Question

Evaluate the definite integral:

`int_0^(pi/2) sin 2x tan^(-1) (sinx) dx`

Answer 1

Let I = `int_0^(pi/2) sin 2x tan^-1 (sin x) dx`

`= 2 int_0^(pi/2) sin x cos x tan^-1 (sin x) dx`

Putting sin x = t, cos x dx = dt

When x = 0, t = sin 0 ⇒ t = 0

And when `x = pi/2, t = sin pi/2`

=> t = 1

∴ `I = 2 int_0^1 t tan^-1 t  dt`

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

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

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

`= [t^2 tan^-1 (t) - t + tan^-1 t]_0^1`

`= tan^-1 (1) - 1 + tan^-1`

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

`= pi/2 - 1`