Question

`int_(pi/4)^(pi/2) sqrt(1-sin 2x)  dx =` ______.

विकल्प

• `sqrt2`

• `sqrt2 - 1`

• `1/(sqrt2-1)`

• `sqrt2 + 1`

Answer 1

`int_(pi/4)^(pi/2) sqrt(1-sin 2x)  dx =underline (sqrt2 - 1)`.

Explanation:

`int_(pi/4)^(pi/2) sqrt(1-sin 2x)  dx`

`=int_(pi/4)^(pi/2) sqrt((sinx - cosx)^2)  dx`

`=int_(pi/4)^(pi/2) (sinx - cosx)  dx`   `...[because x  in (pi/4,pi/2) therefore sinx-cosx>0]`

`=[-cosx-sinx]_(pi/4)^(pi/2)`

`=[-cos(pi/2)-sin(pi/2)]-[-cos(pi/4)-sin(pi/4)]`

`=(0-1)-(-1/sqrt2-1/sqrt2)`

`=sqrt2-1`