∫1/√xcos^2√xdx 1cosxcos2xcos3x
更新时间：2021-12-11 10:56:45 • 作者：TERRY •阅读 6819
请问∫1/√xcos√xdx怎么解？
=2∫cos√xd√x=2sin√x+C或换元法设t=√x求
不定积分 ：∫ xcos^2xdx 求详细过程和答案 拜托大神.
∫ xcos^2xdx
=∫ x(1+cos2x/2)dx
=1/2∫ xdx+1/2∫xcos2xdx
=x²/4+1/4∫xdsin2x
=x²/4+1/4*xsin2x-1/4∫sin2xdx
=x²/4+1/4*xsin2x-1/8∫sin2xd2x
=x²/4+1/4*xsin2x+1/8*cos2x+C
不定积分 ：∫ xcos^2xdx
^∫
=∫ x(1+cos2x/2)dx
=1/2∫ xdx+1/2∫xcos2xdx
=x²/4+1/4∫xdsin2x
=x²/4+1/4*xsin2x-1/4∫sin2xdx
=x²/4+1/4*xsin2x-1/8∫sin2xd2x
=x²/4+1/4*xsin2x+1/8*cos2x+C
xcos^2x的不定积分
本题是不定积分基本练习题，具体步骤如下:
∫ⅹcos^2xdx
=∫x(1+cos2x)dx/2
=(1/2)∫xdx+(1/2)∫xcos2xdx
=(1/2)*(1/2)*x^2+(1/4)∫ⅹcos2xd2x
=(1/2)*(1/2)*x^2+(1/4)∫ⅹdsin2x
=(1/4)ⅹ^2+(1/4)ⅹsin2x-(1/4)∫sin2xdx
=(1/4)ⅹ^2+(1/4)ⅹsin2x-(1/8)∫sin2xd2x
=(1/4)ⅹ^2+(1/4)ⅹsin2x+(1/8)cos2x十C。
本题主要用到分部积分法和三角函数的求导公式。