Integrate x/((1-x^2)^0.5) with respect to x

x = sin(u), dx/du = cos(u), dx = cos(u) * du,[x/(1-x^2)^0.5)] * dx = [sin(u)/((1-(sin(u)^2))^0.5] * cos(u) * du = [sin(u)/(cos(u)^2)^0.5] * cos(u) * du = sin(u) * duIntegral of sin(u) * du = -cos(u) = -(1-sin(u)^2)^0.5 = -(1-x^2)^0.5

Answered by Andrew P. Maths tutor

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