<s>[INST] Evaluate $\int \cos(3x) \sin(2x) \, dx$. 
Evaluate the indefinite integral 
  \begin{align}
\int \cos(3x) \sin(2x) \, dx
\end{align}

Please see my work attempt as my answer below.
 [/INST] Employing trigonometric identities, we have
\begin{align}
\int \cos3x \sin2x \, dx &= \int \cos(2x+x) \sin2x \, dx \\
&= \int [\cos2x\cos x  - \sin2x\sin x ]\sin2x \, dx \\
&= \int \cos2x \cos x  \sin2x - \sin^2 2x \sin x  \, dx \\
&= \int (\cos^2 x - \sin^2 x) \cos x (2 \sin x \cos x) - (2 \sin x \cos x)^2 \sin x \, dx \\
&= \int 2 \cos^4 x \sin x - 2 \cos^2 x \sin^3 x - 4 \sin^3 x \cos^2 x \, dx \\
&= \int 2 \cos^4 x \sin x - 6 \sin^3 x \cos^2 x \, dx \\
&= \int 2 \cos^4 x \sin x \, dx - \int 6 \sin^3 x \cos^2 x \, dx \\
\end{align}
For the first term, we invoke the substitution rule by letting $u = \cos x$. Then $du = -\sin x \, dx$. Thus,
\begin{align}
\int 2 \cos^4 x \sin x \, dx &= \int 2u^4 (-du) \\
&= -\frac{2}{5}u^5 + C \\
&= -\frac{2}{5} \cos^5 x + C
\end{align}
Similarly for the second term, let $v = \cos(x)$. Then $dv = -\sin(x) \, dx$. Thus,
\begin{align}
\int 6 \sin^3 x \cos^2 x \, dx &= \int 6 \sin^2 x \cos^2 x \sin x \, dx \\
&= \int 6 (1 - \cos^2 x) \cos^2 x \sin x \, dx \\
&= \int 6 (1-v^2)v^2 (-dv) \\
&= \int -6v^2 + 6v^4 dv \\
&= -2v^3 + \frac{6}{5}v^5 + C \\
&= -2 \cos^3 x + \frac{6}{5} \cos^5 x + C
\end{align}
Combining the two terms together, we get
\begin{align}
\int \cos 3x \sin 2x \, dx &= \left(-\frac{2}{5} \cos^5 x + C\right) - \left(-2 \cos^3 x + \frac{6}{5} \cos^5 x + C \right) \\
&= -\frac 85 \cos^5 x + 2 \cos^3 x + C
\end{align}
</s>