Question

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

Answer 1

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

Put x = tan t, i.e. t = tan^–1x
∴ dx = sec²t·dt
When x = 0, t = tan–1 0 = 0

When x = 1, t = tan^–1 = `pi/(4)`

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

= `int_0^(pi/4) sin^-1 (sin2t) sec^2t*dt`

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

= `[2t int sec^2t*dt]_0^(pi/4) - int_0^(pi/4) [d/dx (2t) int sec^2t*dt]`

= `[2t tan t]_0^(pi/4) - int_0^(pi/4) 2 tan t *dt`

= `[2* pi/4 tan  pi/4 - 0] - 2log(sec t)]_0^(pi/4)`

= `pi/2 - 2[log(sec  pi/4) - log (sec0)]`

= `pi/2 - 2[log sqrt(2) - log 1]`

= `pi/2 - 2[1/2 log 2 - 0]`

= `pi/(2) - log 2`.