`int_0^(pi/4) sqrt(1 - cos(4 theta)) d theta` Evaluate the integral

`int_0^(pi/4)sqrt(1-cos(4theta))d theta`

Let' first compute the indefinite integral,

`intsqrt(1-cos(4theta))d theta`  

Apply the integral substitution.

Let `theta=u/2`

`=>d theta=1/2du`

`=int1/2sqrt(1-cos2u)du`

`=1/2intsqrt(1-cos2u)du`

use the identity:`cos(2x)=1-2sin^2(x)`

`=1/2intsqrt(1-(1-2sin^2u))du`

`=1/2intsqrt(2sin^2u)du`

`=1/2intsqrt(2)sqrt(sin^2(u))du`

`=sqrt(2)/2intsqrt(sin^2(u))du`

`=(1/sqrt(2))intsin(u)du`

Change the bounds from 0 to pi/4 to lower bound 0 and upper bound pi/2.

`int_0^(pi/2) sin(u) du`

`-cos(u)/sqrt(2)) |_0^(pi/2)`  

`= 0 - -1/sqrt(2) = 1/sqrt(2)`