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Trigonometric + Improper

  1. May 6, 2005 #1
    Here's a integral where I have to use trigonometric substitution but I can't get the right answer.

    [int a=0 b=3] 1/(sqrt[9-x^2]) dx

    I did the limit as t approches 3 from the left.

    Then i did my trigonometric substitution, and it gives me arcsin(x/3).

    Then i computed what i had arcsin(a/3)-arcsin(0/3).

    It gives me 1.57 (estimated) or Pie/2 (real)

    But in the answer sheet, it says 9pie/4...
  2. jcsd
  3. May 6, 2005 #2
    Your integral is correct.

    [tex]\int\frac{1}{\sqrt{9-x^2}}dx = \arcsin{\frac{x}{3}} + C[/tex]

    When you apply the bounds, you get [tex]\arcsin{1} - \arcsin{0}[/tex]

    The arcsin of 0 is 0 and the arcsin of 1 is [tex]\frac{\pi}{2}[/tex]

    I don't see anything wrong with your answer.
  4. May 6, 2005 #3
    I don't know man, maybe the answer sheet is wrong. It says 9pie/4,as I mentionned before.

    I tried, I had pie/2, and test it on graphmatica the program, and it gave something near it.
  5. May 6, 2005 #4


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    Science Advisor

    Well, either your answer sheet is wrong, or you've described the problem incorrectly.
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