Question

一道定积分证明题

求证：
∫0~(π/2) sin^2n xdx=（1/2）*（3/4）*（5/6）*...*[(2n-3)/(2n-2)]*[(2n-1)/2n]*π/2
不胜感激！

Answer 1

设
an=∫ sin^n xdx；(x:0~π/2)
=-(sinx)^(n-1)-(n-1)[an-a(n-2)]
=-(n-1)[an-a(n-2)]
an/a(n-2)=(n-1)/n

a2n/ao
=a2n/a(n-2)*a(n-2)/a(n-4)*……*a2/a0
=（1/2）*（3/4）*（5/6）*...*[(2n-3)/(2n-2)]*[(2n-1)/2n]
ao=π/2
得
∫0~(π/2) sin^2n xdx=（1/2）*（3/4）*（5/6）*...*[(2n-3)/(2n-2)]*[(2n-1)/2n]*π/2

Answer 2

将被积函数展开成无穷乘积，然后积分。或
I(2n)=∫0~(π/2) sin^2n xdx=-sin^(2n-1)xcosx|0~(π/2)+(2n-1)∫0~(π/2) sin^(2n-2) xcos^2 xdx=-(2n-1)I(2n)+(2n-1)I(2n-2),I(2n)=[(2n-1)/(2n)]I(2n-2),I(2n-2)=[(2n-3)/(2n-2)]I(2n-4),...,I(2)=(1/2)I(0),I(0)=∫0~(π/2) dx=x|[0~(π/2)]=π/2。I(2n)=[(2n-1)/(2n)][(2n-3)/(2n-2)]...(3/4)(1/2)π/2