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计算不定积分∫xsin^2xDx

∫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)

∫ e^xsinx dx=(1/2)∫ e^x(1-cos2x) dx=(1/2)e^x - (1/2)∫ e^xcos2x dx (1)下面计算:∫ e^xcos2x dx=∫ cos2x d(e^x)分部积分=e^xcos2x + 2∫ e^xsin2x dx=e^xcos2x + 2∫ sin2x d(e^

∫xsin2xdx=(-1/2)∫xdcos2x=(-1/2)(xcos2x-∫cos2xdx)=(-1/2)(xcos2x-(1/2)sin2x)+C=(1/4)sin2x-(1/2)xcos2x+C

∫xsin2xdx=1/2∫xsin2xd2x=-1/2∫xdcos2x=-1/2xcos2x+1/2∫cos2xdx=-1/2xcos2x+1/4∫cos2xd2x=-1/2xcos2x+1/4cos2x+C

注意sin2x= 1/2 -1/2 cos4x 所以得到 原积分=∫ x*(1/2 -1/2 cos4x)dx= x^2 /4 -∫ x/8 d(sin4x) 使用分部积分法=x^2 /4 - sin4x *x/8 +∫ sin4x *1/8 dx=x^2 /4 - sin4x *x/8 +∫ 1/32 *sin4x d(4x)=x^2 /4 - sin4x *x/8 -1/32 *cos4x +c,c为常数

解答如下:∫xsin2xdx=(-1/2)∫xdcos2x=(-1/2)(xcos2x-∫cos2xdx)=(-1/2)(xcos2x-(1/2)sin2x)+C=(1/4)sin2x-(1/2)xcos2x+C 在微积分中,一个函数f 的不定积分,或原函数,或反导数,是一个导数等于f 的函数 F ,即F ′ = f.不定积分和定积分间的关系

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

∫ e^xsinx dx=(1/2)∫ e^x(1-cos2x) dx=(1/2)e^x - (1/2)∫ e^xcos2x dx (1)下面计算:∫ e^xcos2x dx=∫ cos2x d(e^x)分部积分=e^xcos2x + 2∫ e^xsin2x dx=e^xcos2x + 2∫ sin2x d(e^x)再分部积分=e^xcos2x + 2e^xsin2x - 4∫ e^xcos2x dx将 -4∫ e^xcos2x dx

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

你好∫x^2sin2xdx=-1/2∫x^2d(cos2x)=-1/2[cos2x*x^2-∫2x*cos2xdx]=-1/2[cos2x*x^2-∫xd(sin2x)]=-1/2[cos2x*x^2-(sin2x*x-∫sin2xdx)]=-1/2cos2x*x^2+1/2sin2x*x-1/2∫sin2xdx=-1/2cos2x*x^2+1/2sin2x*x+1/4cos2x+C 为您解答,不理解请追问,理解请及时选为满意回答!(*^__^*)谢谢!

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