I=∫dx/[√(1-x2)-1]

Dear suhas,

I=∫dx/[√(1-x²)-1] 

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/sin²ΦdΦ

-∫cot²ΦdΦ       this can be solved by byparts

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