Answer
$\dfrac{5\pi}{16}$
Work Step by Step
The formula to use is
74. $\displaystyle \int\cos^{n}udu=\frac{1}{n}\cos^{n-1}u\sin u+\frac{n-1}{n}\int\cos^{\mathrm{n}-2}udu$
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Apply the formula
$\displaystyle \int\cos^{6}\theta d\theta=\frac{1}{6}\cos^{5}\theta\sin\theta+\frac{5}{6}\int\cos^{4}\theta d\theta$
Apply the formula again
$=\displaystyle \frac{1}{6}\cos^{5}\theta\sin\theta+\frac{5}{6}\left[\frac{1}{4}\cos^{3}\theta\sin\theta+\frac{3}{4}\int\cos^{2}\theta d\theta \right]$
and, again
$=\displaystyle \frac{1}{6}\cos^{5}\theta\sin\theta+\frac{5}{6}\left[\frac{1}{4}\cos^{3}\theta\sin\theta+\frac{3}{4}\left(\frac{1}{2}\cos\theta\sin\theta+\frac{1}{2}\int d\theta \right)\right]$
$=\displaystyle \frac{1}{6}\cos^{5}\theta\sin\theta+\frac{5}{6}\left[\frac{1}{4}\cos^{3}\theta\sin\theta+\frac{3}{4}\left(\frac{1}{2}\cos\theta\sin\theta+\frac{1}{2}\theta \right)\right]+C$
All the terms except the last are zero when evaluating for $\theta=\pi,$
because $\sin\pi=0.$
All the terms including the last are zero when evaluating for $\theta=0.$
because $\sin 0=0$
$\displaystyle \int_{0}^{\pi}\cos^{6}\theta d\theta=\frac{5}{6}[\frac{3}{4}(\frac{1}{2}\cdot\pi)=\frac{5\pi}{16}$
