Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 44

Determine the integral $\displaystyle \int \cos (\pi x) \cos (4 \pi x) dx$

Let $u = \pi x$, then $du = \pi dx$, so $\displaystyle dx = \frac{du}{\pi}$. Thus


$
\begin{equation}
\begin{aligned}

\int \cos (\pi x) \cos (4 \pi x) dx =& \int \cos u \cos 4 u \cdot \frac{du}{\pi}
&& \text{Apply Trigonometric Identity } \cos A \cos B = \frac{1}{2} [\cos (A - B) + \cos(A + B)]
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\int \cos (\pi x) \cos (4 \pi x) dx =& \frac{1}{\pi} \int \frac{1}{2} [\cos (u - 4u) + \cos (u + 4u)] du
&&
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\\
\int \cos (\pi x) \cos (4 \pi x) dx =& \frac{1}{2 \pi} \int [\cos (-3u) + \cos (5u)] du
&& \text{Apply Even-Odd Identity } \cos (-u) = \cos (u)
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\int \cos (\pi x) \cos (4 \pi x) dx =& \frac{1}{2 \pi} \int [\cos (3u) + \cos (5u)] du
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\int \cos (\pi x) \cos (4 \pi x) dx =& \frac{1}{2 \pi} \int [\cos (3u) du + \frac{1}{2 \pi} \int \cos (5u) du

\end{aligned}
\end{equation}
$


For $\cos (3u)$, let $v = 3u$, then $dv = 3du$, so $\displaystyle du = \frac{dv}{3}$ and for $\cos (5u)$, let $w = 5u$, then $dw = 5du$, so $\displaystyle du = \frac{dw}{5}$. Therefore,


$
\begin{equation}
\begin{aligned}

\frac{1}{2 \pi} \int [\cos (3u) du + \frac{1}{2 \pi} \int \cos (5u) du =& \frac{1}{2 \pi} \int \cos v \cdot \frac{dv}{3} + \frac{1}{2 \pi} \int \cos w \cdot \frac{dw}{5}
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\frac{1}{2 \pi} \int [\cos (3u) du + \frac{1}{2 \pi} \int \cos (5u) du =& \frac{1}{6 \pi} \int \cos v dv + \frac{1}{10 \pi} \int \cos w dw
\\
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\frac{1}{2 \pi} \int [\cos (3u) du + \frac{1}{2 \pi} \int \cos (5u) du =& \frac{1}{6 \pi} \sin v + \frac{1}{10 \pi} \sin w + c
\\
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\frac{1}{2 \pi} \int [\cos (3u) du + \frac{1}{2 \pi} \int \cos (5u) du =& \frac{\sin 3 u}{6 \pi} + \frac{\sin 5 u}{10 \pi} + c
\\
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\frac{1}{2 \pi} \int [\cos (3u) du + \frac{1}{2 \pi} \int \cos (5u) du =& \frac{\sin (3 \pi x)}{6 \pi} + \frac{\sin (5 \pi x)}{10 \pi} + c

\end{aligned}
\end{equation}
$