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%I #30 Jul 18 2021 03:03:23
%S -1,-9,1,-150,25,-3,-3675,735,-147,15,-119070,26460,-6804,1215,-105,
%T -4802490,1143450,-343035,81675,-12705,945,-231891660,57972915,
%U -19324305,5521230,-1171170,159705,-10395,-13043905875,3381753375,-1217431215,395269875,-102477375,19348875,-2338875,135135,-837708621750,223388965800,-85293968760,30462131700,-9112603500,2130219000,-360498600,39054015,-2027025
%N Triangle used for the integral of odd powers of the sine and cosine functions.
%C This entry originated from a proposal by _Ozgur Cem Birler_ for the integral of the fourth power of sin(x). _Peter Bala_ suggested the use of triangles to cover all even and odd powers. See A254932.
%C The signed triangle T(n, k) appears in the formula
%C 2^(2*n)*(2*n+1)!!*int(sin^{2*n+1}x dx) = Sum_{k=0..n} T(n, k)*cos((2*k+1)*x), n >= 0.
%C See the Gradstein - Ryshik reference, 2.513 2., p. 168. Here (2*n+1)!! = A001147(n+1), n >= 0.
%C The unsigned triangle T(n, k) appears in the formula
%C 2^(2*n)* (2*n+1)!!*int(cos^{2*n+1}x dx) = Sum_{k=0..n} |T(n, k)|*sin((2*k+1)*x), n >= 0.
%C See the Gradstein & Ryshik reference, 2.513 4., p. 169.
%D I. S. Gradstein and I. M. Ryshik, Tables of series, products, and integrals, Volume 1, Verlag Harri Deutsch, 1981.
%H Wolfdieter Lang, <a href="/A254932/a254932.pdf">The first 9 rows and more.</a>
%F T(n, k) = (-1)^(k+1)*binomial(2*n+1,n-k)*(2*n+1)!!/(2*k+1), 0 <= k <= n, with the double factorials (2*n+1)!! = A001147(n+1).
%e The triangle T(n, k) begins:
%e n\k 0 1 2 4 5 6...
%e 0: -1
%e 1: -9 1
%e 2: -150 25 -3
%e 3: -3675 735 -147 15
%e 4: -119070 26460 -6804 1215 -105
%e 5: -4802490 1143450 -343035 81675 -12705 945
%e ...
%e For more rows see the link.
%e n=2: 2^4*(1*3*5)*int((sin x)^5 dx) = -150*cos(x) + 25*cos(3*x) - 3*cos(5*x),
%e that is: int((sin x)^5 dx) = -(5/8)*cos(x) + (5/48)*cos(3*x) - (1/80)*cos(5*x).
%e 2^4*(1*3*5)*int((cos x)^5 dx) = +150*sin(x) + 25*sin(3*x) + 3*sin(5*x),
%e that is: int((cos x)^5 dx) = (5/8)*sin(x) + (5/48)*sin(3*x) + (1/80)*sin(5*x).
%t T[n_, k_] := (-1)^(k+1)*Binomial[2*n+1, n-k]*(2*n+1)!!/(2*k+1); Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Feb 18 2015 *)
%Y Cf. A254933 (even powers).
%K sign,easy,tabl
%O 0,2
%A _Wolfdieter Lang_, Feb 15 2015
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