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求下列积分∫xsinx/1+Cos^2 (x)Dx ∫sin^2 (x)ArCtAn...

∫arctan(1/x)dx =∫(x)'arctan(1/x)dx =xarctan(1/x)-∫x*{1/[1+x^(-2)]}*[-1/x^2]dx =xarctan(1/x)+∫1/(x+1/x)dx =xarctan(1/x)+∫x/(x^2+1)dx =xarctan(1/x)+ln(x^2+1)/2+C ∫sin^6(x)cos^2(x)dx =sin^7(x)cosx/7-∫sin^7(x)*(-sinx)/7dx =sin^7(x)...

∫ (x-arctanx)/(1+x²) dx = ∫ x/(1+x²) dx - ∫ arctanx/(1+x²) dx = (1/2)∫ d(x²+1)/(1+x²) - ∫ arctanx d(arctanx) = (1/2)ln(1+x²) - (1/2)(arctanx)² + C L = ∫ e^x * sin(x/2) dx = 2∫ e^x * sin(x/2) d...

本人认为楼上答案有误 如下 更正 f(1-)=-(π/2) f(1+)=π/2 x=1为跳跃间断点 楼上x=2n处无误 但需补充x=0的讨论, 当x=0,易得原式=0,所以x=0为可去间断点

令arctanx=t, 那么x=tant 1+x²=1+(tant)²=1/(cost)²,即√(1+x²)=1/cost 故x/√(1+x²)=cost *tant=sint 而sin(arctanx)就等于sint 所以就得到了证明, sin(arctanx)=x/√(1+x²)

由cos²α=1/(1+tan²α)及cosα>0,得cosα=1/√(1+x²)所以sinα=tanαcosα=x/√(1+x²)即α=arcsin[x/√(1+x²)] (2)从而arctanx..

解析: 令α=arcsin[(1-x²)^½ ,β=arctan[x/(1-x²)^½],其中0

判断间断点的类型还是要从定义出发,肯定不会错的,求解方法是一样的

第一题,直接用万能公式法。即令u=tan(x/2)x=2arctanudx=2/(1+u^2)du,sinx=2u/(1+u^2),cos=(1-u^2)/(1+u^2)原式=∫(1+u^2)/4udu=(1/4)∫(u)^(-1)du+(1/4)∫udu=(1/4)lnu+(1/8)u^2+C=(1/4)ln[tan(x/2)]+(1/8)[tan(x/2)]^2+C第二题,原式=∫(1-sinx)/[...

tan(arctanx1+arctanx2) =(x1+x2)/(1-x1x2) =sin(π/5)/[1-cos(4π/5)] =sin(π/5)/[2sin^2(2π/5)] =sin(4π/5)/[2sin^2(2π/5)] =2sin(2π/5)cos(2π/5)/[2sin^2(2π/5)] =cot(2π/5) 故 arctanx1+arctanx2=π/2-2π/5=π/10

sin[arcsin(2x/1+x^2)]=2x/(1+x^2) ∵cos²x=1/(1+tan²x) ∴(cos(arctanx))²=1/(1+(tan(arctanx))²)=1/(1+x²) cos(arctanx)=1/√(1+x²), 而sin(arctanx)=tan(arctanx)cos(arctanx)=x/√(1+x²...

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