求解一道高数题，要详细过程，谢谢！！！！！

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。

解：∫sin2nx/sinxdx
=∫sin2x/sinxdx
=∫2cosxdx
=-2sinx+C
希望能帮到你！

sin2nx=sin(2n-1)xcosx+cos(2n-1)xsinx
=1/2(sin2nx+sin(2n-2)x)+cos(2n-1)xsinx
∴∫(sin2nx/sinx)dx=1/2∫(sin2nx+sin(2n-2)x)/sinxdx+∫cos(2n-1)xdx
∴1/2∫(sin2nx/sinx)dx=1/2∫(sin(2n-2)x)/sinxdx+∫cos(2n-1)xdx
∴∫(sin2nx/sinx)dx=∫(sin(2n-2)x)/sinxdx+2∫cos(2n-1)xdx
=∫(sin(2n-4)x)/sinxdx+2∫cos(2n-3)xdx+2∫cos(2n-1)xdx
=∫(sin2x)/sinxdx+2∑(1~n)∫cos(2n-1)xdx
=-2sinx+2∑(1~n)[sin(2n-1)/(2n-1)]

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