Answer
$2 \pi -2$
Work Step by Step
Given: $\delta =2-z$
As we know that $ds=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dz}{dt})^2} dt$
or, $ds=\sqrt{(0)^2+( -\sin t )^2+(\cos t)^2} dt$ and $ ds=(1)dt= dt$
Now, $I_x=\int_C (y^2+z^2) \delta ds=\int_C (y^2+z^2) (2-z) ds$
or, $I_x=\int_0^{\pi} (cos^2 t+\sin^2 t)( 2-\sin t) dt$
So, $I_x= \int_0^{\pi} (2-\sin t dt)=2 \pi -2$