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∫sinx^2Dx=

∫x/(sinx)^2dx =∫xd(-cotx) =-xcotx+∫cotxdx =-xcotx+ln|sinx|+C

∫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

令arcsin(x/2)=u,则x=2sinu ∫arcsin(x/2)dx =∫ud(2sinu) =2∫ud(sinu) =2usinu-2∫sinudu =2usinu+2cosu +C =x·arcsin(x/2) +√(4-x²) +C

方法一: ∫[1/(sinx+cosx)^2]dx =(1/2)∫{1/[(1/√2)sinx+(1/√2)cosx]^2}dx =(1/2)∫{1/[sinxcos(π/4)+cosxsin(π/4)]^2}dx =(1/2)∫{1/[sin(x+π/4)]^2}d(x+π/4) =-(1/2)cot(x+π/4)+C 方法二: ∵1...

直接用sinz,显然不满足大圆弧引理的条件

用分步积分 S=∫(0 +∞) (sinx/x)^2 dx =x*(sinx/x)^2(0 +∞) -∫(0 +∞) xd(sinx/x)^2 =-∫(0 +∞) x*2sinx/x*(xcosx-sinx)/x^2dx =-∫(0 +∞) 2sinx/x*(xcosx-sinx)/xdx =∫(0 +∞) 2(sinx/x)^2dx-∫(0 +∞) 2sinx/x*xcosxdx =∫(0 +∞) 2(sinx/x)^2dx-...

x^5sinx^2dx =x^2*sinx^2-1/2*x^4*cosx^2+cosx^2 如果你认可我的回答,敬请及时采纳, ~如果你认可我的回答,请及时点击【采纳为满意回答】按钮 ~~手机提问的朋友在客户端右上角评价点【满意】即可。 ~你的采纳是我前进的动力 ~~O(∩_∩)O,记得好...

∫(arcsinx)² dx = x(arcsinx)² - ∫x d(arcsinx)²,分部积分法第一次第一步 = .. - ∫x * 2(arcsinx) * 1/√(1-x²) dx,分部积分法第一次第二步 = .. - 2∫(x*arcsinx)/√(1-x²) dx = .. - 2∫arcsinx d[-√(1-x²)],分...

题干不详

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