求不定积分∫2dx⼀x√a^2-x^2

2024-11-15 10:24:16
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回答1:

令x = a · sinz,dx = a · cosz dz
∫ 2dx/[x√(a² - x²)]
= ∫ 2(a · cosz)/(a · sinz · a · cosz) dz
= ∫ 2acosz/(a²sinzcosz) dz
= (2/a)∫ cscz dz
= (2/a)ln|cscz - cotz| + C
= (2/a)ln|a/x - √(a² - x²)/x| + C
= (2/a)[ln|a - √(a² - x²)| - ln|x|] + C