`int_0^(pi/6) sqrt(1 + cos(2x)) dx` Evaluate the integral

`int_0^(pi/6)sqrt(1+cos(2x))dx`

Rewrite the integrand by using the identity:`cos(2x)=2cos^2(x)-1`

`=int_0^(pi/6)sqrt(1+2cos^2(x)-1)dx`

`=int_0^(pi/6)sqrt(2cos^2(x))dx`

`=sqrt(2)int_0^(pi/6)|cos(x)|dx`

`=sqrt(2)int_0^(pi/6)cos(x)dx` (as `0<=x<=pi/6=>cos(x)>=0=>|cos(x)|=cos(x)` )

`=sqrt(2)[sin(x)]_0^(pi/6)`

`=sqrt(2)[sin(pi/6)-sin(0)]`

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

`=sqrt(2)/2`

`=1/sqrt(2)`

`int_0^(pi/6)sqrt(1+cos(2x))dx`

Rewrite the integrand by using the identity:`cos(2x)=2cos^2(x)-1`

`=int_0^(pi/6)sqrt(1+2cos^2(x)-1)dx`

`=int_0^(pi/6)sqrt(2cos^2(x))dx`

`=sqrt(2)int_0^(pi/6)|cos(x)|dx`

`=sqrt(2)int_0^(pi/6)cos(x)dx` (as `0<=x<=pi/6=>cos(x)>=0=>|cos(x)|=cos(x)` )

`=sqrt(2)[sin(x)]_0^(pi/6)`

`=sqrt(2)[sin(pi/6)-sin(0)]`

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

`=sqrt(2)/2`

`=1/sqrt(2)`