Question

Integrate the function in `sin^(-1) ((2x)/(1+x^2))`.

Answer 1

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

Put x = tan t

⇒ dx = sec² t dt

∴ `I = int sin^-1 ((2 tan t)/ (1 + tan^2 t)) sec^2 t dt`

`= int sin^-1 (sin 2t) sec^2 t dt`

`= 2t sec^2 t dt = 2 int sec^2 t dt`

`= 2 {t int sec^2 t dt - int [d/dt(t) * int sec^2 t  dt] dt}`

`= 2 [t tant  - int 1 * tan t  dt]`

= 2 t tan t + 2 log |cos t| + C

`= 2 tan^-1 x*x + 2 log |1/ sqrt (1 + x^2)| + C`     `...[∵ cos t = 1/ (sect) = 1/ (sqrt (1 + tan^2 t)) = 1/ (sqrt (1 + x^2))]`

`= 2 x tan^-1 x + 2 log |(1 + x^2)^(1/2)| + C`

`= 2 x tan^-1 x + 2 (- 1/2) log |1 + x^2| + C`

`= 2 x tan^-1 x - log |1 + x^2| + C`