Integration: reduction formula

I(n-2)=∫ (a+bcosx)^2/(a+bcosx)^n dx
=b^2∫(cosx)^2/(a+bcosx)^n dx+2a∫(bcosx+a-0.5a)/(a+bcosx)^ndx
=b^2∫(cosx)^2/(a+bcosx)^n+2a I(n-1)-a^2 I(n)
so, b^2∫(cosx)^2/(a+bcosx)^n dx= a^2 I(n)-2a I(n-1)+I(n-2) -----(A)

b^2∫(sinx)^2/(a+bcosx)^ndx (integration by parts)
=bsinx/[(n-1)(a+bcosx)^(n-1)]-1/(n-1)∫(bcosx+a -a)/(a+bcosx)^(n-1)dx
=bsinx/[(n-1)(a+bcosx)^(n-1)]-1/(n-1) I(n-2)+ a/(n-1) I(n-1)
=p(x)-1/(n-1) I(n-2)+a/(n-1) I(n-1) ----(B)

(A)+(B):
b^2I(n)= a^2I(n)-(2n-3)a/(n-1) I(n-1)+(n-2)/(n-1) I(n-2)+p(x)
so,I(n)=[(2n-3)a/(n-1) I(n-1)- (n-2)/(n-1) I(n-2)-p(x)]/(a^2-b^2)
thus I(n)=[5(2n-3)/(n-1) I(n-1)- (n-2)/(n-1) I(n-2) -p(x)]/16

If I(n)=∫[0~π] 1/(5+4cosx)^n dx, then
I(n)=[5(2n-3)/(n-1) I(n-1)- (n-2)/(n-1) I(n-2)]/16 ----(C)

I(0)=∫[0~π] 1dx=π
I(1)=∫[0~π] 1/(5+3cosx)dx (set t=tan(x/2))
=∫[0~∞] 1/(4+t^2) dt=π/4
by(C): I(2)=[5 I(1)]/16=5π/64
by(C): I(3)=[(15/2) I(2)- I(1)/2]/16= 59π/2^11
by(C): I(4)=[(25/3) I(3)-(2/3) I(2)]/16= 385π/2^15=385π/32768

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