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Series for integral of sin^m(ax)cos^n(ax)...?

i used to have, in my old formula book, a series formula for this that i discovered after writing out one of these questions (at the time, i didn`t know of the reduction formulas).

Unlike reduction formula, this series would allow me to write each term straight down.

So far i have (please don`t laugh): (EDIT)

k=1msm(2k1)cn+k(...?)ak(m+n)k\displaystyle -\sum_{k=1}^m \frac{s^{m-(2k-1)}c^{n+k}(...?)}{a^{k}(m+n)^k} where s= sin, c= cos

for the ?? part, the first term is just:

(m1)(m-1)

the second term has in it:

(m1)(m2)(m-1)(m-2)

3rd term:

(m1)(m2)(m3)(m-1)(m-2)(m-3)

last 3 terms of the m`th part are.....3 x 2 x 1

the trouble is, i don`t know how to bring this into the series notation thingy - i.e. don`t know how to express each term as something to the power of a term in k.

any hints?

thanks for any help/clarity you can give
(edited 11 years ago)
Reply 1
maybe this won`t work, as there are 4 different cases for it, which makes it impractical (m odd n even, m even n odd, both even, both odd)
Reply 2
Original post by Hasufel
i used to have, in my old formula book, a series formula for this that i discovered after writing out one of these questions (at the time, i didn`t know of the reduction formulas).

Unlike reduction formula, this series would allow me to write each term straight down.

So far i have (please don`t laugh): (EDIT)

k=1msm(2k1)cn+k(...?)ak(m+n)k\displaystyle -\sum_{k=1}^m \frac{s^{m-(2k-1)}c^{n+k}(...?)}{a^{k}(m+n)^k} where s= sin, c= cos

for the ?? part, the first term is just:

(m1)(m-1)

the second term has in it:

(m1)(m2)(m-1)(m-2)

3rd term:

(m1)(m2)(m3)(m-1)(m-2)(m-3)

last 3 terms of the m`th part are.....3 x 2 x 1

the trouble is, i don`t know how to bring this into the series notation thingy - i.e. don`t know how to express each term as something to the power of a term in k.

any hints?

thanks for any help/clarity you can give


(m1)!(mk1)!\frac{(m-1)!}{(m-k-1)!} from k=1 to m-1

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