Answer
$\dfrac{\pi^6}{3}+2 \pi$
Work Step by Step
$I=\int_{C}(x^2(x^2)+\sin x)(2x dx)=\int_{C}2x^{5}+2x \sin x dx$
$=\int_{0}^{\pi} x(2x^4+2\sin x) dx$
$=\int_{0}^{\pi}\dfrac{2x^6}{5}-2x \cos x-\int (2x^5/5)-2 \cos x dx$
$=\int_{0}^{\pi}\dfrac{x^6}{5}-2x \cos x+2 \sin x+k$
$=[\dfrac{x^6}{3}-2x \cos x+2 \sin x]_{0}^{\pi}$
$= \dfrac{\pi^6}{3}+2 \pi$
