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I was able to obtain some indefinite integrals for the trigonometric function manually,

$\Gamma^{pq}_n = \int \frac {\cos^p x \sin^q x} {(1+A \cos x)^n} dx$

for $(p,q,n)\geq 0$ in recursive form. For example,

$\Gamma^{11}_n=\frac{1}{A}(\Gamma^{01}_{n-1}-\Gamma^{01}_{n}),$ for $n>2$

$\Gamma^{21}_n=\frac{1}{A^2}(\Gamma^{01}_{n}-2\Gamma^{01}_{n-1}+\Gamma^{01}_{n-2}),$ for $n>2$.

How can I generate similar recursive integrals for the trigonometric function

$\Delta^{pq}_n = \int \frac {\cos^p x \sin^q x} {(1+A \cos (x-a))^n} dx$

for $(p,q,n)\geq 0$ using Mathematica?

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  • $\begingroup$ ...your 2nd recursive form isn't true! $\endgroup$ Dec 31, 2017 at 14:12
  • $\begingroup$ A typo in the second recursive form. Included a minus sign. $\endgroup$
    – Harish
    Dec 31, 2017 at 15:32
  • $\begingroup$ I can't verify your second recursion form ( Simplify[\[Gamma][2, 1, n , A] == -1/ A^2 (\[Gamma][0, 1, n , A] - 2 \[Gamma][0, 1, n - 1 , A] + \[Gamma][0, 1, n - 2, A] ) /. a -> 0] ) $\endgroup$ Dec 31, 2017 at 16:55
  • $\begingroup$ Gam[p_, q_, n_] := (Cos[x]^p Sin[x]^q)/(1 + A Cos[x])^n; Simplify[Gam[2, 1,n] == (Gam[0, 1, n] - 2 Gam[0, 1, n - 1] + Gam[0, 1, n - 2])/A^2] This is verifying. $\endgroup$
    – Harish
    Dec 31, 2017 at 17:34

1 Answer 1

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Lets have a look at the integrand

γ[p_, q_, n_, A_] := ((Cos[x]^p Sin[x]^q)/(1 + A Cos[x - a])^n)

If we expand numerator and denominator by (1 + A Cos[x - a]) (as you probably did to achieve your recursion) keeping in mind

Cos[x - a] == Cos[a] Cos[x] + Sin[a] Sin[x]

you'll get the recursion

Simplify[ γ[p, q, n, A] == γ[p, q, n + 1, A] + 
                           A Cos[a] γ[p + 1, q, n + 1, A] +  
                           A Sin[a] γ[p , q + 1, n + 1, A]]
(* True*)

This result also holds for the integral you're considering!

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