Anonymous asked in Science & MathematicsMathematics · 1 decade ago

I need to calculate the indefinite integral of sqrt(4-x^2). Is there an easy trick to doing this?

2 Answers

  • 1 decade ago
    Favorite Answer

    ok, you need to do a trig substitution.

    let x = 2cos theta

    dx = -2sin theta dtheat

    so you get:

    [integral] -2sin theta sqrt(4-4cos^2 theta) dtheat

    you can factor out a sqrt(4) = 2, and the -2

    then 1-cos^2 theta = sin^2 theta

    so sqrt(1-cos^2 theta) = sin theta

    so you get [integral] -4sin^2 theta d theta

    solve that then plug back in for x

  • Mαtt
    Lv 6
    1 decade ago


    Rewrite it as ∫√1-(x/2)² dx

    Let sin u = x/2 and du/dx cos u = 1/2

    dx = 2 cos u

    Our integral is now

    2∫√1-sin² u cos u du = 2∫cos² u du

    2∫cos² u du = ∫ 1 + cos(2u) du

    = u + sin(2u)/2 = u + cos u sin u

    from our org. subsitutions

    sin u = x/2

    u = arcsin (x/2)

    cos u = √4-x² --> draw a triangle if you do not get this one

    so the final answer is:

    arcsin(x/2) + (x/2)√4-x²

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