记该积分为Jn,则由楼上的公式
sin2nx = sin(2n-1)xcosx+cos(2n-1)xsinx
= (1/2)[sin2nx+sin2(n-1)x]+cos(2n-1)xsinx,
有
Jn = ∫(sin2nx/sinx)dx
= (1/2)∫[sin2nx+sin2(n-1)x]/sinxdx + ∫cos(2n-1)xdx
= (1/2)[Jn+J(n-1)] + sin(2n-1)x/(2n-1)
= J(n-1) + 2sin(2n-1)x/(2n-1),
这是一个
递推公式,递推计算,可得
Jn = J(n-2) + 2sin(2n-3)x/(2n-3)+ 2sin(2n-1)x/(2n-1)
= …
= J1 + 2Σ(k=2~n)sin(2k-1)x/(2k-1)
= ∫(sin2x)/sinxdx + 2Σ(k=2~n)sin(2k-1)x/(2k-1)
= 2∫cosxdx + 2Σ(k=2~n)sin(2k-1)x/(2k-1)
= 2sinx + 2Σ(k=2~n)sin(2k-1)x/(2k-1) + C
= 2Σ(k=1~n)sin(2k-1)x/(2k-1) + C。