[sin（x^2）sin(1/x)]/[ln(1+2x)]的极限（X趋向于0）

傅里叶级数作图f（x）=2sin[x] - sin[2x] + 2/3sin[3x] - 1/2sin[4x]我用mathematica输入程序Plot[{2sin[x],-2sin[x],2sin[x] - sin[2x],-2sin[x] + sin[2x],2sin[x] - sin[2x] + 2/3sin[3x],-2sin[x] + sin[2x] - 2/3sin[3x],2sin[x] - sin[2x] + 2/3si 1/sin^2x的不定积分谢谢!（是1/sin x * sin x） x*（1+sin^2 x )/sin^2x 不定积分 化简(sin x+cos x-1)(sin x-cos x+1)/sin 2x (sin 1/x) /x (1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x 化简[1-(sin^4 x-sin^2 xcos^2 x+cos^4 x)]/(sin^2 x)+3sin^2 x 化简(sin^2 x/sin x-cosx)-(sin x+cosx/tan^2 x-1) sin(1/X)求导sin（1/X）怎么求导, ∫sin^2x/(1+sin^2x )dx求解, sin根号下（x+1）-sin根号下x 求不定积分【sin(x)^5 + sin(x)^3】^(1/2) 1-2sinx*sin(60+x)sin(60-x) sin(x^2)/sin^2x 求导 sin（x／2）等于 sin（x／2）等于 （xdx）/sin^2x 化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x