All binomial are
A: I=∫xMdx(xN+k)P
We do
x=th
dx=thM th-1t=dt·th(M+1)-1 I=∫dt·t(h(M+1)-1[tNh+k]P we do hN=2 I=∫dt·t2(M+1)/N-1)[t2+k]p we do interest there, is exponent 2 and when k/A2 will be 1
I=∫dwchw·shqw·ch2pw
Now this two factors remind sh·ch t=Achw)
when we do t=Ashw
if k>0 (if not we do
I=∫dw·ch[2(M+1)/N-1]w·sh2P+1
Cfr. Wikipedia series of shmchn when m & n>0, nm=2P+1 n=2(M+1)/N-1
B: similarly [k+tNh]P
sinmcosn, when
then t=Ashw.
I=∫dw·sinqw·cos2p+1w P can be ½, 3/2, 5/2.. etc
I=∫dt·t[h(M+1)N-1][k-tNh]P
t=Asinw iffis
I=∫dwshqw·ch2P+1w q=2(M+1)/N-1 n=2p+1 m=2(M+1)/N both enters but also 2(M+1)/N
(M+1)/N can be multiple of ½ like P.
Much more than Chebishev!!! A) HIPERBOLICS k>0 shmwchnw, they are integrated in wikipedia.
For products of hyperbolics see at the end of the anex: I(m,n)= ∫du·shmu·chnu=solution when m & n >0 (both enters) =shm-1u·chn+1u/(m+n)-(m-1)/(n+1) ∫shm-2u·chnu·du h=2/N Mh+h-1=(2M/N+2/N-N/N=2(M+1)/N-1 I=∫d(shw)sh[2(M+1)N-1]w[sh2w+1]P, if k<0 ;,we would do t=Achw I=∫dwchwsh[2M+1)/N-1]w·ch2Pw=∫dwsh[2M+1)/M-1]w·ch2P+1w
wikipedia solves
I=∫dz·zM(z2-1)P the lists of wikipedia only ask for enter exponents. 2P exponent for ch, and M for sh (or when ∫dz·zM(z2+1)P then M is for sh and 2P for ch
If one exponent is negative we have other formulas:
And the circulars
The condition for being integrable is than n and m were enters. M &N enters.
m and p can be, besides enters, a half of an enter.
(Chebyshev believed to have demonstrated that there was no more cases, than his three enters)