这是我们学过的微积分里面推导的两个公式,第一个的第一个用换元可以证明但第二个就不会了。。。还有下面的也不会求大神给出证明过程,!!!!!!!!!
还有下面的呢个。貌似要用分部积分的呢两个。。。。求详细。。。!!!
追答首先来证明下J(0,pi/2) sin^n(x) dx = J(0,pi/2) cos^n(x) dx,令t = pi/2 - x ,则
J(0,pi/2) sin^n(x) dx = J(pi/2,0) sin^n(pi/2 - t) d(pi/2 - t) = J(pi/2,0) cos^n(t) -dt = J(0, pi/2) cos^n(t) -dt = J(0,pi/2) cos^n(x) dx
令An = J(0,pi/2) sin^n(x) dx = J(0,pi/2) cos^n(x) dx,则
当n>=2时
An
= J(0,pi/2) sin^n (x) dx = J(0,pi/2) sin^(n-1) (x) -dcosx
= -sin^(n-1) (x) *cos(x) |(0,pi/2)- J(0,pi/2) cosx d(-sin^(n-1) (x))
= 0 - J(0,pi/2) (-(n-1)sin^(n-2) (x)) cos^2 (x) dx
= (n-1)J(0,pi/2) sin^(n-2) (x) (1- sin^2 (x)) dx
= (n-1)(An-2 - An)
An = (n-1)/n An-2
= (n-1)/n * (n-3)/(n-2) An-4
=...
=(n-1)/n * (n-3)/(n-2) * ... * (n- 2k + 1)/(n- 2k + 2) *An-2k
当n为奇数, n- 2k的 最小自然数为1时
An = (n-1)/n An-2
= (n-1)/n * (n-3)/(n-2) An-4
=...
=(n-1)/n * (n-3)/(n-2) * ... * 2/3 * A1
当n为偶数,n-2k的最小自然数为0时
An = (n-1)/n An-2
= (n-1)/n * (n-3)/(n-2) An-4
=...
=(n-1)/n * (n-3)/(n-2) * ... * 1/2 * A0
A1 = J(0,pi/2) sinx dx = 1
A0 = J(0,pi/2) dx = pi/2
代入An得证