phmg.net
当前位置:首页 >> 2CosxCos2xDx >>

2CosxCos2xDx

∫(-π/2~π/2)[cosx-2cosx(sinx)^2 ]dx=∫(-π/2~π/2)coxdx-∫(-π/2~π/2)2cosx(sinx)^2dx=sinx(-π/2~π/2)-2∫(-π/2~π/2)(sinx)^2dsinx=2-2/3[(sinx)^3](-π/2~π/2)=2-4/3=2/3

∫[0→π/2] 2xcos2x dx =∫[0→π/2] x d(sin2x) =xsin2x - ∫[0→π/2] sin2x dx =xsin2x + (1/2)cos2x |[0→π/2] =-(1/2) - (1/2) =-1 希望可以帮到你,不明白可以追问,如果解决了问题,请点下面的"选为满意回答"按钮,谢谢。

∫xcos²xdx=∫x(1+cos2x)/2dx=1/2(∫xdx+∫xcos2xdx) =1/2(1/2x²+∫xcos2xdx) =1/2(1/2x²+1/2∫xdsin2x) =1/2(1/2x²+1/2(xsin2x-∫sin2xdx)) =1/2(1/2x²+1/2xsin2x+1/4cos2x)+C

欢迎采纳,不要点错答案哦╮(╯◇╰)╭

∫(0->π/2) xcos2xdx =(1/2)∫(0->π/2) xdsin2x =(1/2)[ x.sin2x]|(0->π/2) -(1/2)∫(0->π/2) sin2x dx =0-(1/2)∫(0->π/2) sin2x dx =(1/4)[ cos2x ]|(0->π/2) =-1/2

答: ∫ xcos(x/2) dx =2∫ xcos(x/2) d(x/2) =2∫ x d [sin(x/2)] =2xsin(x/2)-2∫ sin(x/2) dx =2xsin(x/2) +4cos(x/2)+C

xsin2xdx=1/2∫x(-cos2x)′dx=1/2(-xcos2x+∫(x)′cos2xdx)=-x/2cos2x+1/4sin2x+c1)∫kdx=kx+c2)∫x^udx=(x^(u+1))/(u+1)+c3)∫1/xdx=ln|x|+c4)∫a^xdx=(a^x)/lna+c5)∫e^xdx=e^x+c6)∫sinxdx=-cosx+c7)∫cosxdx=sinx+c8)∫1/(cosx)^2dx=tanx+c...

用两次分部积分,结果等于π/2(见图片)

有一个地方打错了,在“证明”后面,是i是虚数单位,不是n.

这里就是用来凑微分的办法, 显然求导得到(cos2x)'= -2sin2x 所以就有 d(cos2x)= -2sin2x dx 于是就得到了 ∫ x sin2xdx= -1/2 *∫ xd(cos2x)

网站首页 | 网站地图
All rights reserved Powered by www.phmg.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com