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求不定积分∫xCosx 2Dx

∫ (cosx)/x dx= ∫ cosx d(- 1/x)= - (cosx)/x + ∫ 1/x d(cosx)= - (cosx)/x - ∫ (sinx)/x dx= - (cosx)/x - Si(x) + C Si(x)是正弦积分,无法用初等函数表示的. 或者用级数表示也行.∫ (sinx)/x dx= ∫ 1/x ∑(k=0→∞) (- 1)^k x^(1 + 2k)/(1 + 2k)! dx= ∑(k=0→∞)

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

∫xcos^2 x dx=∫x(cos2x+1)/2 dx=1/2*∫xcos2xdx+1/2*∫xdx=1/4∫xcos2xd2x+1/4∫dx^2=1/4∫xdsin2x +x^2/4=1/4 *xsin2x-1/4∫sin2xdx +x^2/4=xsin2x/4+x^2/4-1/8∫sin2xd2x=xsin2x/4+x^2/4+1/8∫dcos2x=xsin2x/4+x

∫xcosx^2dx=(1/2)∫cosx^2dx^2=(1/2)sinx^2+C

∫xdx/cosx= ∫xdtanx= xtanx - ∫tanxdx = xtanx - ∫sinxdx/cosx= xtanx + ∫d(cosx)/cosx + c= xtanx + ln┃cosx┃ + c (c为常数)

原式=(1/2)∫x^2(1+cos2x)dx =(1/2)∫x^2dx+(1/2)∫x^2cos2xdx =(1/6)x^3+(1/4)∫x^2d(sin2x) =(1/6)x^3+(1/4)x^2sin2x-(1/4)∫sin2xd(x^2) =(1/6)x^3+(1/4)x^2sin2x-(1/2)∫xsin2xdx =(1/6)x^3+(1/4)x^2sin2x+(1/4)∫xd(cos2x) =(1/6)x^3+(1/4)x^2sin2x+(1/4)xcos2x-(1/4)∫cos2xdx =(1/6)x^3+(1/4)x^2sin2x+(1/4)xcos2x-(1/8)sin2x+c

∫x.cosx/2dx=2∫xdsinx/2=2xsinx/2-2∫sinx/2dx= 2xsinx/2+4cosx/2+C

^∫xcosx/2dx=2xsin(x/2)+4cos(x/2)+C.C为积分常数.解答过程如下du:zhi∫xcosx/2dx=2∫xdsin(x/2)=2xsin(x/2)-2∫sin(x/2)dx=2xsin(x/2)-4∫sin(x/2)d(x/2)=2xsin(x/2)+4cos(x/2)+C扩展资料:分部积分:(uv)'=u'v+uv'得:u'v=(uv)'-uv'两边积分得:∫ u'v

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