$\mathbf c$ is a differentiable path in $\mathbb R^3$. If $\mathbf F (\mathbf c(t))$ is parallel to $\mathbf c$'(t), show that $$\int_{\mathbf c} \mathbf F \cdot \mathrm d \mathbf s = \int_{\mathbf c} \lvert \mathbf F \rvert \mathrm d \mathrm s$$
I believe that $\mathbf F (\mathbf c(t))$ being parallel to $\mathbf c$'(t) means that $\mathbf F (\mathbf c(t))$ = $\alpha \ $$\mathbf c$'(t) where {$\alpha \in \mathbb R | \alpha \gt 0$}, and I know that $$\int_{\mathbf c} \mathbf F \cdot \mathrm d \mathbf s = \int_{\mathbf c} \mathbf F(\mathbf c(t)) \cdot \frac{\mathrm d \mathbf c}{\mathrm dt}\mathrm dt$$ and I was thinking if I substituted $\frac{\mathrm d \mathbf c}{\mathrm dt}$ = $\beta \ \mathbf F(\mathbf c(t))$, where $\beta = \frac{1}{\alpha}$, that the integral would become $$\int_{\mathbf c} \beta \ \mathbf F(\mathbf c(t)) \cdot \mathbf F(\mathbf c(t)) \mathrm dt = \int_{\mathbf c} \beta \ \lvert \mathbf F \rvert \mathrm dt$$ but I know this isn't the way to go and any help would be appreciated, cheers guys :)
Hint:
Going with your idea, letting $\mathbf{c}’=\beta \mathbf{F}$ and parametrizing the curve in terms of $t$, where $\beta(t)$ is positive (or better nonnegative) everywhere, we get,
$$\int_{c} \mathbf{F} \cdot \mathbf{ds}=\int_{c} \beta |\mathbf{F}|^2 dt$$
Now do you recognize $\beta |\mathbf{F}| dt$? Note since $\beta$ is nonegative, $|\beta|=\beta$ and $\beta|\mathbf{F}|=|\beta \mathbf{F}|$.