If I ₁ =∫ sin ^-1 x dx and I ₂ = ∫ sin ^-1 √1-x ² dx then show that I ₁ +I ₂ = pie x / 2

we shld know that

arc.sin x +arc.sin y = arc.sin ( x.(1-y²)^1/2 + y.(1-x²)^1/2 )  for -1 < x,y < 1and x²+y²<=1

here put y = (1-x²)^1/2  and it satisfies that for -1 < x < 1 , 0< (1-x²)^1/2 < 1  

                                                                                           and  x²+1-x²=1

so I₁ + I₂ = integral ( arc.sin ( x² + (1-x²)^1/2.(1-x²)^1/2 ))

                = integral ( arc.sin ( 1 ))

               = integral ( Pi/2 )

              = x.Pi/2 

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IIT Kgp - 2005 batch