slrt.net
当前位置:首页 >> ∫xCos(x/2)Dx >>

∫xCos(x/2)Dx

cos2x=2cos^2x-1 所以 1+cos2x=2cos²x

∫xcos(x²)dx =1/2∫cos(x²)dx² =1/2*sin(x²)+C

如图

∫xcos2xdx =(1/2)∫xdsin2x =(1/2)x.sin2x -(1/2)∫sin2xdx =(1/2)x.sin2x +(1/4)cos2x + C

∫cos²(x/2)dx =∫[(1+cosx)/2]dx =1/2∫dx+1/2∫cosxdx =(x+sinx)/2+C

分部积分法

b21c8701a18b87d6a983e2eb0c0828381e30fd05 参考

∫(x/cos²x)dx =∫xd(tanx) =xtanx-∫tanxdx =xtanx+ln|cosx|+C 所以 ∫(x/cos²x)dx=xtanx+ln|cosx|+C

(x+sin(x))/(4cos(x)-4) 首先改写降次: xcos^4(x/2)/sin^3(x) = 1/8 xcot(x/2)csc^2(x/2) ... 注:csc(x)=1/sin(x) 换元积分:令u=x/2: dx=2du, ∫ 1/8 xcot(u)csc^2(u) 2du =1/8 * 2 * 2 ∫ ucot(u)csc^2(u) du 然后分部积分, =1/2 u(-1/2cot^2...

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