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∫2x(sinx)^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...

I=∫2xsin(x^2)dx =∫sin(x^2)d(x^2) =-cos(x^2) + C

这个题还行

直接用sinz,显然不满足大圆弧引理的条件

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=...

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=

sinx/2)平方,的不定积分 =2积分:(sinx/2)^2dx/2 =2*1/2(x/2-sinx/2*cosx/2)+C =x/2-sinx/2cosx/2+C ---------------

∫[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+(...

I = ∫e^x(sinx)^2dx = (1/2)∫e^x(1-cos2x)dx = (1/2)e^x - (1/2)∫e^xcos2xdx 其中 J = ∫e^xcos2xdx = ∫cos2xde^x = e^xcos2x + 2∫sin2xe^xdx = e^xcos2x + 2e^xsin2x - 2∫cos2xe^xdx = e^x(cos2x + 2sin2x) - 2J, 则 J = (1/3)e^x(cos2x + 2sin2...

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