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求∫xsin²xDx

显然sin²x= -0.5cos2x+0.5 所以得到 原积分 =∫ x *(-0.5cos2x+0.5) dx =0.25x² - 0.25∫ x d(sin2x) =0.25x² - 0.25x *sin2x + ∫0.25sin2x dx =0.25x² - 0.25x *sin2x -0.125cos2x +C,C为常数

本题主要求y=x2极坐标程,即rsinθ=r2cos2θ,整理:r=sinθ/cos2θ 则∫(0->1)dx∫(x^2->x)(x^2+y^2)^(-1/2)dy =∫[0->π/4]dθ∫[0->sinθ/cos2θ] (1/r)*rdr =∫[0->π/4]dθ∫[0->sinθ/cos2θ] 1dr =∫[0->π/4] sinθ/cos2θdθ =-∫[0->π/4] 1/cos2θd(cosθ) =1/cosθ...

公式:∫[0→π] xf(sinx) dx = (π/2)∫[0→π] f(sinx) dx ∫[0→π] x(sinx)⁶(cosx)⁴ dx 由公式: =(π/2)∫[0→π] (sinx)⁶(cosx)⁴ dx =(π/2)∫[0→π/2] (sinx)⁶(cosx)⁴ dx + (π/2)∫[π/2→π] (sinx)⁶(cosx)⁴...

换元法+分部积分

D={(x,y)|0≤x≤π,0≤y≤π)} 区域D的面积 σ =π² 0≤sin²xsin²y≤1,当(x,y)∈D时,则 0* σ≤∫D∫sin²xsin²ydσ ≤1*σ,即 0≤∫D∫sin²xsin²ydσ≤π²

因为xsin²x是奇函数,cos²x是偶函数, 所以 原式=∫(-π,π)cos²xdx =2∫(0,π)cos²xdx =∫(0,π)(1+cos2x)dx =(x+1/2sin2x)|(0,π) =π

∫ e^xsin²x dx =(1/2)∫ e^x(1-cos2x) dx =(1/2)e^x - (1/2)∫ e^xcos2x dx (1) 下面计算: ∫ e^xcos2x dx =∫ cos2x d(e^x) 分部积分 =e^xcos2x + 2∫ e^xsin2x dx =e^xcos2x + 2∫ sin2x d(e^x) 再分部积分 =e^xcos2x + 2e^xsin2x - 4∫ e^xcos...

解: ∫1/(cos²xsin²x) dx =∫4/[(2cosxsinx)²]dx =4∫1/sin²(2x) dx =2∫csc²(2x)d(2x) =-2cot(2x) +C

如下 ∫xsin²xdx =1/2∫x(1-cos2x)dx =x^2/4-1/2∫xcos2xdx =x^2/4-1/4∫xdsin2x =x^2/4-1/4xsin2x+1/4∫sin2xdx =x^2/4-1/4xsin2x-1/8cos2x+C

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