Question

Evaluate the following : `int_0^1 (1/(1 + x^2))sin^-1((2x)/(1 + x^2))*dx`

Answer 1

Let I = `int_0^1 (1/(1 + x^2))sin^-1((2x)/(1 + x^2))*dx`

Put x = tan t, i.e. t = tan^–1x
∴ dx = sec²t dt

When x = 1, t = tan^–11 = `pi/(4)`

When x = 0, t = tan–1 0 = 0

∴ I = `int_0^(pi/4) (1/(1 + tan^2t))sin^-1 ((2tan t)/(1 + tan^2t))sec^2t*dt`

= `int_0^(pi/4) (1)/(sec^2t) sin^-1 (sin 2t) sec^2t*dt`

= `int_0^(pi/4) 2t*dt`

= `2int_0^(pi/4)t*dt`

= `2[(t^2)/2]_0^(pi/4)`

= `2[pi/(32) - 0]`

= `pi^2/(16)`.