Integration by parts

I(2n)
=∫x^2n sinx dx
=-∫x^2n dcosx
=-x^2n cosx+∫cosx d x^2n
=-x^2n cosx+2n∫cosx x^(2n-1)dx
=-x^2n cosx+2n∫x^(2n-1)dsinx
=-x^2n cosx+2nx^(2n-1) sinx - 2n∫sinx dx^(2n-1)
=-x^2n cosx+2nx^(2n-1) sinx - 2n(2n-1)∫sinx x^(2n-2)dx
=-x^2n cosx+2nx^(2n-1) sinx - I(2n-2)
=-x^2n cosx+2nx^(2n-1) sinx -[ -x^(2n-2)cosx + (2n-2)x^(2n-3)sinx - I(2n-4) ]
=-x^2n cosx+2nx^(2n-1) sinx + x^(2n-2)cosx - (2n-2)x^(2n-3)sinx + I(2n-4)
......
=cosx[-x^(2n)+2n(2n-1)x^(2n-2)-2n(2n-1)(2n-2)(2n-3)x^(2n-4)+2n(2n-1)...(2n-5)x^(2n-6)+...—2n(2n-1)...3x^2]+sinx[2nx^(2n-1)-2n(2n-1)(2n-2)x^(2n-3)+...＋2n(2n-1)(2n-2)...2x]—2n!I(2) when n=2k+1 (k is N)
or
=cosx[-x^(2n)+2n(2n-1)x^(2n-2)-2n(2n-1)(2n-2)(2n-3)x^(2n-4)+2n(2n-1)...(2n-5)x^(2n-6)+...+2n(2n-1)...3x^2]+sinx[2nx^(2n-1)-2n(2n-1)(2n-2)x^(2n-3)+...-2n(2n-1)(2n-2)...2x]+2n!I(2) when n=2k (k is N)
=自己再計埋I（2）

第2題情況差不多，只不過有少少不同，所以在此不長列出來，因為打出來太麻煩啦...

maths is fun!!

Just an indefinite integral

∫_[0~π/2] x^n sinx dx ?

I(2n) = -x^2n cosx+2nx^(2n-1) sinx - I(2n-2)
I(2n)+I(2n-2) = -x^2n cosx+2nx^(2n-1) sinx = f(x)

2011-11-17 08:54:45 補充：
Let J(n)=[(-1)^n][I(2n)]
=> [J(n)]/[(-1)^n]+[J(n-1)]/[(-1)^(n-1)] = f(x)
=> J(n) - J(n-1) = (-1)^n[f(x)]
=> ∑(n=n,n=1) J(n) - J(n-1) = ∑(n=n,n=1) [f(x)][(-1)^(n)]
=> J(n) - J(0) =∑(n=n,n=1) [f(x)]/[(-1)^(n)]
=> J(n) - I(0) =∑(n=n,n=1) [f(x)]/[(-1)^(n)]
(Since J(0)=(-1^(0))(I(0))=I(0))