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I=∫dx/[√(1-x2)-1]
Dear suhas,
now rationalise it multiply √(1-x2) + 1 with denominator and numerator we will get
I = ∫[{√(1-x2) + 1}/-x2]
I = ∫√(1-x2)/- x2 dx + ∫(1/-x2)dx
∫(1/-x2)dx = 1/x
∫√(1-x2)/- x2 dx
take x = sin Φ dx = cos Φ dΦ
∫[√(1-sinΦ2)/- sinΦ2 ] cosΦ dΦ
-∫cos2Φ/sin2