![求不定積分∫[1(sin^2 cos^2(x)]dx,第1張](data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw== "求不定積分∫[1(sin^2 cos^2(x)]dx,第1張")
求不定積分∫[1/(sin^2 cos^2(x)]dx網友廻答：
- 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爲任意常數).
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爲任意常數).
= = = = = = = = =
以上計算可能有誤,你最好檢查一下.
正負是個大問題.
注意：
sec x =1/ cos x,
csc x =1/ sin x.
(tan x)' =(sec x)^2,
(cot x)' = -(csc x)^2.
admin![求不定積分∫[1(sin^2 cos^2(x)]dx,第2張](/statics/default/images/niming.jpg "求不定積分∫[1(sin^2 cos^2(x)]dx,第2張")
0條評論