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प्रश्न
Evaluate: `int_0^1 (1/(1 + x^2)) sin^-1 ((2x)/(1 + x^2)) "d"x`
उत्तर
Let I = `int_0^1 (1/(1 + x^2)) sin^-1 ((2x)/(1 + x^2)) "d"x`
Put x = tan θ
∴ dx = sec2θ dθ
When x = 0, θ = 0 and when x = 1, θ = `pi/4`
∴ I = `int_0^(pi/4)(1/(1 + tan^2 theta)) sin^-1((2tan theta)/(1 + tan^2theta)) sec^2theta "d"theta`
= `int_0^(pi/4) (1/(sec^2 theta)) sin^-1 (sin 2theta) sec^2theta "d"theta`
= `int_0^(pi/4) 2theta "d"theta`
= `2[theta^2/2]_0^(pi/4)`
= `(pi/4)^2 - 0`
∴ I = `pi^2/16`
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