∫[(1-sin(2x))/(sin^2(x)-x)]dx 怎么解

来源：学生作业帮助网编辑：作业帮时间：2024/05/12 09:52:37

∫[(1-sin(2x))/(sin^2(x)-x)]dx 怎么解
∫[(1-sin(2x))/(sin^2(x)-x)]dx 怎么解

∫[(1-sin(2x))/(sin^2(x)-x)]dx 怎么解
∫[(1-sin(2x))/(sin^2(x)-x)]dx
=∫[(1-2sinxcosx)/(sin^2(x)-x)]dx
=∫1/[sin^2(x)-x]d[x-sin^2(x)]
=-∫1/[x-sin^2(x)]d[x-sin^2(x)]
=C-ln[x-sin^2(x)]

靠,很简单啊
1-sin(2x)可以放到dx里面去
即原式等于∫1/(sin^2(x)-x)]d(x-sin^2(x))
接下来的该会算了吧 !!
比上面的简单多了

∫sin^2x/(1+sin^2x )dx求解, 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 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 (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(6x)sin(2x)dx=? 化简(sin^2 x/sin x-cosx)-(sin x+cosx/tan^2 x-1) 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 ∫(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 ∫x/sin^2xdx