Answer
$I_x=k( \dfrac{\pi}{2}-\dfrac{4}{3}) $ and $I_y=k( \dfrac{\pi}{2}-\dfrac{2}{3}) $
Work Step by Step
Here, $I_x=\int_C y^2 \rho(x,y) ds=\int_0^{\pi} \sin ^2 t[k(1-\sin t)] dt=k\int_0^{\pi} (\sin ^2 t-\sin^3 t) dt $
or, $=k\int_0^{\pi} (\dfrac{1-\cos 2t}{2})-(\dfrac{3 \sin t-\sin 3t}{4}) dt $
or, $=k\int_0^{\pi} \dfrac{2-2 \cos 2t-3 \sin t+\sin 3t}{4} dt $
or, $=k( \dfrac{\pi}{2}-\dfrac{4}{3}) $
Now, $I_y=\int_C x^2 \rho(x,y) ds=\int_0^{\pi} \cos ^2 t[k(1-\sin t)] dt=k\int_0^{\pi} (\cos^2 t-\cos^2 t \sin t) dt $
or, $=k\int_0^{\pi} (\dfrac{1+\cos 2t}{2})-(\cos t)^2(-\sin t dt)$
or, $=k\int_0^{\pi} \dfrac{1+ \cos 2t}{2} dt+k[\dfrac{\cos^3 t}{3}]_0^{\pi} $
or, $=k( \dfrac{\pi}{2}-\dfrac{2}{3}) $
Hence, $I_x=k( \dfrac{\pi}{2}-\dfrac{4}{3}) $ and $I_y=k( \dfrac{\pi}{2}-\dfrac{2}{3}) $
