Answer
$\int A(t)dt=\begin{bmatrix}
\frac{e^{2t} }{2}& \frac{-\cos2t}{2}\\
\frac{t^3}{3}-5t & te^t-e^t\\
\tan t & \frac{3t^2}{2}+\cos t
\end{bmatrix}$
Work Step by Step
Given: $A(t)=\begin{bmatrix}
e^{2t} & \sin 2t\\
t^2-5 & te^t\\
\sec^2t & 3t-\sin t
\end{bmatrix}$
The antiderivative of the matrix function is given by:
$\int^b_aA(t)dt=\int^b_aa_{ij}(t)dt$
Hence here, $\int A(t)dt=\int\begin{bmatrix}
e^{2t} & \sin 2t\\
t^2-5 & te^t\\
\sec^2t & 3t-\sin t
\end{bmatrix}=\begin{bmatrix}
\frac{e^{2t} }{2}& \frac{-\cos2t}{2}\\
\frac{t^3}{3}-5t & te^t-e^t\\
\tan t & \frac{3t^2}{2}+\cos t
\end{bmatrix}$
