Question

Evaluate the integral by using substitution.

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

Answer 1

Let  `int_0^1  sin^-1 ((2x)/(1 + x^2))  dx`

Substituting x = tan θ

`dx = sec^2 theta  d  theta`

And `(2 tan theta)/(1 + tan^2 theta) = sin 2 theta`

When x = 0

⇒ θ = 0

or x = 1

`=> theta = pi/4` 

Hence, `int_0^(pi/4)  sin^-1 (sin 2 theta) xx sec^2 theta  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] - 2 [log cos theta]_0^(pi/4)`

`= pi/4 + 2 [log cos  pi/4 - log cos 0]`

`= pi/2 + 2 [log  1/sqrt2 - log 1]`

`= pi/2 - log 2`