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∫2x(sinx)^2Dx

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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...

∫[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+(...

是0,因为后面的求导对象是一个定积分,对于求导的变量x来说是一个常数(积分值取决于被积分函数和积分限),相当于对常数求导,至于后面积分变量是x其实和前面求导的x没有关系,换成任何字母都一样,其值取决于积分限a、b(被积分函数固定了,...

这个题还行

直接用sinz,显然不满足大圆弧引理的条件

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...

原式=∫(cscx)^2dx =-∫-(cscx)^2dx =-cotx+C

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