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प्रश्न
Prove the following:
`int_0^1sin^(-1) xdx = pi/2 - 1`
बेरीज
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उत्तर
Let `I = int_0^1 sin^-1 x dx`
`= int_0^1 sin^-1 x * 1 dx`
Integrating by parts,
`= [sin^-1 x * x]_0^1 - int_0^1 1/sqrt(1 - x^2) * x dx`
`= [x sin^-1 x]_0^1 + 1/2 int_0^1 (-2x)/sqrt(1 - x^2) dx`
`= {x sin^-1 x + 1/2 [(1 - x^2)^(1/2)/(1/2)]}_0^1`
`= [(sin^-1) + 1/2 xx 2/1 (0 - 1)]`
`= pi/2 - 1`
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