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∫(0→π/2)xsinx^2Dx

若是 ∫ xsin(x^2)dx 则 ∫ xsin(x^2)dx = (1/2)∫ sin(x^2)d(x^2) = - (1/2)[cosx^2] = (1/2)[1-cos(π^2/4)] 若是 ∫ x(sinx)^2dx 则 ∫ x(sinx)^2dx = (1/2) ∫ x(1-cos2x)dx = (1/2) ∫ xdx - (1/2) ∫ xcos2xdx = (1/4) [x^2] - (1/4) ∫ xdsin2x = π...

1、本题的积分方法是: A、运用余弦二倍角公式; B、凑微分; C、分部积分。 . 2、具体解答如下,若有疑问,欢迎追问,有问必答; . 3、若点击放大,图片更加清晰。 .

因为(xsinx)2=x2(1?cos2x)2,所以∫π0(xsinx)2dx =∫π0x22dx-∫π0x2cos2x2dx.利用分部积分法可得,∫π0x2cos2x2dx=x2sin2x4|π0-∫π0xsin2x2dx=0-(?xcos2x4|π0+∫π0cos2x4dx)=-π4-sin2x8|π0=-π4,又因为 ∫π0x22dx=x36|π0=π36,所以∫π0(xsinx)2dx=

这个题还行

为简化起见,以不定积分的方式,从第三个等号前开始计算。 ∫x²dsin2x =x²sin2x-∫sin2xdx² =x²sin2x-2∫xsin2xdx =x²sin2x-∫xsin2xd2x =x²sin2x+∫xdcos2x =x²sin2x+xcos2x-∫cos2xdx =x²sin2x+xcos2x-(1/2...

∫[0,π]x^2(sinx^2)dx =∫[0,π]x^2(1/2)(1-cos2x)dx =∫[0,π](1/2)x^2dx-∫[0,π](1/2)x^2cos2xdx =(1/6)π^3 -(1/4)∫[0,π]x^2dsin2x =(1/6)π^3+(1/2)∫[0,π]xsin2xdx =(1/6)π^3+(-1/4)∫[0,π]xdcos2x =(1/6)π^3+(-1/4)π+(1/4)∫[0,π]cos2xdx =(1/6)π^3+(...

若是 I = ∫ e^x(sinx)^2dx, 则 I = (1/2) ∫ e^x(1-cos2x)dx = (1/2) ∫ e^xdx - (1/2) ∫ e^xcos2xdx = (1/2)e^x-(1/2)J 其中 J = ∫ e^xcos2xdx = ∫ cos2xde^x = e^xcos2x+2 ∫ e^xsin2xdx = e^xcos2x+2 ∫ sin2xde^x = e^xcos2x+2e^xsin2x-4 ∫ e^xc...

应用两次施笃兹定理lim an/n^2变为(0,+∞)∫xsin[(3n-3)x]sin[(n-1)x]/(sinx)^2dx+(0,+∞)∫xcos[(4n-4)x]dx=(0,+∞)∫xsin[(3n-3)x]sin[(n-1)x]/(sinx)^2dx=(0,+∞)∫x{cos[(2n-2)x]-cos[4(n-1)x]}/(sinx)^2dx(sinx)^2=-(cotx)'洛朗级数展开得(sinx)^2=...

∫x(sinx)^2dx =(1/2)∫x(1-cos2x)dx =(1/4)x^2-(1/2)∫xcos2xdx =(1/4)x^2-(1/4)∫xdsin2x =(1/4)x^2-(1/4)xsin2x +(1/4)∫sin2x dx =(1/4)x^2-(1/4)xsin2x -(1/8)cos2x + C

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