∫(1 + (sin x)^2)^(1/2) dx ∫1 / (1 - x^2) dx

∫sin^2x/(1+sin^2x )dx求解, 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 ∫(1-sin^2( x/2))dx ∫1/(sin^2xcos^2x) ∫(cosx／1+sin^2x)dx ∫sin^2x(1+tanx)dx ∫cosx/(1+4sin^2x) ∫1/sin(x/2)dx ∫[(1-sin(2x))/(sin^2(x)-x)]dx 怎么解 (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) ∫sin²x(1-sin²x)dx是多少? ∫sin(1/x)dx ∫1/sin^22x (sin 1/x) /x