原式= ∫{ [(sin x)^2 +(cos x)^2 ] /[(sin x)^2 (cos x)^2 ] }dx
= ∫[ (sec)^2 ]dx +∫[ (csc)^2 ]dx
= tan x -cot x +C
= sin x /cos x -cos x /sin x +C
= [ (sin x)^2 -(cos x)^2 ] / (cos x sin x) +C
= -cos 2x / [ (1/2)sin 2x ] +C
= -2 cot 2x +C,(C為任意常數(shù)).
解法二：原式= ∫dx / [(1/4) (sin 2x)^2]
= 4 ∫[ (csc 2x)^2 ] dx
= 2 ∫[ (csc 2x)^2 ] d(2x)
= -2 cot 2x +C,(C為任意常數(shù)).
= = = = = = = = =
以上計(jì)算可能有誤,你最好檢查一下.
正負(fù)是個(gè)大問題.
注意：
sec x =1/ cos x,
csc x =1/ sin x.
(tan x)' =(sec x)^2,
(cot x)' = -(csc x)^2.