Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 10

Determine the integral $\displaystyle \int^{\pi}_0 \cos^6 \theta d \theta$


$
\begin{equation}
\begin{aligned}

\int^{\pi}_0 \cos^6 \theta d \theta =& \int^{\pi}_0 (\cos^2 \theta)^3 d \theta
\qquad \text{Apply the half angle formula } \cos (2 \theta) = 2 \cos^2 \theta - 1
\\
\\
\int^{\pi}_0 \cos^6 \theta d \theta =& \int^{\pi}_0 \left( \frac{\cos 2 \theta + 1}{2} \right)^3 d \theta
\\
\\
\int^{\pi}_0 \cos^6 \theta d \theta =& \int^{\pi}_0 \left( \frac{\cos^3 2 \theta + 3 \cos^2 2 \theta + 3 \cos 2 \theta + 1}{8} \right) d \theta
\\
\\
\int^{\pi}_0 \cos^6 \theta d \theta =& \frac{1}{8} \int^{\pi}_0 (\cos^3 2 \theta + 3 \cos^2 2 \theta + 3 \cos 2 \theta + 1) d \theta
\\
\\
\int^{\pi}_0 \cos^6 \theta d \theta =& \frac{1}{8} \int^{\pi}_0 \cos^3 2 \theta d \theta + \frac{1}{8} \int^{\pi}_0 3 \cos^2 2 \theta d \theta + \frac{1}{8} \int^{\pi}_0 3 \cos 2 \theta d \theta + \frac{1}{8} \int^{\pi}_0 1 d \theta
\\
\\
\int^{\pi}_0 \cos^6 \theta d \theta =& \frac{1}{8} \int^{\pi}_0 + \cos^3 2 \theta d \theta + \frac{3}{8} \int^{\pi}_0 \cos^2 2 \theta d \theta + \frac{3}{8} \int^{\pi}_0 \cos 2 \theta d \theta + \frac{1}{8} \int^{\pi}_0 1 d \theta

\end{aligned}
\end{equation}
$


We let $u = 2 \theta$, then $du = 2 d \theta$, so $\displaystyle d \theta = \frac{du}{2}$. When $\theta = \theta, u = 0$ and when $\displaystyle \theta = \pi, u = 2 \pi$. We will integrate the equation term by term

@ 1st term


$
\begin{equation}
\begin{aligned}

\frac{1}{8} \int^{\pi}_0 \cos^3 2 \theta d \theta =& \frac{1}{8} \int^{2 \pi}_0 \cos^3 u \cdot \frac{du}{2}
\\
\\
\frac{1}{8} \int^{\pi}_0 \cos^3 2 \theta d \theta =& \frac{1}{16} \int^{2 \pi}_0 \cos^3 u du
\\
\\
\frac{1}{8} \int^{\pi}_0 \cos^3 2 \theta d \theta =& \frac{1}{16} \int^{2 \pi}_0 \cos^2 u du \cos u du
\qquad \text{Apply Trigonometric Identity } \cos^2 x = 1 - \sin^2 x
\\
\\
\frac{1}{8} \int^{\pi}_0 \cos^3 2 \theta d \theta =& \frac{1}{16} \int^{2 \pi}_0 (1 - \sin^2 u) \cos u du

\end{aligned}
\end{equation}
$


Let $v = \sin u$, then $dv = \cos u du$. When $u = 0, v = 0$ and when $u = 2 \pi, v = 0$. Thus,


$
\begin{equation}
\begin{aligned}

\frac{1}{6} \int^{2 \pi}_0 (1 - \sin^2 u) \cos udu =& \frac{1}{6} \int^0_0 (1 - v^2) dv
\\
\\
\frac{1}{6} \int^{2 \pi}_0 (1 - \sin^2 u) \cos udu =& \frac{1}{16} \left[ v - \frac{v^{2 + 1}}{2 + 1} \right]^0_0
\\
\\
\frac{1}{6} \int^{2 \pi}_0 (1 - \sin^2 u) \cos udu =& \frac{1}{16} \left[ v - \frac{v^3}{3} \right]^0_0
\\
\\
\frac{1}{6} \int^{2 \pi}_0 (1 - \sin^2 u) \cos udu =& \frac{1}{16} (0)
\\
\\
\frac{1}{6} \int^{2 \pi}_0 (1 - \sin^2 u) \cos udu =& 0

\end{aligned}
\end{equation}
$


@ 2nd term


$
\begin{equation}
\begin{aligned}

\frac{3}{8} \int^{\pi}_0 \cos^2 2 \theta d \theta =& \frac{3}{8} \int^{2 \pi}_0 \cos^2 u \cdot \frac{du}{2}
\\
\\
\frac{3}{8} \int^{\pi}_0 \cos^2 2 \theta d \theta =& \frac{3}{16 } \int^{2 \pi}_0 \cos^2 u du
\qquad \text{Apply half-angle formula } \cos 2a = 2 \cos^2 a - 1
\\
\\
\frac{3}{8} \int^{\pi}_0 \cos^2 2 \theta d \theta =& \frac{3}{16} \int^{2 \pi}_0 \left( \frac{\cos 2 u + 1}{2} \right) du
\\
\\
\frac{3}{8} \int^{\pi}_0 \cos^2 2 \theta d \theta =& \frac{3}{32} \int^{2 \pi}_0 (\cos 2u + 1) du

\end{aligned}
\end{equation}
$


Let $v = 2u$, then $dv = 2 du$, so $\displaystyle du = \frac{dv}{2}$. When $u = 0, v = 0$ and when $u = 2 \pi, v = 4 \pi$. Therefore,


$
\begin{equation}
\begin{aligned}

\frac{3}{32} \int^{2 \pi}_0 (\cos 2 u + 1) du =& \frac{3}{32} \int^{4 \pi}_0 (\cos v + 1) \cdot \frac{dv}{2}
\\
\\
\frac{3}{32} \int^{2 \pi}_0 (\cos 2 u + 1) du =& \frac{3}{64} \int^{4 \pi}_0 (\cos v + 1) dv
\\
\\
\frac{3}{32} \int^{2 \pi}_0 (\cos 2 u + 1) du =& \frac{3}{64} \left[ \sin v + v \right]^{4 \pi}_0
\\
\\
\frac{3}{32} \int^{2 \pi}_0 (\cos 2 u + 1) du =& \frac{3}{64} (\sin 4 \pi + 4 \pi - \sin 0 - 0)
\\
\\
\frac{3}{32} \int^{2 \pi}_0 (\cos 2 u + 1) du =& \frac{3}{64} (0 + 4 \pi - 0 - 0)
\\
\\
\frac{3}{32} \int^{2 \pi}_0 (\cos 2 u + 1) du =& \frac{3}{64} (4 \pi)
\\
\\
\frac{3}{32} \int^{2 \pi}_0 (\cos 2 u + 1) du =& \frac{3 \pi}{16}

\end{aligned}
\end{equation}
$


@ 3rd term


$
\begin{equation}
\begin{aligned}

\frac{3}{8} \int^{\pi}_0 \cos 2 \theta d \theta =& \frac{3}{8} \int^{2 \pi}_0 \cos u \cdot \frac{du}{2}
\\
\\
\frac{3}{8} \int^{\pi}_0 \cos 2 \theta d \theta =& \frac{3}{16} \int^{2 \pi}_0 \cos u du
\\
\\
\frac{3}{8} \int^{\pi}_0 \cos 2 \theta d \theta =& \frac{3 }{16} \left[ \sin u \right]^{2 \pi}_0
\\
\\
\frac{3}{8} \int^{\pi}_0 \cos 2 \theta d \theta =& \frac{3}{16} (\sin 2 \pi - \sin 0)
\\
\\
\frac{3}{8} \int^{\pi}_0 \cos 2 \theta d \theta =& \frac{3}{16} (0)
\\
\\
\frac{3}{8} \int^{\pi}_0 \cos 2 \theta d \theta =& 0

\end{aligned}
\end{equation}
$


@ 4th term


$
\begin{equation}
\begin{aligned}

\frac{1}{8} \int^{\pi}_0 1 d \theta =& \frac{1}{8} \left[ \theta \right]^{\pi}_0
\\
\\
\frac{1}{8} \int^{\pi}_0 1 d \theta =& \frac{1}{8} (\pi - 0)
\\
\\
\frac{1}{8} \int^{\pi}_0 1 d \theta =& \frac{\pi}{8}

\end{aligned}
\end{equation}
$


We add all the results from integrating term by term.


$
\begin{equation}
\begin{aligned}

\int^{\pi}_0 \cos^6 \theta d \theta =& 0 + \frac{3 \pi}{16} + 0 + \frac{\pi}{8}
\\
\\
\int^{\pi}_0 \cos^6 \theta d \theta =& \frac{0 + 3 \pi + 0 2 \pi}{16}
\\
\\
\int^{\pi}_0 \cos^6 \theta d \theta =& \frac{5 \pi}{16}

\end{aligned}
\end{equation}
$