Question

求不定积分∫[1/(sin^2 cos^2(x)]dx

Answer 1

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为任意常数).

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为任意常数).

= = = = = = = = =

以上计算可能有误,你最好检查一下.

正负是个大问题.

注意：

sec x =1/ cos x,

csc x =1/ sin x.

(tan x)' =(sec x)^2,

(cot x)' = -(csc x)^2.