∫sin^2x/(1+sin^2x )dx求解,

∫sin^2x/(1+sin^2x )dx求解, ∫(1/sin^2x+1)d(sinx)= x*（1+sin^2 x )/sin^2x 不定积分 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 化简(sin x+cos x-1)(sin x-cos x+1)/sin 2x sin(x^2)/sin^2x 求导 S(1-sin^2x)d(sinx)为什么=sinx-1/3sin^3x? (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(6x)sin(2x)dx=? 求不定积分【sin(x)^5 + sin(x)^3】^(1/2) 1-2sinx*sin(60+x)sin(60-x) ∫x/sin^2xdx 求不定积分 ∫(sin^5x sin^2x)dx抱歉是∫(sin^5x+sin^2x)dx 化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x 换元积分法这是我的做法∵d( sin(x/2) ) = cos(x/2)/2 dx∴cos(x/2)dx = 2d( sin(x/2) )∴原式=∫ 2cosx d( sin(x/2) ) =2∫ 1-2sin^2(x/2) d( (sin(x/2) ) =2sin(x/2) - 2/3sin^3(x/2) + C 哪里错了...不好意思 打错了 1/sin^2x的不定积分谢谢!（是1/sin x * sin x）