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12 October, 11:56

Indefinate integral of sin-1 (x? i know you have to use integral by parts: u = sin-1 (x v = x

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  1. 12 October, 14:00
    0
    Yes

    you let u = sin-1 x and dv = 1 (so v = x)

    and use the rule

    INT udv = uv - INT vdu

    = x sin^-1 x - INT x sin-1 x

    u = sin-1 x so x = sin u

    dx/du = cos u, du/dx = 1 / cos u = 1 / sqrt (1 - sin^2) = 1 / sqrt (1-x) ^2

    so substituting we have

    INT sin-1 (x) = x sin-1x - INT (x / sqrt (1 - x^2)

    = x sun-1 x - ( - sqrt (1 - x^2)

    = x sin-1 x + sqrt (1 - x^2) + C
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