求解一道高数题,要详细过程,谢谢!!!!!

补充图片

第1个回答  2013-02-15
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,

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。
第2个回答  2013-02-15
解:∫sin2nx/sinxdx
=∫sin2x/sinxdx
=∫2cosxdx
=-2sinx+C
希望能帮到你!追问

∫sin2nx/sinxdx
=∫sin2x/sinxdx

这步时你忘了n!!!!!!!!!!!!!!!!

追答

正弦函数是周期函数
所以sin2nπ=sin2π
希望帮到你

第3个回答  2013-02-15
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)]
第4个回答  2014-01-21
答案已经发送,请您注意查收,谢谢!!!!!本回答被提问者采纳
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