A254932 - OEIS

%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