問積分sec(x)三次方的問題

(1) 利用恆等式sec^2(x)=1+tan^2(x)w=∫sec^3(x)dx=∫sec(x)*[1+tan^2(x)]*dx=∫sec(x)dx+∫sec(x)*tan^2(x)dx=w1+w2w1=∫sec(x)dx=∫dx/cos(x)=∫cos(x)dx/cos^2(x)=∫d(sin(x))/[1-sin^2(x)]=∫dy/(1-y^2);y=sin(x)=0.5∫(1/(1+y)+1/(1-y))dy=0.5*ln[(1+y)(1-y)]=ln[√(1-y^2)]=ln{√[1-sin^2(x)]}=ln{√cos^2(x)}=ln[cos(x)]w2=∫sec(x)*tan^2(x)dx=∫tan(x)d(sec(x))=tan(x)*sec(x)-∫sec(x)*sec^2(x);部份積分=tan(x)*sec(x)-∫sec^3(x)dx=tan(x)*sec(x)-w代入上式裡面:w=w1+w2=ln[cos(x)]+tan(x)*sec(x)-w2w=ln[cos(x)]+tan(x)*sec(x)w=0.5*ln[cos(x)]+0.5*tan(x)*sec(x)+c=ln[√cos(x)]+0.5*tan(x)*sec(x)+c
(2) 使用部份分式,即版主喜歡用1-u^2來做w=∫sec^3(x)dx=∫dx/cos(x)=∫cos(x)dx/cos^4(x)=∫d(sin(x))/[1-sin^2(x)]^2=∫dy/(1-y^2)^2;y=sin(x)=0.25∫[1/(1+y)+1/(1+y)^2+1/(1-y)+1/(1-y)^2]dy=0.25*ln[(1+y)(1-y)]-0.25/(1+y)+0.25/(1-y)=0.25*ln(1-y^2)+0.25[1/(1-y)-1/(1+y)]=0.25*ln[cos^2(x)]+0.25*2y/(1-y^2)=ln[√cos(x)]+0.5*sin(x)/cos^2(x)+c=ln[√cos(x)]+0.5*tan(x)*sec(x)+c=兩者答案相同

2013-08-08 08:02:19 補充：
修正:w=∫sec^3(x)dx=∫dx/cos^3(x)

到下面的網址看看吧

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I recommend integrating by part twice.

你的公式和代換很混亂，

應該是 tan^2(x) + 1 = sec^2(x)，你記反了，

所以sec(x)換成(1-u^2)也有問題