∫（1+sinx ）/（sin3x +sinx ）d x 怎么求

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∫（1+sinx ）/（sin3x +sinx ）d x 怎么求
∫（1+sinx ）/（sin3x +sinx ）d x 怎么求

∫（1+sinx ）/（sin3x +sinx ）d x 怎么求
∫ (1 + sinx)/(sin3x + sinx) dx
= ∫ (1 + sinx)/[2sin(3x + x)/2 * cos(3x - x)/2]
= (1/2)∫ dx/(sin2xcosx) + (1/2)∫ sinx/(sin2xcosx) dx
= (1/2)∫ dx/(2sinxcos²x) + (1/2)∫ sinx/(2sinxcos²x) dx
= (1/4)∫ cscxsec²x dx + (1/4)∫ sec²x dx
= (1/4)∫ cscx(1 + tan²x) dx + (1/4)tanx
= (1/4)∫ cscx dx + (1/4)∫ secxtanx dx + (1/4)tanx
= (1/4)ln| cscx - cotx | + (1/4)(secx + tanx) + C

∫（1+sinx ）/（sin3x +sinx ）d x 怎么求求数学积分∫（sinx*sin2x*sin3x）dx sinx+cosx-1/sin3x-x2（平方）求极限x趋向0 sinx+sin2x+sin3x 求最值 sinx*sin2x*sin3x=? sin3x/sinx+cos3x/cosx 求极限limx→0（sin3x-sinx）/x x->1时(sin5x-sin3x)/sinx的极限是多少求证sinx+sin3x+sin2x=1+cos2x+cosx 解方程；SinX+Sin2X+Sin3X = 1+CosX+Cos2X. (sin3x/2/sinx/2)-1的最小正周期 sinx+0.5sin2x+(1/3)sin3x的最小正周期高数 limx→π /2 cos3x*sin3x/（cosx*sinx）=?根据极限四则运算 =lim x→π /2 cos3x/cosx * lim x→π /2 sin3x/sinx再用洛必达法则 =lim x→π /2 -3sin3x/-sinx * lim x→π /2 sin3x/sinx=-3-1=-4哪错了...答案里写的是=-lim x sinx+sin5x+sin3x=0 sin3x-sin2x+sinx=0 sinx+sin2x+sin3x的最大值 (sin3x+sinx)/(cos3x+cosx)= sin3x/sinx-cos3x/cosx化简