Quick Integration with Vectors Question So I was given the following prompt:

The instantaneous rate of change of the vector-valued function $g(t)$ is given by $r(t)=\langle\sqrt{t^2+1}, \sin(t^2)\rangle$. If $g(7)=\langle\sqrt{2},\pi\rangle$, find $g(0)$.

I guess I'm a bit confused about the integration that would be involved with this problem. I understand that I'm going to have to integrate both given functions, which I would then plug $0$ in to, but I'm a bit confused about what this integration would look like. Some clarification would be appreciated!
Edit: I understand what the integral formula would look like here, it's the actual integration that's confusing to me, I'm not coming up with a normal integral when I integrate.
 A: You have
$$g(7) - g(0) = \int_0^7 r(t) dt.$$ hence
$$g(0)  = g(7) - \int_0^7 r(t) dt = \left(\sqrt 2 - \int_0^7 \sqrt{t^2+1} dt,\pi - \int_0^7\sin t^2 dt\right)$$
A: Take the indefinite integral of $r(t)$ to get $g(t)+c$, where $c$ is a constant vector. If you know $g(7)$, then you know $c$, and then you can find $g(0)$.
A: Hint
$$\begin{align*}\vec g(t)=\int{{\vec r\left( t \right)\,dt}}  & = \left\langle {\int{{f\left( t \right)\,dt}},\int{{g\left( t \right)\,dt}}} \right\rangle \color{red}{+\vec C}\end{align*}$$
To integrate $\sqrt{t^2+1}$, use trig-substitution, then Integration by Parts.
$\int{\sin{t^2}}dt$ is not an elementary integral; it requires a $u$-substitution that leads to a Fresnel Integral.
A: $$\int \left(\sqrt{t^2+1},\sin \left(t^2\right)\right) \, dt=\left(a+\frac{1}{2} \left(t \sqrt{t^2+1}+\text{arcsinh} (t)\right),b+\sqrt{\frac{\pi }{2}} S\left(\sqrt{\frac{2}{\pi }} t\right)\right)$$
Where $S(x)$ is the Fresnel integral.
To get the two constant we substitute $(\sqrt 2,\pi)$ so we get
$$\small g(t)=\left(\frac{1}{2} \left(t \sqrt{t^2+1}+\text{arcsinh} (t)\right)+\sqrt{2}+\frac{1}{2} \left(-35 \sqrt{2}-\text{arcsinh} (7)\right),\sqrt{\frac{\pi }{2}} S\left(\sqrt{\frac{2}{\pi }} t\right)-\sqrt{\frac{\pi }{2}} S\left(7 \sqrt{\frac{2}{\pi }}\right)+\pi \right)$$
Plugging $t=0$ we have
$$g(0)=\left(\sqrt{2}+\frac{1}{2} \left(-35 \sqrt{2}-\text{arcsinh}(7)\right),\pi -\sqrt{\frac{\pi }{2}} S\left(7 \sqrt{\frac{2}{\pi }}\right)\right)\approx (-24.6566,2.53571)$$
