∫[1,-1]sin^3x+x^2/1+x^2dx=?

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^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 ∫sin^2x/(1+sin^2x )dx求解, 求不定积分【sin(x)^5 + sin(x)^3】^(1/2) (sin 1/x) /x 化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x 化简(sin x+cos x-1)(sin x-cos x+1)/sin 2x x*（1+sin^2 x )/sin^2x 不定积分 s = 2*sin(x)-sin(2*x)+2/3*sin(3*x)-1/2*sin(4*x)+2/5*sin(5*x)用matlab画图,求教啊 求极限 （（sin(x^3+x^2-x)+sin x) /x x→0 已知lim sinx/x=1 化简 1/2cos x-根号3/2sin x 根3sin x+cos x (x^3-x+1)sin^2x的不定积分 化简sin(2x-pi)sin[x+(3/2)pi]/cos[x+(5/2)pi][1-sin(pi+x)] 三角等式求证：cos^6x+sin^6x=1-3sin^2x+3sin^4x 化简(sin^2 x/sin x-cosx)-(sin x+cosx/tan^2 x-1) ∫3/(x*x*x+1)dx ∫1/(2+sin x)dx 和 ∫1/(2sin x-cos x+5)dx的解法 用换元法求下列积分∫x^3sin^x/(x^4+1)dx∫x^3sin^2x/(x^4+1)dx