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xsinx2的不定积分

用分部积分法 ∫xsin^2x dx =1/2∫x(1-cos2x)dx=1/2(∫xdx -∫xcos2x dx)=1/2(1/2*x^2-1/2∫x dsin2x) =1/4(x^2-xsin2x+∫sin2x dx )=1/4(x^2-xsin2x-1/2cos2x)+c

∫xsin^2xdx=1/4∫2xsin^2xd2x令t=2x=1/4∫tsin^tdt=1/4(sint-tcost)因此∫xsin^2xdx=1/4(sin2x-2xcos2x)

∫sin(x^2)d(x^2)/2=-cos(x^2) 就给出结果就行啦 带入上下限=1-cos(π^2/4)

cos^2 x=(cos2x+1)/2∫ xcos^2 xdx =∫ x(cos2x+1)dx/2+C=(∫xcos2xdx+∫xdx)/2+C=(∫xdsin2x+x^2)/4+C=(xsin2x-∫sin2xdx+x^2)/4+C=(2xsin2x+cos2x+2x^2)/8+C

请问下sinx2是什么意思.假如是sinx^2的话,就简单了 ∫xsinx^2dx=∫sinx^2d(x^2)/2=-COSX^2/2

∫x sin x2dx=1/2 ∫sin x2 dx2=- 1/2 ∫d cos x2=- 1/2 cos x2

∫xsin^2xdx=∫xcsx^2xdx=-∫xd(cotx)=-xcotx-∫cotxdx=-xcotx-∫cosxdx/sinx=-xcotx-∫d(sinx)/sinx=-xcotx-lnsinx+c.

∫ xsin2x dx= -(1/2)∫ xdcos2x=-(1/2)xcos2x + (1/2)∫ cos2x dx=-(1/2)xcos2x + (1/4) sin2x + C

一楼的是对的: 1/2∫x(1-cos2x)dx是怎么得出来的? cos2x=1-2sin^2x sin^2x=(1-cos2x)/2

∫xsin2xdx,运用分部积分法吧=(-1/2)∫xd(cos2x)=(-1/2)(xcos2x-∫cos2xdx)=(-xcos2x)/2+(1/2)∫cos2xdx=(-xcos2x)/2+(1/2)*(1/2)sin2x+C=(1/4)(sin2x)-(1/2)(xcos2x)+C

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