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∫ 0 π x sinx 2Dx

∫[0---->π] √(sinx-sinx) dx=∫[0---->π] √[sinx(1-sinx)] dx=∫[0---->π] √[sinx(cosx)] dx=∫[0---->π] |cosx|√(sinx) dx 在这一步你做错了,cosx在[0---->π]有正有负,因此这里要加绝对值=∫[0---->π/2] cosx√(sinx) dx-∫[π/2---->π] cosx√(sin

∫[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+(-1/4)π

先将(sinx)^2降次,如下:原式=∫x^2*(1/2-cos2x/2)dx再将x^2看成u,括号里的看成v',就有:=x^2*(x/2-sin2x/4)-∫2x(x/2-sin2x/4)dx,再把被减数化简,减数求不定积分,如下:=x^3/2-xsin2x/8-∫x^2-xsin2x/2dx接着对∫xsin2x/2dx再用一

根据题意,可得∫ π0 (x?sinx)dx=(1 2 x2+cosx)| π0=(1 2 *π2+cosπ)-(1 2 *02+cos0)=π2 2 ?2故答案为:π2 2 ?2

Im=∫[0,π]x(sinx)^mdx =∫[0,π]x(sinx)^(m-1)d(-cosx) =-cosx*x(sinx)^(m-1)|[0,π]+∫[0,π]cosxdx(sinx)^(m-1) =∫[0,π]cosx*sinx^(m-1)dx+(m-1)∫[0,π]x(cosx)^2(sinx)^(m-2)dx =∫[0,π]sinx^(m-1)dsinx +(m-1)∫[0,π]x(1-sinx)^2(sinx)^(m-2)dx =(m-1)∫[0,π]x(sinx)^(m-2)

若是 ∫ <0→π/2> xsin(x^2)dx 则 ∫ <0→π/2> xsin(x^2)dx = (1/2)∫ <0→π/2> sin(x^2)d(x^2)= - (1/2)[cosx^2] <0→π/2> = (1/2)[1-cos(π^2/4)] 若是 ∫ <0→π/2> x(sinx)^2dx 则 ∫ <0→π/2> x(sinx)^2dx = (1/2) ∫ <0→π/2> x(1-cos2x)dx= (1/2) ∫ <0→π/2> xdx - (1/2)

解:∵当0<x<π时,sinx>0,则|sinx|=sinx 当π<x<2π时,sinx<0,则|sinx|=-sinx ∴∫<0.2π>x|sinx|dx=∫<0.π>xsinxdx+∫<π.2π>x(-sinx)dx =∫<0.π>xsinxdx-∫<π.2π>xsinxdx =-∫<0.π>xd(cosx)+∫<π.2π>xd(cosx) =-(xcosx-sinx)│<0.π>+(xcosx-sinx)│<π.2π> (应用分部积分法) =-π(-1)+(2π+π) =4π.

(1) ∫[0,π](xsinx)^2dx =1/2∫[0,π]x^2(1-cos2x)dx =1/2∫[0,π]x^2dx-1/4∫[0,π]x^2dsin2x =1/6x^3|[0,π]-1/4x^2 sin2x|[0,π]+1/2∫[0,π]xsin2xdx =π^3/6-1/4∫[0,π]xdcos2x =π^3/6-1/4xcos2x|[

∫x*sinxdx=-∫xdcosx=-xcosx+∫cosxdx=sinx-xcosx0,π 带入,除2=-π/2

∫ xsinx^2 dx=1/2 ∫ sinx^2 dx^2=-1/2*cosx^2=-1/2*(cosπ-cos0)=1 有好多符号不好打就省略了,将就看一下 = =

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