Question

工科数学 根号下的(1+cscx)dx的不定积分

如题如题
求详解

Answer 1

解：∫√(1+cscx)dx=∫√(1+1/cosx)dx
=∫√[(1+cosx)/cosx]dx
=∫√[(2cos²(x/2))/(1-2sin²(x/2))]dx (应用余弦倍角公式)
=√2∫[cos(x/2)/√(1-2sin²(x/2))]dx
=2√2∫d(sin(x/2))/√(1-2sin²(x/2))
=2∫costdt/cost (令√2sin(x/2)=sint，则t=arcsin(√2sin(x/2)))
=2∫dt
=2t+C (C是积分常数)
=2arcsin(√2sin(x/2))+C。

Answer 2

令y=1+cscx => dy=-cscxcotx dx
cscx=y-1，sinx=1/(y-1)，cosx=√[1-1/(y-1)²]=√y√(y-2)/(y-1)
tanx=1/[√y√(y-2)]，cotx=√y√(y-2)
∫√(1+cscx) dx
= -∫√y * 1/(cscxcotx) dy
= -∫√y * 1/[(y-1)*√y√(y-2)] dy
= -∫1/[(y-1)√(y-2)] dy
令z²=y-2 => 2zdz=dy
= -∫1/[(z²+2-1)√(z²+2-2)] * 2zdz
= -∫1/[(z²+1)*z] * 2zdz
= -2∫1/(z²+1) dz
= -2arctan(z) + C
= -2arctan√(y-2) + C
= -2arctan√(1+cscx-2) + C
= -2arctan√(cscx-1) + C