Question

Evaluate: `int_0^1 (1/(1 + x^2)) sin^-1 ((2x)/(1 + x^2))  "d"x`

Answer 1

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`