Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 20

Determine the integral $\displaystyle \int \cos^2 x \sin 2x dx$


$
\begin{equation}
\begin{aligned}

\int \cos^2 x \sin 2x dx =& \int \cos^2 x (2 \sin x \cos x) dx \qquad \text{Apply Trigonometric Identity } \sin 2x = 2 \sin x \cos x
\\
\\
\int \cos^2 x \sin 2x dx =& 2 \int \cos^3 x \sin x dx

\end{aligned}
\end{equation}
$


Let $u = \cos x$, then $du = - \sin x dx$, so $\sin x dx = -du$. Thus,


$
\begin{equation}
\begin{aligned}

2 \int \cos^3 x \sin x dx =& 2 \int u^3 \cdot -du
\\
\\
2 \int \cos^3 x \sin x dx =& -2 \int u^3 du
\\
\\
2 \int \cos^3 x \sin x dx =& -2 \left( \frac{u^{3 + 1}}{3 + 1} \right) + c
\\
\\
2 \int \cos^3 x \sin x dx =& \frac{-2u^4}{4} + c
\\
\\
2 \int \cos^3 x \sin x dx =& \frac{-u^4}{2} + c
\\
\\
2 \int \cos^3 x \sin x dx =& \frac{-(\cos x)^4}{2} + c
\\
\\
2 \int \cos^3 x \sin x dx =& \frac{-\cos^4 x}{2} + c


\end{aligned}
\end{equation}
$