## Problem Statement

Calculate the definite integral:

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

$$
\begin{aligned}
& \int_{0}^{\pi} 2^{4} \cdot \cos ^{8}\left(\frac{x}{2}\right) d x=\left|\cos ^{2}\left(\frac{x}{2}\right)=\frac{1}{2}(1+\cos x)\right|=\int_{0}^{\pi} 2^{4} \cdot \frac{1}{2^{4}}(1+\cos x)^{4} d x= \\
& =\int_{0}^{\pi}\left(1+4 \cos x+6 \cos ^{2} x+4 \cos ^{3} x+\cos ^{4} x\right) d x= \\
& =\int_{0}^{\pi}(1+4 \cos x) d x+\int_{0}^{\pi} 6 \cos ^{2} x d x+\int_{0}^{\pi} 4 \cos ^{3} x d x+\int_{0}^{\pi} \cos ^{4} x d x= \\
& =\left|\cos ^{2} x=\frac{1}{2}(1+\cos 2 x), \cos ^{2} x=1-\sin ^{2} x, d(\sin x)=\cos x d x\right|= \\
& =\left.(x+4 \sin x)\right|_{0} ^{\pi}+\int_{0}^{\pi} 6 \cdot \frac{1}{2}(1+\cos 2 x) d x+\int_{0}^{\pi} 4\left(1-\sin ^{2} x\right) d(\sin x)+ \\
& +\int_{0}^{\pi} \frac{1}{2^{2}}(1+\cos 2 x)^{2} d x= \\
& =(\pi+4 \sin \pi)-(0+4 \sin 0)+\left.3\left(x+\frac{1}{2} \sin 2 x\right)\right|_{0} ^{\pi}+\left.4\left(\sin x-\frac{1}{3} \sin ^{3} x\right)\right|_{0} ^{\pi}+ \\
& +\int_{0}^{\pi} \frac{1}{2^{2}}\left(1+2 \cos 2 x+\cos ^{2} 2 x\right) d x= \\
& =\pi-0+3\left(\pi+\frac{1}{2} \sin 2 \pi\right)-3\left(0+\frac{1}{2} \sin (2 \cdot 0)\right)+4\left(\sin \pi-\frac{1}{3} \sin ^{3} \pi\right)-4\left(\sin 0-\frac{1}{3} \sin ^{3} 0\right) \\
& +\frac{1}{2^{2}} \cdot \int_{0}^{\pi}(1+2 \cos 2 x) d x+\int_{0}^{\pi} \cos ^{2} 2 x d x= \\
& =\pi+3 \pi-0+0-0+\left.\frac{1}{2^{2}} \cdot\left(x+2 \cdot \frac{1}{2} \sin 2 x\right)\right|_{0} ^{\pi}+\frac{1}{2^{2}} \cdot \int_{0}^{\pi} \cos ^{2} 2 x d x= \\
& =\left|\cos ^{2} 2 x=\frac{1}{2}(1+\cos 4 x)\right|= \\
& =4 \pi+\frac{1}{2^{2}} \cdot\left(\pi+2 \cdot \frac{1}{2} \sin 2 \pi\right)-\frac{1}{2^{2}} \cdot\left(0+2 \cdot \frac{1}{2} \sin (2 \cdot 0)\right)+\frac{1}{2^{2}} \cdot \int_{0}^{\pi} \frac{1}{2}(1+\cos 4 x) d x= \\
& =4 \pi+\frac{\pi}{4}-0+\left.\frac{1}{2^{3}}\left(x+\frac{1}{4} \sin 4 x\right)\right|_{0} ^{\pi}=4 \pi+\frac{\pi}{4}+\frac{1}{2^{3}}\left(\pi+\frac{1}{4} \sin 4 \pi\right)-\frac{1}{2^{3}}\left(0+\frac{1}{4} \sin (4 \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 $\% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_10-5$ »

Categories: Kuznetsov's Integral Problems Problem $10 \mid$ Integrals

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- Last edited: 17:37, 17 May 2009.
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## Problem Kuznetsov Integrals 10-6

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