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Integ 1 / sqr root(a^2 - x^2 )

  1. Jan 19, 2006 #1
    does anybody knows how to do integ 1 / sqr root(a^2 - x^2 )......
    pls help..........
     
  2. jcsd
  3. Jan 19, 2006 #2
    Think inverse trig.
     
  4. Jan 19, 2006 #3
    ya i thought of using diff of arcsin ax = a / sqr root (1 - (ax)^2 )
    but if i do so the numerator will be 1/a^2 rite??

    pls help
     
  5. Jan 19, 2006 #4
    Close, it would be 1/a, like this: [tex]\int\frac{du}{\sqrt{a^2-u^2}} = arcsin \frac{u}{a} + C[/tex]
     
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