General

[quote=BBoongBBoong]I think im close
I(n)=(integral 2pi to 0) (1+cosQ)^n dQ let Q=theta
Required to prove I(n+1)=((2n+1)/(n+1))I(n)
I(n+1)=(integral 2pi to 0)(1+cosQ)^n(1+cosQ) dQ
[b]I(n+1)=(integral 2pi to 0)(1+cosQ)^n+cosQ(cosQ)^n dQ[/b]
seperate integrals
I(n+1)=(integral 2pi to 0)(1+cosQ)^n dQ + (integral 2pi to 0) cosQ(cosQ)^n dQ
I(n+1)=I(n)+ (integral 2pi to 0) cosQ(cosQ)^n dQ
let (integral 2pi to 0) cosQ(cosQ)^n dQ be A - then integrate by parts
dv/dQ=cosQ --- v= sinQ
u=(1+cosQ)^n --- du/dQ= n(1+cosQ)^(n-1)*-sinQ
A= [sinQ*((1+cosQ)^n)]- (integral 2pi to 0) sinQ*n(1+cosQ)^(n-1)*-sinQ dQ
but sinQ^2=1-cosQ^2 and [sinQ*((1+cosQ)^n)]=0
so I(n+1)= I(n) + n(integral 2pi to 0)sinQ^2(1+cosQ)^(n-1) dQ
I(n+1)= I(n) + n(integral 2pi to 0)(1-cosQ^2)(1+cosQ)^(n-1) dQ let this be B
so in B I(n+1)=I(n)+n(integral 2pi to 0)(((1-cosQ^2)(1+cosQ)^n)/(1+cosQ)) dQ
but (1-cosQ^2)= (1+cosQ)(1-cosQ)
simplify into I(n+1)=I(n)+n(integral 2pi to 0)(1-cosQ)(1+cosQ)^n dQ
I(n+1)=I(n)+n(integral 2pi to 0)(1+cosQ)^n-cosQ(1+cosQ)^n dQ
I(n+1)=I(n)+n[(I(n)-(integral 2pi to 0)cosQ(1+cosQ)^n dQ]
look familiar. because (integral 2pi to 0)cosQ(1+cosQ)^n dQ=A
so we have recurring series
this was about as far as I went. maybe try limiting sum of series?[/quote]

Only read up to the bolded line but that line looks wrong. Please explain

EDIT: OP, @unnecessary, I think you might be able to use mathematical induction to solve this. It's 2am and I'm too tired to do it but try that first

Did you not try Google, but Basil BEFORE Google?