求不定积分∫(dx)/√[(x-a)(b-x)] , (a<x<b)书上写的设我看得都茫然了设 x=acos²t+bsin²t (0<t<π/2),咋设成这样呀?看不懂
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![求不定积分∫(dx)/√[(x-a)(b-x)] , (a<x<b)书上写的设我看得都茫然了设 x=acos²t+bsin²t (0<t<π/2),咋设成这样呀?看不懂](/uploads/image/z/7077209-41-9.jpg?t=%E6%B1%82%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%E2%88%AB%28dx%29%2F%E2%88%9A%5B%28x-a%29%28b-x%29%5D+%2C+%28a%EF%BC%9Cx%EF%BC%9Cb%29%E4%B9%A6%E4%B8%8A%E5%86%99%E7%9A%84%E8%AE%BE%E6%88%91%E7%9C%8B%E5%BE%97%E9%83%BD%E8%8C%AB%E7%84%B6%E4%BA%86%E8%AE%BE+x%3Dacos%26sup2%3Bt%2Bbsin%26sup2%3Bt+%280%EF%BC%9Ct%EF%BC%9C%CF%80%2F2%29%2C%E5%92%8B%E8%AE%BE%E6%88%90%E8%BF%99%E6%A0%B7%E5%91%80%3F%E7%9C%8B%E4%B8%8D%E6%87%82)
求不定积分∫(dx)/√[(x-a)(b-x)] , (a<x<b)书上写的设我看得都茫然了设 x=acos²t+bsin²t (0<t<π/2),咋设成这样呀?看不懂
求不定积分∫(dx)/√[(x-a)(b-x)] , (a<x<b)
书上写的设我看得都茫然了
设 x=acos²t+bsin²t (0<t<π/2),
咋设成这样呀?看不懂
求不定积分∫(dx)/√[(x-a)(b-x)] , (a<x<b)书上写的设我看得都茫然了设 x=acos²t+bsin²t (0<t<π/2),咋设成这样呀?看不懂
令x-(a+b)/2=[(b-a)sint/2],t=arcsin{[x-(a+b)/2]/[(b-a)/2]}=arcsin[(2x-a-b)/(b-a)]
d[x-(a+b)/2]=[(b-a)/2]dsint
∫1/√(x-a)(b-x)dx
=∫1/√(x-a)(b-x)dx
=∫1/√[-x²+(a+b)x-ab]dx
=∫1/√{-[x-(a+b)/2]²+(a+b)²/4-ab}dx
=∫1/√{[(b-a)/2]²-[x-(a+b)/2]²}dx
=∫1/√{[(b-a)/2]²-[x-(a+b)/2]²}d[x-(a+b)/2]
=[(b-a)/2]∫1/√{[(b-a)/2]²-[(b-a)sint/2]²}dsint
=[(b-a)/2]∫1/√[(b-a)cost/2]²dsint
=[(b-a)/2]∫cost/[(b-a)cost/2]dt
=∫dt
=t+C
=arcsin[(2x-a-b)/(b-a)]+C
首先把被积函数经过代数变形转化为∫1/√(a²-x²)dx的形式,再作换元x=asint