## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{2 \pi} \cos ^{8} x d x
$$
## Solution

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

$$
\begin{aligned}
& =\frac{\pi}{2}+\frac{\pi}{2^{5}}-0+\left.\frac{1}{2^{7}}\left(x+\frac{1}{8} \sin 8 x\right)\right|_{0} ^{2 \pi}=\frac{\pi}{2}+\frac{\pi}{2^{5}}+\frac{1}{2^{7}}\left(2 \pi+\frac{1}{8} \cdot 0\right)-\frac{1}{2^{7}}\left(0+\frac{1}{8} \cdot 0\right)= \\
& =\frac{\pi}{2}+\frac{\pi}{2^{5}}+\frac{\pi}{2^{6}}=\frac{32 \pi+2 \pi+\pi}{2^{6}}=\frac{35 \pi}{2^{6}}=\frac{35 \pi}{64}
\end{aligned}
$$

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 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+10-15$ " Categories: Kuznetsov's Problem Book Integrals Problem $10 \mid$ Integrals

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- Last edited: 11:28, 26 May 2009.
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## Problem Kuznetsov Integrals 10-16

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