这题不定积分怎么做
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这题不定积分怎么做
![](http://img.wesiedu.com/upload/2/a1/2a12e4930a94b9ba33fcf0f162a51801.jpg)
![](http://img.wesiedu.com/upload/2/a1/2a12e4930a94b9ba33fcf0f162a51801.jpg)
![这题不定积分怎么做](/uploads/image/z/14575115-11-5.jpg?t=%E8%BF%99%E9%A2%98%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%E6%80%8E%E4%B9%88%E5%81%9A)
∫(1-x^2)^(-3/2) dx
= ∫ (1/x )d(1/√(1-x^2) )
= 1/[x√(1-x^2)] + ∫ dx/[x^2√(1-x^2)]
let
x= siny
dx = cosy dy
∫ dx/[x^2√(1-x^2)]
=∫ (cscy)^2 dy
=-coty + C'
=-√(1-x^2)/x + C'
∫(1-x^2)^(-3/2) dx
= 1/[x√(1-x^2)] + ∫ dx/[x^2√(1-x^2)]
=1/[x√(1-x^2)] -√(1-x^2)/x + C
=x/√(1-x^2) + C
= ∫ (1/x )d(1/√(1-x^2) )
= 1/[x√(1-x^2)] + ∫ dx/[x^2√(1-x^2)]
let
x= siny
dx = cosy dy
∫ dx/[x^2√(1-x^2)]
=∫ (cscy)^2 dy
=-coty + C'
=-√(1-x^2)/x + C'
∫(1-x^2)^(-3/2) dx
= 1/[x√(1-x^2)] + ∫ dx/[x^2√(1-x^2)]
=1/[x√(1-x^2)] -√(1-x^2)/x + C
=x/√(1-x^2) + C