## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{2 \pi} \sin ^{6}\left(\frac{x}{4}\right) \cos ^{2}\left(\frac{x}{4}\right) d x
$$
## Solution

$$
\begin{aligned}
& \int_{0}^{2 \pi} \sin ^{6}\left(\frac{x}{4}\right) \cos ^{2}\left(\frac{x}{4}\right) d x=\left|\sin \left(\frac{x}{4}\right) \cos \left(\frac{x}{4}\right)=\frac{1}{2} \sin \left(\frac{x}{2}\right), \sin ^{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^{2}} \sin ^{2}\left(\frac{x}{2}\right) \cdot \frac{1}{2^{2}}\left(1-\cos \left(\frac{x}{2}\right)\right)^{2} d x=\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi} \sin ^{2}\left(\frac{x}{2}\right)\left(1-2 \cos \left(\frac{x}{2}\right)+\cos ^{2}\left(\frac{x}{2}\right)\right) d x= \\
& =\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi} \sin ^{2}\left(\frac{x}{2}\right) d x-\frac{1}{2^{3}} \cdot \int_{0}^{2 \pi} \sin ^{2}\left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right) d x+\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi} \sin ^{2}\left(\frac{x}{2}\right) \cos ^{2}\left(\frac{x}{2}\right) d x= \\
& =\left|\sin ^{2}\left(\frac{x}{2}\right)=\frac{1}{2}(1-\cos x), \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)=\frac{1}{2} \sin x, 2 d\left(\sin \left(\frac{x}{2}\right)\right)=\cos \left(\frac{x}{2}\right) d x\right|= \\
& =\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi} \frac{1}{2}(1-\cos x) d x-\frac{1}{2^{3}} \cdot \int_{0}^{2 \pi} \sin ^{2}\left(\frac{x}{2}\right) \cdot 2 d\left(\sin \left(\frac{x}{2}\right)\right)+\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi} \frac{1}{2^{2}} \sin ^{2} x d x= \\
& =\left|\sin ^{2} x=\frac{1}{2}(1-\cos 2 x)\right|= \\
& =\left.\frac{1}{2^{5}} \cdot(x-\sin x)\right|_{0} ^{2 \pi}-\left.\frac{1}{2^{2}} \cdot \frac{1}{3} \sin ^{3}\left(\frac{x}{2}\right)\right|_{0} ^{2 \pi}+\frac{1}{2^{6}} \cdot \int_{0}^{2 \pi} \frac{1}{2}(1-\cos 2 x) d x= \\
& =\frac{1}{2^{5}}(2 \pi-\sin 2 \pi)-\frac{1}{2^{5}}(0-\sin 0)-\frac{1}{12} \sin ^{3} 2 \pi+\frac{1}{12} \sin ^{3} 0+\left.\frac{1}{2^{7}} \cdot\left(x-\frac{1}{2} \sin 2 x\right)\right|_{0} ^{2 \pi}= \\
& =\frac{1}{2^{5}}(2 \pi-0)-\frac{1}{2^{5}}(0-0)-\frac{1}{12} \cdot 0+\frac{1}{12} \cdot 0+\frac{1}{2^{7}} \cdot\left(2 \pi-\frac{1}{2} \sin ^{2 \pi}\right)-\frac{1}{2^{7}} \cdot\left(0-\frac{1}{2} \sin (2 \cdot 0)\right)= \\
& =\frac{\pi}{16}+\frac{1}{2^{7}} \cdot(2 \pi-0)-\frac{1}{2^{7}} \cdot(0-0)=\frac{\pi}{16}+\frac{\pi}{64}=\frac{5 \pi}{64}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\� \%D1\% $83 \% \mathrm{D} 0 \% \mathrm{~B} 7 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \mathrm{H} 0 \% 98 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3$ $\% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+10-22$ "

Categories: Kuznetsov's Problem Book Integrals Problem 10 | Integrals

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## Problem Kuznetsov Integrals $10-23$

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