Answer
$r(t)=(t-\sin t) i+(3-2 \cos t)j+(t^3-t-4)k$
Work Step by Step
As we are given that $a(t)=\sin t i+2 \cos tj+6tk$ and $v(0)=-k$
Since, $v(t)=\int a(t)$ and $r(t)=\int v(t)$
Thus, $v(t)=\int (\sin t i+2 \cos tj+6tk) dt$
$\implies v(t)=(1-\cos t) i+2 \sin tj+(3t^2-1)k$
Now, $r(t)=\int [(1-\cos t) i+2 \sin tj+(3t^2-1)k] dt=(t-\sin t) i-2 \cos tj+(t^3-t)k+c$
Here, $c$ represents a constant of integration.
That is, $r(0)=j-4k \implies c= j-4k$
$r(t)=(t-\sin t) i-2 \cos tj+(t^3-t)k+j-4k$
Hence, $r(t)=(t-\sin t) i+(3-2 \cos t)j+(t^3-t-4)k$
