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

Question

If I 1 =&int; sin -1 x dx and I 2 = &int; sin -1 &radic;1-x 2 dx then show that I 1 +I 2 = pie x / 2

Ramesh V · Answer

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)^1/2and 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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Naga Ramesh
IIT Kgp - 2005 batch

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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

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

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