## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{2 \pi} \cos ^{8}\left(\frac{x}{4}\right) d x
$$
## Solution

$$
\begin{aligned}
& \int_{0}^{2 \pi} \cos ^{8}\left(\frac{x}{4}\right) d x=\left|\cos ^{2}\left(\frac{x}{4}\right)=\frac{1}{2}\left(1+\cos \left(\frac{x}{2}\right)\right)\right|=\int_{0}^{2 \pi} \frac{1}{2^{4}}\left(1+\cos \left(\frac{x}{2}\right)\right)^{4} d x= \\
& =\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi}\left(1+4 \cos \left(\frac{x}{2}\right)+6 \cos ^{2}\left(\frac{x}{2}\right)+4 \cos ^{3}\left(\frac{x}{2}\right)+\cos ^{4}\left(\frac{x}{2}\right)\right) d x= \\
& =\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi}\left(1+4 \cos \left(\frac{x}{2}\right)\right) d x+\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi} 6 \cos ^{2}\left(\frac{x}{2}\right) d x+\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi} 4 \cos ^{3}\left(\frac{x}{2}\right) d x+\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi} \cos \\
& =\left\lvert\, \cos ^{2}\left(\frac{x}{2}\right)=\frac{1}{2}(1+\cos x)\right., \cos ^{2}\left(\frac{x}{2}\right)=1-\sin ^{2}\left(\frac{x}{2}\right), 2 d\left(\sin \left(\frac{x}{2}\right)\right)=\cos \left(\frac{x}{2}\right) \\
& =\left.\frac{1}{2^{4}} \cdot\left(x+8 \sin \left(\frac{x}{2}\right)\right)\right|_{0} ^{2 \pi}+\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi} 6 \cdot \frac{1}{2}(1+\cos x) d x+\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi} 4\left(1-\sin ^{2}\left(\frac{x}{2}\right)\right) \cdot 2 d(\sin \\
& +\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi} \frac{1}{2^{2}}(1+\cos x)^{2} d x= \\
& =\frac{1}{2^{4}} \cdot(2 \pi+8 \sin \pi)-\frac{1}{2^{4}} \cdot(0+8 \sin 0)+\left.\frac{1}{2^{4}} \cdot 3(x+\sin x)\right|_{0} ^{2 \pi}+\frac{1}{2^{4}} \cdot 8\left(\sin \left(\frac{x}{2}\right)-\frac{1}{3} \sin ^{3}(\right. \\
& +\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi} \frac{1}{2^{2}}\left(1+2 \cos x+\cos ^{2} x\right) d x= \\
& =\frac{\pi}{2^{3}}-0+\frac{3}{2^{4}} \cdot(2 \pi+0)-\frac{3}{2^{4}} \cdot(0+0)+\frac{1}{2} \cdot\left(0-\frac{1}{3} \cdot 0^{3}\right)-\frac{1}{2} \cdot\left(0-\frac{1}{3} \cdot 0^{3}\right)+ \\
& +\frac{1}{2^{6}} \cdot \int_{0}^{2 \pi}(1+2 \cos x) d x+\frac{1}{2^{6}} \cdot \int_{0}^{2 \pi} \cos ^{2} x d x= \\
& =\frac{\pi}{2^{3}}+\frac{3 \pi}{2^{3}}+\left.\frac{1}{2^{6}} \cdot(x+2 \sin x)\right|_{0} ^{2 \pi}+\frac{1}{2^{6}} \cdot \int_{0}^{2 \pi} \cos ^{2} x d x=\left|\cos ^{2} x=\frac{1}{2}(1+\cos 2 x)\right|= \\
& =\frac{\pi}{2}+\frac{\pi}{2^{5}}-0+\frac{1}{2^{6}} \cdot(2 \pi+2 \cdot 0)-\frac{1}{2^{6}} \cdot(0+2 \cdot 0)+\frac{1}{2^{6}} \cdot \int_{0}^{2 \pi} \frac{1}{2}(1+\cos 2 x) d x=
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{\pi}{2}+\frac{\pi}{2^{5}}+\frac{\pi}{2^{6}}=\frac{32 \pi+2 \pi+\pi}{2^{6}}=\frac{35 \pi}{2^{6}}=\frac{35 \pi}{64}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_ \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+10-10 »$ Categories: Kuznetsov's Integral Problems 10-11

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- Last edited: 11:25, May 26, 2009.
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## Problem Kuznetsov Integrals 10-11

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