## Problem Statement

Calculate the definite integral:

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

$$
\begin{aligned}
& \int_{0}^{2 \pi} \sin ^{2}\left(\frac{x}{4}\right) \cos ^{6}\left(\frac{x}{4}\right) d x=\int_{0}^{2 \pi} \frac{1-\cos \frac{x}{2}}{2}\left(\frac{1+\cos \frac{x}{2}}{2}\right)^{3} d x=\int_{0}^{2 \pi} \frac{1-\cos \frac{x}{2}\left(1+\cos \frac{x}{2}\right)^{3}}{2} d x= \\
& =\frac{1}{2^{4}} \int_{0}^{2 \pi}\left(1-\cos \frac{x}{2}\right)\left(1+\cos \frac{x}{2}\right)\left(1+\cos \frac{x}{2}\right)^{2} d x=\frac{1}{2^{4}} \int_{0}^{2 \pi}\left(1-\cos ^{2} \frac{x}{2}\right)\left(1+2 \cos \frac{x}{2}+\cos ^{2} \frac{x}{2}\right) d x= \\
& =\frac{1}{2^{4}} \int_{0}^{2 \pi}\left(1+2 \cos \frac{x}{2}+\cos ^{2} \frac{x}{2}-\cos ^{2} \frac{x}{2}-2 \cos ^{3} \frac{x}{2}-\cos ^{4} \frac{x}{2}\right) d x=\frac{1}{2^{4}} \int_{0}^{2 \pi}\left(1+2 \cos \frac{x}{2}-2 \cos ^{3} \frac{x}{2}-\cos ^{4} \frac{x}{2}\right) d x= \\
& =\frac{1}{2^{4}}\left[\int_{0}^{2 \pi} d x+2 \int_{0}^{2 \pi} \cos \frac{x}{2} d x-2 \int_{0}^{2 \pi} \cos ^{2} \frac{x}{2}\left(\cos \frac{x}{2} d x\right)-\int_{0}^{2 \pi}\left(\cos ^{2} \frac{x}{2}\right)^{2} d x\right]= \\
& =\frac{1}{2^{4}}\left[\left.x\right|_{0} ^{2 \pi}+\int_{0}^{2 \pi} \cos \frac{x}{2} d \frac{x}{2}-2 \int_{0}^{2 \pi}\left(1-\sin ^{2} \frac{x}{2}\right) d \sin \frac{x}{2}-\int_{0}^{2 \pi}\left(\frac{1+\cos x}{2}\right)^{2} d x\right]= \\
& =\frac{1}{2^{4}}\left[2 \pi+\left.\sin \frac{x}{2}\right|_{0} ^{2 \pi}-2\left(\int_{0}^{2 \pi} d \sin \frac{x}{2}-\int_{0}^{2 \pi} \sin ^{2} \frac{x}{2} d \sin \frac{x}{2}\right)-\int_{0}^{2 \pi} \frac{1+2 \cos x+\cos ^{2} x}{4} d x\right]= \\
& =\frac{1}{2^{4}}\left[2 \pi-2\left(\left.\sin \frac{x}{2}\right|_{0} ^{2 \pi}-\left.\frac{\sin ^{3} \frac{x}{2}}{3}\right|_{0} ^{2 \pi}\right)-\frac{1}{4} \int_{0}^{2 \pi}\left(1+2 \cos x+\cos ^{2} x\right) d x\right]= \\
& =\frac{1}{2^{4}}\left[2 \pi-\frac{1}{4}\left(\int_{0}^{2 \pi} d x+2 \int_{0}^{2 \pi} \cos x d x+\int_{0}^{2 \pi} \frac{1+\cos 2 x}{2} d x\right)\right]= \\
& =\frac{1}{2^{4}}\left[2 \pi-\frac{1}{4}\left(\left.x\right|_{0} ^{2 \pi}+\left.2 \sin x\right|_{0} ^{2 \pi}+\int_{0}^{2 \pi}\left(\frac{1}{2}+\frac{1}{2} \cos 2 x\right) d x\right)\right]= \\
& =\frac{1}{2^{4}}\left[2 \pi-\frac{1}{4}\left(2 \pi+\frac{1}{2} \int_{0}^{2 \pi} d x+\frac{1}{4} \int_{0}^{2 \pi} \cos 2 x d 2 x\right)\right]= \\
& =\frac{1}{2^{4}}\left[2 \pi-\frac{1}{4}\left(2 \pi+\left.\frac{1}{2} x\right|_{0} ^{2 \pi}+\left.\frac{1}{4} \sin 2 x\right|_{0} ^{2 \pi}\right)\right]=\frac{1}{2^{4}}\left[2 \pi-\frac{1}{4}\left(2 \pi+\frac{1}{2} \cdot 2 \pi\right)\right]= \\
& =\frac{1}{2^{4}}\left[2 \pi-\frac{1}{4}(2 \pi+\pi)\right]=\frac{1}{2^{4}}\left[\frac{8}{4} \pi-\frac{3}{4} \pi\right]=\frac{1}{2^{4}} \cdot \frac{5}{4} \pi=\frac{5}{64} \pi
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\� \%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5\%D0\%B3\%D1\%80\%D0\%B0 $\% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_10-4$ >

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

- Last edited: 19:56, 18 May 2009.
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## Problem Kuznetsov Integrals 10-5

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