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 

--

Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and
we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best.

Regards,
Naga Ramesh
IIT Kgp - 2005 batch