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Integral of x^2 arcsin(x) dx?

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  • 2 months ago
    Favourite answer

    ∫x²arcsin(x) dx =

    by parts

    finite f(x) = arcsinx ⇒ f'(x) = 1/√(1-x²)

    differ. g'(x) = x² ⇒ g(x) = x³/3

    = x³*arcsin(x)/3 - (1/3)∫x³/√(1-x²) dx =

    separately evaluate the last integral

    ∫x³/√(1-x²) dx = 

    by substitution

    y=√(1-x²) ⇒ y² = 1-x² ⇒ x² = 1-y² ⇒ x = √(1-y²) ⇒ x³=(1-y²)*√(1-y²)

    dy = x/√(1-x²) dx = √(1-y²)/y dx ⇒ y/√(1-y²) dy = dx 

    = ∫[(1-y²)*√(1-y²)/y]*y/√(1-y²) dy =

    = ∫(1-y²) dy =

    = y³/3 - y +c =

    = (1-x²)*√(1-x²)/3 - √(1-x²) + c =

    = √(1-x²)*(1/3-x²/3-1) +c =

    = √(1-x²)*(-x²/3-2/3) + c =

    = -(1/3)*(x²+2)√(1-x²) +c 

    back to the integral

    = x³*arcsin(x)/3 + (1/9)*(x²+2)√(1-x²) +c

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