Question

Evaluate: `int_0^(π/2) sin 2x tan^-1 (sin x) dx`.

Answer 1

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

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

Let sin x = t

cos x dx = dt

When x = 0, t = 0 and x = `π/2`

`\implies` t = 1

`2int_0^1 t tan^-1 t  dt`

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

= `(π/4 . 1 - 0) - int_0^1 (t^2 + 1 - 1)/(1 + t^2)dt`

= `π/4 - int_0^1 (1 - 1/(1 + t^2))dt`

= `π/4 - [t - tan^-1 t]_0^1`

= `π/4 - [1 - π/4 - 0]`

= `π/4 - 1 + π/4`

= `π/2 - 1`.