## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\pi} 2^{4} \cdot \cos ^{8} x d x
$$
## Solution

$$
\begin{aligned}
& \int_{0}^{\pi} 2^{4} \cdot \cos ^{8} x d x=\left|\cos ^{2} x=\frac{1}{2}(1+\cos 2 x)\right|=\int_{0}^{\pi} 2^{4} \cdot \frac{1}{2^{4}}(1+\cos 2 x)^{4} d x= \\
& =\int_{0}^{\pi}\left(1+4 \cos 2 x+6 \cos ^{2} 2 x+4 \cos ^{3} 2 x+\cos ^{4} 2 x\right) d x= \\
& =\int_{0}^{\pi}(1+4 \cos 2 x) d x+\int_{0}^{\pi} 6 \cos ^{2} 2 x d x+\int_{0}^{\pi} 4 \cos ^{3} 2 x d x+\int_{0}^{\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} \cdot d(\sin 2 x)=\cos 2 x d x\right|= \\
& =\left.(x+2 \sin 2 x)\right|_{0} ^{\pi}+\int_{0}^{\pi} 6 \cdot \frac{1}{2}(1+\cos 4 x) d x+\int_{0}^{\pi} 4\left(1-\sin ^{2} 2 x\right) \cdot \frac{1}{2} \cdot d(\sin 2 x)+ \\
& +\int_{0}^{\pi} \frac{1}{2^{2}}(1+\cos 4 x)^{2} d x= \\
& =(\pi+2 \sin 2 \pi)-(0+2 \sin 0)+\left.3\left(x+\frac{1}{4} \sin 4 x\right)\right|_{0} ^{\pi}+\left.2\left(\sin 2 x-\frac{1}{3} \sin ^{3} 2 x\right)\right|_{0} ^{\pi}+ \\
& +\int_{0}^{\pi} \frac{1}{2^{2}}\left(1+2 \cos 4 x+\cos ^{2} 4 x\right) d x= \\
& =\pi-0+3\left(\pi+\frac{1}{4} \sin 4 \pi\right)-3\left(0+\frac{1}{4} \sin (4 \cdot 0)\right)+2\left(\sin 2 \pi-\frac{1}{3} \sin ^{3} 2 \pi\right)-2\left(\sin 0-\frac{1}{3} \sin ^{3}\right. \\
& +\frac{1}{2^{2}} \cdot \int_{0}^{\pi}(1+2 \cos 4 x) d x+\frac{1}{2^{2}} \cdot \int_{0}^{\pi} \cos ^{2} 4 x d x= \\
& =\pi+3 \pi-0+0-0+\left.\frac{1}{2^{2}} \cdot\left(x+2 \cdot \frac{1}{4} \sin 4 x\right)\right|_{0} ^{\pi}+\frac{1}{2^{2}} \cdot \int_{0}^{\pi} \cos ^{2} 4 x d x= \\
& =\left|\cos ^{2} 4 x=\frac{1}{2}(1+\cos 8 x)\right|= \\
& =4 \pi+\frac{1}{2^{2}} \cdot\left(\pi+2 \cdot \frac{1}{4} \sin 4 \pi\right)-\frac{1}{2^{2}} \cdot\left(0+2 \cdot \frac{1}{4} \sin (4 \cdot 0)\right)+\frac{1}{2^{2}} \cdot \int_{0}^{\pi} \frac{1}{2}(1+\cos 8 x) d x= \\
& =4 \pi+\frac{\pi}{4}-0+\left.\frac{1}{2^{3}}\left(x+\frac{1}{8} \sin 8 x\right)\right|_{0} ^{\pi}=4 \pi+\frac{\pi}{4}+\frac{1}{2^{3}}\left(\pi+\frac{1}{8} \sin 8 \pi\right)-\frac{1}{2^{3}}\left(0+\frac{1}{8} \sin (8 \cdot 0)\right)= \\
& =\frac{17 \pi}{4}+\frac{\pi}{8}-0=\frac{35 \pi}{8}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\� \%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5\%D0\%B3 $\% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_10-20$ »

Categories: Kuznetsov's Problem Book Integrals Problem $10 \mid$ Integrals

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- Last edited: 06:21, May 30, 2010.
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## Problem Kuznetsov Integrals 10-21

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