Answer
$\int A(t)dt=\begin{bmatrix}
-\cos t & \sin t & 0\\
-\sin t & -\cos t & \frac{t^2}{2}\\
0 & \frac{3t^2}{2} & t
\end{bmatrix}$
Work Step by Step
Given: $A(t)=\begin{bmatrix}
\sin t & \cos t & 0\\
-\cos t & \sin t & t \\
0 & 3t & 1
\end{bmatrix}$
The antiderivative of the matrix function is given by:
$\int^b_aA(t)dt=\int^b_aa_{ij}(t)dt$
Hence here, $\int A(t)dt=\int\begin{bmatrix}
\sin t & \cos t & 0\\
-\cos t & \sin t & t \\
0 & 3t & 1
\end{bmatrix}dt=\begin{bmatrix}
-\cos t & \sin t & 0\\
-\sin t & -\cos t & \frac{t^2}{2}\\
0 & \frac{3t^2}{2} & t
\end{bmatrix}$
