integral 0 to sqrt(3) dx/sqrt(4-x^2)
I don't understand why the answer is pi/6. I get pi/3. Thanks for the help!

u=x/2 du=1/2dx 2du=dx

sin^-1(sqrt(3)/2)= pi/3

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  1. Integral of dx/sqrt(4-x^2
    Let x = 2u and that becomes
    2 du/sqrt(4 - 4u^2)
    Integral of du/(sqrt(1-u^2)
    = sin^1 u = sin^-1(x/2)

    OK so far.. sin-1(sqrt3/2) - sin^-1 0
    = pi/3

    I agree with your answer

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