Answer
$I_x=k( \dfrac{\pi}{2}-\dfrac{4}{3}) $ and $I_y=k( \dfrac{\pi}{2}-\dfrac{2}{3}) $
Work Step by Step
$I_x=\int_0^{\pi} \sin ^2 t[k(1-\sin t)] dt=k\int_0^{\pi} (\sin ^2 t-\sin^3 t) dt =k\int_0^{\pi} (\dfrac{1-\cos 2t}{2})- k\int_0^{\pi}(\dfrac{3 \sin t-\sin 3t}{4}) dt=k( \dfrac{\pi}{2}-\dfrac{4}{3}) $
Now, we have $I_y=\int_0^{\pi} \cos ^2 t[k(1-\sin t)] dt=(k) \times \int_0^{\pi} (\cos^2 t-\cos^2 t \sin t) dt =(k) \times \int_0^{\pi} \dfrac{1+ \cos 2t}{2} dt+k\int_0^{\pi} [\dfrac{\cos^3 t}{3}]_0^{\pi} =k( \dfrac{\pi}{2}-\dfrac{2}{3}) $
Hence, we have
$I_x=k( \dfrac{\pi}{2}-\dfrac{4}{3}) ; I_y=k( \dfrac{\pi}{2}-\dfrac{2}{3}) $
