Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 12

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


$
\begin{equation}
\begin{aligned}

\int x \cos^2 x dx =& \int x \left( \frac{\cos 2x + 1}{2} \right) dx
\qquad \text{Apply half-angle formula } \cos 2x = 2 \cos^2 x - 1
\\
\\
\int x \cos^2 x dx =& \frac{1}{2} \int (x \cos 2x + x ) dx
\\
\\
\int x \cos^2 x dx =& \frac{1}{2} \int x \cos 2x dx + \frac{1}{2} \int x dx

\end{aligned}
\end{equation}
$


We integrate term by term

@ 1st term

$\displaystyle \frac{1}{2} \int x \cos 2x dx \qquad$ Using Formula of Integration by parts $\int u dv = uv - \int v du$

Let $u = x$, then $du = dx$, and $dv = \cos 2x dx$, then $\displaystyle v = \frac{1}{2} \sin 2 x$. Thus,


$
\begin{equation}
\begin{aligned}

\frac{1}{2} \int x \cos 2x dx =& \frac{1}{2} \left[ (x) \left( \frac{1}{2} \sin 2x \right ) - \frac{1}{2} \int \sin 2x dx \right]
\\
\\
\frac{1}{2} \int x \cos 2x dx =& \frac{x \sin 2 x}{4} - \frac{1}{4} \left( \frac{- \cos 2x}{2} \right) + c
\\
\\
\frac{1}{2} \int x \cos 2x dx =& \frac{x \sin 2x}{4}+ \frac{\cos 2x}{8} + c

\end{aligned}
\end{equation}
$


@ 2nd term


$
\begin{equation}
\begin{aligned}

\frac{1}{2} \int x dx =& \frac{1}{2} \left( \frac{x^{1 + 1}}{1 + 1} \right) + c
\\
\\
\frac{1}{2} \int x dx =& \frac{1}{2} \left( \frac{x^2}{2} \right) + c
\\
\\
\frac{1}{2} \int x dx =& \frac{x^2 }{4} + c

\end{aligned}
\end{equation}
$


Add the results of integration term by term, we have

$\displaystyle \int x \cos^2 x dx = \frac{x \sin 2x}{4} + \frac{\cos 2x}{8} + \frac{x^2}{4} + c$