Put the equation in standard form by replacing
\(\sec\theta\) with
\(\frac{1}{\cos\theta}\) and moving the
\(A'\) term to the left:
\begin{equation*}
\frac{1}{\cos\theta} A' - A = \frac{\theta e^{\large\theta^2 + \sin\theta}}{\cos \theta}\text{.}
\end{equation*}
Multiplying this by \(\cos\theta\) allows us to identify \(P\) and compute \(\mu\text{:}\)
\begin{equation*}
A' + \us{P}{\boxed{- \cos\theta}}\ A = \theta e^{\large \theta^2 + \sin\theta}
\quad \implies \quad
\mu = e\vphantom{\large|}^{\large\int -\cos\theta d\theta} = e^{\large -\sin\theta}
\end{equation*}
Multiply by \(e^{1/x}\) and complete the product rule on the left side.
\begin{gather*}
e^{\large -\sin\theta}A' - \cos\theta e^{\large -\sin\theta}A = \theta e^{\large\theta^2 + \sin\theta} \\
\frac{d}{d\theta}\left[ e^{\large -\sin\theta}A \right] = \theta e^{\large\theta^2} \text{.}
\end{gather*}
Integrate both sides and use
\(u\)-substitution:
\begin{gather*}
\int \frac{d}{d\theta}\left[ e^{\large -\sin\theta}A \right] d\theta = \int \theta e^{\large\theta^2} d\theta\\
e^{\large -\sin\theta}A = \frac{1}{2}e^{\large\theta^2} + c\\
A = \frac{1}{2}e^{\large\theta^2 + \sin\theta} + ce^{\large\sin\theta}
\end{gather*}
\begin{align*}
u \amp = \theta^2 \\
du \amp = 2\theta\ d\theta
\end{align*}
Finally, we apply the initial condition to determine the constant \(c\text{.}\)
\begin{align*}
-\frac{1}{2} = A(0) \amp = \frac{1}{2}e^{\large 0^2 + \sin(0)} + ce^{\large\sin(0)} \\
\amp = \frac{1}{2}e^0 + ce^0 \\
\amp = \frac{1}{2} + c\quad \implies \quad c = -1
\end{align*}
We can then write the particular solution to the initial value problem.
\begin{equation*}
A(\theta) = \frac{1}{2}e^{\large\theta^2 + \sin\theta} - e^{\large\sin\theta}
\end{equation*}
