急！微积分题目

∫sin(x)dx/{cos(x)[5-4cos(x)]^(1/2)}

= (1/4)∫d[5 -4cos(x)]/{cos(x)[5-4cos(x)]^(1/2)}

= (1/2)∫d[5 -4cos(x)]^(1/2)/cos(x)

= 2∫d[5 -4cos(x)]^(1/2)/{5 - [5 - 4cos(x)]}

= (5)^(-1/2)∫d[5 -4cos(x)]^(1/2)/{5^(1/2) - [5 - 4cos(x)]^(1/2)} + (5)^(-1/2)∫d[5 -4cos(x)]^(1/2)/{5^(1/2) + [5 - 4cos(x)]^(1/2)}

= 5^(-1/2){ln|5^(1/2) + [5-4cos(x)]^(1/2)| - ln|5^(1/2) - [5-4cos(x)]^(1/2)|} + C.

C = const.

先令t=cosx

积分化为: int ( -dt/tsqrt(5-4t) )

再令y=sqrt(5-4t)
于是t=(5-y^2)/4, dt=-2ydy, 代入积分有

int ( -dt/tsqrt(5-4t) )
= int( 2ydy/y(5-y^2)/4)
= int (8dy/(5-y^2))

利用公式∫1/(a^2-x^2)dx=(1/2a)ln|(a+x)/(a-x)|+c

得 =(4/5)ln|(5+y)/(5-y)|+c

原式=-∫{1/[cosx(5-4cosx)^0.5]}d(cosx)
设(5-4cosx)^0.5=t
=∫8dt/(5-t^2)
=0.8√5ln{[√5+(5-4cosx)^05]/[√5-(5-4cosx)^0.5]}+C
{}是绝对值符号，打不出来才用的{}

我只公布标准答案，供答题者参考，楼主不要给分

(ln[Sqrt[5] + Sqrt[5