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Evaluate the indefinite integral

$$I(a,b)=\int \sin(ax) \sin^b(x)\mathrm{d}x \hspace{40pt} a,b\in\mathbb{N}$$

How do we evaluate the above indefinite integral?

Here is a question with $a=2015$ and $b=2013$, I was thinking of generalising this integral , so I tried using complex numbers by letting $z=\cos x+i\sin x$ , therefore $\mathrm{d}z=i ~z \mathrm{d}x$ and $\sin(ax)=\Im{(z^a)}$, so our integral converts to $$\Im{\left[\int z^a \left(\frac{z^2-1}{2iz}\right)^b \frac{\mathrm{d}z}{iz}\right] }=\frac{1}{2^{b}}\Im{\left[\frac{1}{i^{b+1}}\int z^{a-b-1}(z^2-1)^b\mathrm{d}z\right]} $$

Should we proceed further by binomial theorem?

Or possibly, a recurrence relation can be made...

Or it could be that it is not possible to do by hand...

V.G
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  • This may be useful: https://math.stackexchange.com/questions/456899/integrating-int-0-pi-2-cosax-cosbx-dx – VIVID Jul 28 '20 at 16:10
  • @VIVID, I am asking for Indefinite Integral not a definite integral. – V.G Jul 28 '20 at 16:11
  • According to Wolf, it involves the hypergeometric function. – Ty. Jul 28 '20 at 16:52
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    @Ty. , but wolframalpha assumes $a$ and $b$ to be reals, maybe there is a nice result for natural $a$ and $b$... – V.G Jul 28 '20 at 16:56
  • I would definite try $$\sin^n(x)=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^n=\frac{i}{2^n}\sum_{k=0}^{n}\binom{n}{k}\cos\left((n-2k)(x-\tfrac{\pi}{2})\right).$$ – Sangchul Lee Jul 28 '20 at 17:18
  • @SangchulLee, sir that is definitely helpful. Converting an exponential to a linear function indeed does the job. Also, on a different note, I wanted to ask you that if in a particular problem, I want your opinion/answer , where can I ask you? or if I want an answer from a particular person...should I just comment anywhere where they will see....like I did now...or something else... – V.G Jul 28 '20 at 17:26
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    Although I understand your passion to math, please be understandable that I do not want to accept any private requests. My activity here is purely for my recreation, and I think that peer-to-peer requests would harm the very reason I enjoy this community (as well as it is contrary to what this community is really for). – Sangchul Lee Jul 28 '20 at 17:37
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    Sure, no problem sir, I understand :) – V.G Jul 28 '20 at 17:40

1 Answers1

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Case $1$: $b$ is even

Then $\int\sin ax\sin^{2n}x~dx$

$=\int\dfrac{C_n^{2n}\sin ax}{4^n}~dx+\int\sum\limits_{k=1}^n\dfrac{(-1)^kC_{n+k}^{2n}\cos2kx\sin ax}{2^{2n-1}}~dx$

(according to https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Power-reduction_formulae)

$=\int\dfrac{(2n)!\sin ax}{4^n(n!)^2}~dx+\int\sum\limits_{k=1}^n\dfrac{(-1)^k(2n)!\sin((2k+a)x)}{4^n(n+k)!(n-k)!}~dx-\int\sum\limits_{k=1}^n\dfrac{(-1)^k(2n)!\sin((2k-a)x)}{4^n(n+k)!(n-k)!}~dx$

$=\sum\limits_{k=1}^n\dfrac{(-1)^k(2n)!\cos((2k-a)x)}{4^n(n+k)!(n-k)!(2k-a)}-\sum\limits_{k=1}^n\dfrac{(-1)^k(2n)!\cos((2k+a)x)}{4^n(n+k)!(n-k)!(2k+a)}-\dfrac{(2n)!\cos ax}{4^n(n!)^2a}~dx+C$

Case $2$: $b$ is odd

Then $\int\sin ax\sin^{2n+1}x~dx$

$=\int\sum\limits_{k=0}^n\dfrac{(-1)^kC_{n+k+1}^{2n+1}\sin((2k+1)x)\sin ax}{4^n}~dx$

(according to https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Power-reduction_formulae)

$=\int\sum\limits_{k=0}^n\dfrac{(-1)^k(2n+1)!\cos((2k-a+1)x)}{2^{2n+1}(n-k)!(n+k+1)!}~dx-\int\sum\limits_{k=0}^n\dfrac{(-1)^k(2n+1)!\cos((2k+a+1)x)}{2^{2n+1}(n-k)!(n+k+1)!}~dx$

$=\sum\limits_{k=0}^n\dfrac{(-1)^k(2n+1)!\sin((2k-a+1)x)}{2^{2n+1}(n-k)!(n+k+1)!(2k-a+1)}-\sum\limits_{k=0}^n\dfrac{(-1)^k(2n+1)!\sin((2k+a+1)x)}{2^{2n+1}(n-k)!(n+k+1)!(2k+a+1)}+C$

Harry Peter
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