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

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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} $