sin n次方 x 的降幂公式

Posted 2021-01-28 likeghee

A(n) = ∫ sin?x dx
= ∫ sin??1xsinx dx
= - ∫ sin??1x d(cosx)
= - sin??1xcosx + ∫ cosx ? d(sin??1)
= - sin??1xcosx + (n - 1)∫ cosx ? sin??2x ? cosx dx
= - sin??1xcosx + (n - 1)∫ sin??2x ? (1 - sin2x) dx
= - sin??1xcosx + (n - 1)A(n - 2) - (n - 1)A(n)
[1 + (n - 1)]A(n) = - sin??1xcosx + (n - 1)A(n - 2)
A(n) = (- 1/n)sin??1xcosx + [(n - 1)/n]A(n - 2),这就是让sin?x降幂的公式
∴
∫ sin?x dx
= (- 1/6)sin?xcosx + (5/6)∫ sin?x dx
= (- 1/6)sin?xcosx + (5/6)[(- 1/4)sin3xcosx + (3/4)∫ sin2x dx]
= (- 1/6)sin?xcosx - (5/24)sin3xcosx + (15/24)[(- 1/2)sinxcosx + (1/2)∫ dx]
= (- 1/6)sin?xcosx - (5/24)sin3xcosx - (15/48)sinxcosx + 15x/48 + C
特别地,当下限是0,上限是π/2时,有
∫(0→π/2) sin?x dx = ∫(0→π/2) cos?x dx = 
{ (n - 1)!/n! ,n是正奇数
{ (n - 1)!/n! ? π/2,n是正偶数
是Wallis公式