## Problem Statement

Calculate the definite integral:

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

$$
\begin{aligned}
& \int_{0}^{2 \pi} \sin ^{4}\left(\frac{x}{4}\right) \cos ^{4}\left(\frac{x}{4}\right) d x=\int_{0}^{2 \pi}\left(\sin \left(\frac{x}{4}\right) \cos \left(\frac{x}{4}\right)\right)^{4} d x=\int_{0}^{2 \pi}\left(\frac{\sin \left(\frac{x}{2}\right)}{2}\right)^{4} d x= \\
& =\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi} \sin ^{4}\left(\frac{x}{2}\right) d x=\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi} \frac{1}{2^{2}} \cdot(1-\cos x)^{2} d x=\frac{1}{2^{6}} \cdot \int_{0}^{2 \pi}\left(1-2 \cos x+\cos ^{2} x\right) d x= \\
& =\frac{1}{2^{6}} \cdot \int_{0}^{2 \pi}\left(1-2 \cos x+\frac{1+\cos 2 x}{2}\right) d x=\frac{1}{2^{7}} \cdot \int_{0}^{\pi}(2-4 \cos x+1+\cos 2 x) d x= \\
& =\frac{1}{2^{7}} \cdot \int_{0}^{2 \pi}(3-4 \cos x+\cos 2 x) d x=\left.\frac{1}{2^{7}} \cdot\left(3 x-4 \sin x+\frac{\sin 2 x}{2}\right)\right|_{0} ^{2 \pi}=\frac{3 \pi}{2^{6}}=\frac{3 \pi}{64}
\end{aligned}
$$

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 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+10-28$ »

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

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## Problem Kuznetsov Integrals $10-29$

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