Answer
$\dfrac{83}{3}$
Work Step by Step
$I=\int_{C_1}x^2 dx+y^2 dy+\int_{C_2}x^2 dx+y^2 dy$
$=8\int_{0}^{\pi/2} \sin^2 (2t) \cos t-\cos^2 t \sin t dt+\int_{0}^{1} (4t)^2 (4 dt) +(2+t)^2 dt$
Plug $a=\sin t; b=\cos t $ and $da=\cos t dt; db=-\sin t dt$
$=8 \int_{0}^{\pi/2} a^2 da+8\int b^2 db+\int_0^1 (64t^2+4+4t+t^2) dt$
$=\dfrac{8}{3} \sin^3(t)]_0^{\pi/2}+\dfrac{8}{3}(\cos^3 t)_0^{\pi/2}+(64(t^3/3)+4t+4(t^2/2)+(t^3/3)) _0^1$
$= \dfrac{8}{3}(1-0+0-1)+\dfrac{83}{3}$
$=\dfrac{83}{3}$
