Question

`int_0^1 sin^-1 ((2x)/(1 + x^2))"d"x` = ______.

Options

• `pi/2 + log 2`

• `pi/2 - log 2`

• `pi/2 + 1/2 log 2`

• `pi/2 - 1/2 log 2`

Answer 1

`int_0^1 sin^-1 ((2x)/(1 + x^2))"d"x` = `pi/2 - log 2`.

Explanation:

Let I = `int_0^1 sin^-1 ((2x)/(1 + x^2))"d"x`

Put x = θ

⇒ dx = sec²θ dθ

∴ I  = `int_0^(pi/4) sin^1 ((2tantheta)/(1 ++ tan^2theta))sec^2theta "d"theta`

= `int_0^(pi/4) sin^-1(sin 2theta) sec^2theta  "d"theta`

= `int_0^(pi/4) 2theta sec^3theta  "d"theta`

= `2 int_0^(pi/4) theta sec^2 theta  "d"theta`

= `2[[theta tan theta]_0^(pi/4) - int_0^(pi/4) 1*tan theta  "d'theta]`

= `2[(pi/4 tan  pi/4 - 0) - [log|sectheta|]_0^(pi/4)]`

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

= `2[pi/4 - (log sqrt(2) - log 1)]`

= `2(pi/4 - 1/2 log 2)`

∴ I = `pi/2 - log 2`