根号(1+cos x2)积分
来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/15 22:40:09
根号(1+cos x2)积分
根号(1+cos x2)积分
根号(1+cos x2)积分
∫√(1+cosx^2)dx
=∫√(cosx^4+cosx^2)dx/cosx^2
=∫√(cosx^4+cosx^2)dtanx
=∫√(1+1/cosx^2)(cosx^2)dtanx
=∫√(2+tanx^2)dtanx/(1+tanx^2)
tanx=u
=∫√(2+u^2)du/(1+u^2)
=∫√(2/u^2+1)du/[u^2(1+1/u^2)]
= -∫√(2/u^2+1)d(1/u)/[(1/u)(1+1/u^2)]
1/u=t t=cotx
=-∫√(2t^2+1)dt/[t(1+t^2)]
=-∫√(2t^2+1)(1+t^2-t^2)dt/[t(1+t^2)]
=-∫√(2t^2+1)dt/t+∫√(2t^2+1)tdt/(1+t^2)
=-∫√(2+1/t^2)dt +(1/2)∫√(2t^2+1)d(t^2+1)/(1+t^2)
=(√2/2)ln|(1/2)tanx+√(1+(1/4)tanx^2)| +(-√2/2)√(1+4cotx^2)
+ln|√(2cscx^2-1)+cscx^2| +(1/2) √(2cscx^2-1)/cscx^2 +C
∫√(2+1/t^2)dt
t=(1/2)cotv
=∫√2secv*(-1/2)dv/sinv^2
=(-√2/2)∫cosvdv/(cosv^2sinv^2)
=(-√2/2)∫dsinv/(1-sinv^2)sinv^2
=(-√2/2)∫dsinv/(1-sinv^2)+(-√2/2)∫dsinv/sinv^2
=(-√2/4)ln|1+sinv|/|1-sinv|+(√2/2)(1/sinv)
=(-√2/2)ln|1/cosv+tanv|+(√2/2)(1/sinv)
=(-√2/2)ln|1/(2t)+√(1+1/4t^2)|+(√2/2)(√1+4t^2)
=(-√2/2)ln|(1/2)tanx+√(1+(1/4)tanx^2)| +(√2/2)√(1+4cotx^2)
∫√(2t^2+1)d(t^2+1)/(t^2+1)
t^2+1=tanm
=∫secm^3dm/tanm
=∫dm/sinmcosm^2
=∫dsinm/(sinm^2)(1-sinm^2)
=ln|1+sinm|/|1-sinm|+1/sinm
=2ln|1/cosm+tam|+1/sinm
=2ln|√(2t^2+1)+t^2+1|+√(2+t^2)/(1+t^2)
=2ln|√(2cscx^2-1)+cscx^2| +√(2cscx^2-1)/cscx^2