration in this way always? – emonHR Nov 24 '18 at 09:05
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    Yes, we can always do so once we find the appropriate $f_m$. For example, for $\int \cos^n x~dx$, we may choose $f_n=\cos^{n-1}x \sin x$ and differentiate it to cancel $\sin x$ out. – Tianlalu Nov 24 '18 at 09:11