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    Hi,

    just unsure on tackling this qu:

    a
    ∫√(a^2 - x^2) dx = (a^2)/2 . (Pi/3 - √3/4)
    0.5a

    Prove that this is true using the substitution x = asinθ

    many thanks

    streety
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    Use the substitution x=asinθ
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    It comes down to integrating cos^2(theta), which you can do by using cos(2theta) = 2cos^2(theta) - 1.
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    (Original post by streetyfatb)
    Hi,

    just unsure on tackling this qu:

    a
    ∫√(a^2 - x^2) dx = (a^2)/2 . (Pi/3 - √3/4)
    0.5a

    Prove that this is true using the substitution x = asinθ

    many thanks

    streety
    ∫√(a^2 - x^2) dx = a∫√(1 - (x/a)^2) dx

    substitute: x = asinθ
    =>dx/dθ = acosθ

    Hence, integral becomes:

    a^2∫(Costheta)^2 dθ

    Using a well known identity, we get:

    a^2∫ (Cos2θ + 1)/2 dθ

    => a^2[ (Sin2theta)/4 + θ/2 ] (between a, and a/2)
    => a^2[ (x√(1 - (x/a)^2))/2a + arcsin(x/a)/2 ] (between a, and a/2)
    => a^2[ Pi/4 - √3/8 - Pi/12 ]
    => a^2[ Pi/6 - √3/8 ]
    => (a^2)/2[ Pi/3 - √3/4 ]

    which is the required result.

    Euclid
 
 
 
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