$\vec{a}, \vec{b}$ and $\vec{c}$ are constant linearly independent vectors and $\vec{r}$ is the position vector. Determine the integral $$\int_L \vec{A} \dot{} d\vec{r}$$ with $$\vec{A} = \vec{a}(\vec{b} \dot{} \vec{r})+\vec{b}(\vec{r} \dot{} \vec{a})+\vec{r}(\vec{a} \dot{} \vec{b})$$ and $L$ as the curve $$\vec{r(t)} = \vec{a}cost + \vec{b}sint + \vec{c}sin2t$$ where $0 \leq t \leq \pi/2$.
Attemped solution
Obviously I could evaluate $$\int_{0}^{\pi/2} [\vec{a}(\vec{b} \dot{}( \vec{a}cost + \vec{b}sint + \vec{c}sin2t))+\vec{b}((\vec{a}cost + \vec{b}sint + \vec{c}sin2t) \dot{} \vec{a})+(\vec{a}cost + \vec{b}sint + \vec{c}sin2t)(\vec{a} \dot{} \vec{b})] \dot{} (-\vec{a}sint + \vec{b}cost + 2\vec{c}cos2t) dt$$ but I rather not. Is there a simpler way?
According to the definition of $A$ you have $$A\cdot r'=\bigl((a\cdot r)(b\cdot r)\bigr)'+{1\over2}\bigl(|r|^2\bigr)'(a\cdot b)\ .$$ Now use $\int_p^q f'(t)\>dt=f(q)-f(p)$. In the end there will be no integral to compute.