1. integrate (sin^-1)dx/((1-x^2)^3/2) sin^-1 aka arcsin 2. integrate dx/((1-x^2)^3/2) by using 1/z Any and all help will be appreciated!
#2 dx/(1-x^2)^(3/2) If you let x = 1/z dx = -1/z^2 dz 1-x^2 = 1 - 1/z^2 = (z^2-1)/z^2 Now you have -dz/z^2 z^3/(z^2-1)^(3/2) = -z dz/(z^2-1)^(3/2) and the integral is simply 1/√(z^2-1) = x/√(1-x^2)
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