Trouble calculating the line Integral of a vector field

Question

Let $\gamma$ be the curve in $\mathbb{R}^3$ defined by $$ \gamma(t)=\begin{pmatrix} 1-\frac{\cos(t)}{2}\\ \sin(t)\\ \sin(t) \end{pmatrix} $$ for $t\in [0,1]$, and define the vector field $F$ by $$ F(x,y,z)=\begin{pmatrix} 2x(y+z)\\ x^2+y^2\\ x^2+2yz \end{pmatrix} $$ for $(x,y,z)\in \mathbb{R}^3$. First, we see that $$ F(\gamma(t))=\begin{pmatrix} 2\left ( 1-\frac{1}{2}\cos(t) \right )2\sin( t)\\ \left ( 1-\frac{1}{2}\cos( t) \right )^2+\sin^2(t)\\ \left ( 1-\frac{1}{2}\cos( t) \right )^2+2\sin^2( t) \end{pmatrix} $$ then it takes really long time to calculate $F(\gamma(t))\cdot\gamma'(t)$, and then determining $\int_{0}^{1}F(\gamma(t))\cdot\gamma'(t)\,\mathrm{d}t$. I am asking if there is a shorter way to do that, or some lemma to get past the tedious part. Using the calculator, the integral will be zero.