In=∫(1-x^2)^n/2dx 分部积分
=[x*(1-x^2)^n/2]-∫x*n*(-2x)*(1-x^2)^(n-1)/2dx
前面一项代入0和1都为0,则
In=n*∫x^2*(1-x^2)^(n-1)dx=2n*∫x^2*(1-x^2)^(n-1)/2dx
那么,
(2n+1)In= 2n*In+In
=2n*∫(1-x^2)^n/2dx + 2n*∫x^2*(1-x^2)^(n-1)/2dx
=2n*∫(1-x^2)*(1-x^2)^(n-1)/2dx + 2n*∫x^2*(1-x^2)^(n-1)/2dx
=2n*∫(1-x^2)^(n-1)/2dx
=2n*I(n-1)
所以,In=I(n-1)*2n/(2n+1)
I(n-1)=I(n-2)*(2n-2)/(2n-1)
… = …
I2=I1*4/5
I1=I0*2/3
可得 In=I0*(2n)!/(2n+1)!
而I0=∫(1-x^2)^0/2dx=1/2
故In=(1/2)*(2n)!/(2n+1)!或(1/2)*4^n*(n!)^2/(2n+1)!
I99= (1/2)*198!/199!