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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

换元法+分部积分

分部积分:原式=-1/k*∫xd(coskx) =-1/kxcoskx+1/k∫coskxdx =-1/kxcoskx+1/k^2sinkx+C

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...

你好,答案如下: 需用分部积分法 ∫ x sin(100x) dx = -1/100*∫ x d(cos(100x)) = -1/100*xcos(100x)+1/100*∫ cos(100x) dx = -1/100*xcos(100x)+1/100^2*sin(100x) + C 很高兴能回答您的提问,您不用添加任何财富,只要及时采纳就是对我们最好...

∫√x.sin√x dx let y=√x dy = [1/(2√x)] dx dx = 2y dy ∫√x.sin√x dx =2∫y^2. siny dy =-2∫y^2. dcosy =-2y^2. cosy +4∫y cosy dy =-2y^2. cosy +4∫y dsiny =-2y^2. cosy +4ysiny -4∫siny dy =-2y^2. cosy +4ysiny +4cosy + C =-2x.cos√x = 4√x.s...

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...

sin(x+a)-xcos(x+a)+C 解: ∫xsin(x+a)dx =-∫xdcos(x+a) =-[xcos(x+a)-cos(x+a)d x] =-[xcos(x+a)-sin(x+a)]+C =sin(x+a)-xcos(x+a)+C

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