Question

`int_(-2)^2 sqrt((2 - x)/(2 + x))` = ?

Options

• `pi/4`

• `pi/2`

• `pi`

• 2`pi`

Answer 1

2`pi`

Explanation:

`int_(-2)^2 sqrt((2 - x)/(2 + x))"d"x = int_(-2)^2 sqrt((2 - x)/(2 + x) * (2 - x)/(2 - x))`dx

= `int_(-2)^2 (2 - x)/(sqrt(4 - x^2))`dx

= 2`int_(-2)^2 "dx"/(sqrt(4 - x^2)) - int_(-2)^2 x/(sqrt(4 - x^2))`dx

= `2 [sin(x/2)]_-2^2 + [sqrt(4 - x^2)]_-2^2`

= 2 [sin^-1 1 - sin^-1 (- 1)] + 0

= `2 ((2pi)/2)` = 2π