Question

Evaluate: `int_0^(pi/2) sin2x*tan^-1 (sinx)*dx`

Answer 1

Let I = `int_0^(pi/2) sin2x*tan^-1 (sinx)*dx`

= `int_0^(pi/2) tan^-1 (sinx)*(2sinx cosx)*dx`

Put sinx  = t

∴ cos x·dx = dt

When x = 0, t = sin 0 = 0

When x = `pi/(2), t = sin  pi/(2)` = 1

∴ I = `int_0^1(tan^-1 t)(2t)*dt`

= `[tan^-1 t int 2t  dt]_0^1 - int_0^1(d/dt (tan^-1 t) int 2t  dt)*dt`

= `[tan^-1 t int (t)^2 ]_0^1 - int_0^1 1/(1 + t^2)*t^2*dt`

= `[t^2 tan^-1  t]_0^1 - int_0^1 ((1 + t^2) - 1)/(1 + t^2)*dt`

= `[1*tan^-1 -  0] -int_0^1 (1 - 1/(1 + t^2))*dt`

= `pi/(4) - [t - tan^-1 t]_0^1`

= `pi/(4) - [(1 - tan^-1 1) - 0]`

= `pi/(4) - 1 + pi/(4)`

= `pi/(2) - 1`