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Math Help - evaluate the integral with inverse trig func.

  1. #1
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    Post evaluate the integral with inverse trig func.

    4ds/ (√4-s 2 )

    (integrate from 0 to 1

    i tried moving the top 4 outside (as 1/4):
    [1/4 sin^-1 (s/2) ] from 0 to 1
    1/4 sin^/1 (1/2) = 1/4 (pi/6)= pi/24

    but the answer is 2pi/3
    where did i go wrong?
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by genlovesmusic09 View Post
    4ds/ (√4-s 2 )

    (integrate from 0 to 1

    i tried moving the top 4 outside (as 1/4):
    [1/4 sin^-1 (s/2) ] from 0 to 1
    1/4 sin^/1 (1/2) = 1/4 (pi/6)= pi/24

    but the answer is 2pi/3
    where did i go wrong?
    \int_0^1\frac{4}{\sqrt{4-s^2}}ds=4\int_0^1\frac{1}{\sqrt{4-s^2}}ds=4\int_0^1\frac{1}{\sqrt{(2)^2-s^2}}ds.

    The formula for this type of integral is

    \int\frac{1}{\sqrt{a^2-x^2}}dx=\arcsin{\frac{x}{a}}+C

    Can you see awhat needs to be done?
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  3. #3
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    yes thanks
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