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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
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
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