plz solve the below integration problem! thanks a lot
{ ( 2/[cot(x/2) - tan(x/2)] )^2 dx
p.s. since i don't know how to type the integration symbol , i use { to denote
thz
amaths ... integration
2008-12-01 4:03 am
回答 (3)
2008-12-01 5:27 am
✔ 最佳答案
{ ( 2/[cot(x/2) - tan(x/2)] )^2 dx= { (2/(1-tan^2(x/2))/tan(x/2))^2 dx
= { (2tan(x/2)/(1-tan^2(x/2))^2 dx [by Double-angle formula of tangle
= { tan^2(x) dx
= { (sec^2(x -1) dx
= tanx -x +c
2008-12-02 6:02 am
good
2008-12-01 5:07 am
[cot(x/2)-tan(x/2)]/2
=[cos(x/2)/sin(x/2)-sin(x/2)/cos(x/2)]/2
=[cos^2 (x/2)-sin^2 (x/2)]/[2sin(x/2)cos(x/2)]
=cos x/sin x=cot x
Thus
(2/[cot(x/2)-tan(x/2)])^2=(1/cot x)^2=tan^2 x=sec^2 x-1
Hence
∫(2/[cot(x/2)-tan(x/2)])^2 dx
=∫(sec^2 x-1) dx
=tan x-x+C
=[cos(x/2)/sin(x/2)-sin(x/2)/cos(x/2)]/2
=[cos^2 (x/2)-sin^2 (x/2)]/[2sin(x/2)cos(x/2)]
=cos x/sin x=cot x
Thus
(2/[cot(x/2)-tan(x/2)])^2=(1/cot x)^2=tan^2 x=sec^2 x-1
Hence
∫(2/[cot(x/2)-tan(x/2)])^2 dx
=∫(sec^2 x-1) dx
=tan x-x+C
參考: ME
收錄日期: 2021-04-29 15:43:10
原文連結 [永久失效]:
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