J(a,b)f(x)dx为函数f(x)从[a,b]的积分,pi为圆周率
sinx 在[0,1]之间,当x在[0,pi]之间 所以f(sinx)在[0,pi]之间连续,因此,f(sinx)在[0,pi]上可积.
同理可证f(cosx)在[0,pi/2]上可积.
xf(sinx)在[0,pi]上连续,因此xf(sinx)也可积.
1.令x = pi/2 - t
J(0,pi/2) f(sinx) dx = J(pi/2,0) f(sin(pi/2 - t)) d(pi/2 - t) = J(pi/2,0) f(cos t) -1 dt
= J(0,pi/2)f(cost)dt = J(0,pi/2)f(cosx)dx
2.令x = pi - t,则
J(0,pi) xf(sinx) dx = J(pi,0) (pi - t)f(sin(pi - t))d(pi - t) = J(pi,0) (pi - t)f(sint) -dt
= J(0,pi) (pi - t)f(sint)dt = J(0,pi) pi * sint dt - J(0,pi) t sint dt
=pi * J(0,pi) sinx dx - J(0,pi) x sinx dx
将等式右边 J(0,pi) x sinx dx,移到左边可得
2 J(0,pi) x sinx dx = pi * J(0,pi) sinx dx
J(0,pi) x sinx dx = pi/2 * J(0,pi) sinx dx
再问： 还有下面的呢个。貌似要用分部积分的呢两个。。。。求详细。。。！！！
再答： 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
