www.5615.net > ∫xsinx2Dx

∫xsinx2Dx

∫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)

∫xsin(x2+1)dx= 1/2 ∫sin(x2+1)dx2=1/2 ∫sin(x2+1)d(x2-1)=1/2 ∫sinudu=-1/2cosu=-1/2cos(x2-1)

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

∫xcos(2x)dx = (1/2)∫xdsin(2x) = (1/2)xsin2x - (1/2)∫sin2xdx = (1/2)xsin2x +(1/4)cos2x+C.

分步积分就可以了 ∫ xsin(2x) dx=1/2*∫ xsin(2x) d(2x)=-1/2*∫ xdcos(2x) =-1/2*xcos(2x)+1/2*∫ cos(2x)dx=-1/2*xcos(2x)+1/4*∫ cos(2x)d(2x)=-1/2*xcos(2x)+1/4*sin(2x)+常数

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

不定积分怎么会有值? ∫xsin(x^2+1)dx 因为d(x^2)=2xdx 所以上面式子 =(1/2)*∫sin(x^2+1)d(x^2) =(1/2)*∫sin(x^2+1)d(x^2+1) =(1/2)*(-cos(x^2+1)) =-[cos(x^2+1)]/2

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

∫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

网站地图

All rights reserved Powered by www.5615.net

copyright ©right 2010-2021。
www.5615.net内容来自网络,如有侵犯请联系客服。zhit325@qq.com