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x 2Cosx的积分

∫cos2xcosxdx =∫[1-2(sinx)^2]d(sinx) =∫d(sinx)-2∫(sinx)^2d(sinx) =sinx-(2/3)(sinx)^3+C =(1/3)sinx[3-2(sinx)^2]+C =(1/3)sinx(2+cos2x)+C =(2/3)sinx+(1/3)sinxcos2x+C.

这题采用分部积分法,具体过程如下:∫ x^2 cosx dx= ∫ x^2 dsinx= x^2 sinx - ∫ sinx dx^2= x^2 sinx - 2∫ x sinx dx= x^2 sinx - 2∫ x d(-cosx)= x^2 sinx + 2x cosx - 2∫ cosx dx= x^2 sinx + 2x cosx - 2sinx + C

原式=∫xdsinx=xsinx-∫sinxdx=xsinx-2∫xsinxdx=xsinx+2∫xdcosx=xsinx+2xcosx-2∫cosxdx=xsinx+2xcosx-2sinx+C

∫x^2cosxdx=∫x^2d(sinx) =x^2*sinx-∫sinxd(x^2) =x^2*sinx-2∫xsinxdx =x^2*sinx+2∫xd(cosx) =x^2*sinx+2[xcosx-∫cosxdx] =x^2*sinx+2xcosx-2sinx+C

∫ xcosx dx= ∫ x dsinx= xsinx - ∫ sinx dx= xsinx - 2∫ xsinx dx= xsinx + 2∫ x dcosx= xsinx + 2xcosx - 2∫ cosx dx= xsinx + 2xcosx - 2sinx + C

x^2*(cosx)^2的积分为1/6 x+1/4x *sin2x+1/4cos2x-1/8sin2x+C 解: ∫( x cosx)dx= ∫( x (cos2x+1)/2)dx=1/2∫( x cos2x+x)dx=1/2∫xdx+1/2∫ x cos2xdx=1/6 x+1/2∫ x cos2xdx=1/6 x+1/4∫ x dsin2x=1/6 x+1/4x *sin2x-1/4∫

=0.5∫sec(x^2)d(x^2) =0.5ln|tan(x^2/2+π/4)|+C

你好!这题采用分部积分法,具体过程如下: ∫ x^2 cosx dx= ∫ x^2 dsinx= x^2 sinx - ∫ sinx dx^2= x^2 sinx - 2∫ x sinx dx= x^2 sinx - 2∫ x d(-cosx)= x^2 sinx + 2x cosx - 2∫ cosx dx= x^2 sinx + 2x cosx - 2sinx + C满意请采纳,谢谢~

∫xcos(x/2)dx=2∫xcos(x/2)d(x/2)=2∫ xdsin(x/2)=2xsin(x/2)-2∫ sin(x/2)dx=2xsin(x/2)+4cos(x/2)+C

[代表积分号,[x^2cosxdx=[x^2dsinx=x^2sinx-[2xsinxdx=x^2sinx+[2xdcosx=x^2sinx+2xcosx-[2cosxdx=(x^2-2)sinx+2xcosx+C

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