## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{2 \pi} \sin ^{4} 3 x \cos ^{4} 3 x d x
$$
## Solution

$$
\begin{aligned}
& \int_{0}^{2 \pi} \sin ^{4} 3 x \cos ^{4} 3 x d x=\int_{0}^{2 \pi} \frac{1}{2^{4}} \cdot(2 \sin 3 x \cdot \cos 3 x)^{4} d x=\int_{0}^{2 \pi} \frac{1}{2^{4}} \cdot \sin ^{4} 6 x d x= \\
& =\frac{1}{16} \cdot \int_{0}^{2 \pi} \sin ^{4} 6 x d x=\frac{1}{16} \cdot \int_{0}^{2 \pi} \frac{1}{4} \cdot(1-\cos 12 x)^{2} d x=\frac{1}{64} \cdot \int_{0}^{2 \pi}\left(1-2 \cos 12 x+\cos ^{2} 12 x\right) d x= \\
& =\frac{1}{64} \cdot \int_{0}^{2 \pi}\left(1-2 \cos 12 x+\frac{1+\cos 24 x}{2}\right) d x=\frac{1}{2^{7}} \cdot \int_{0}^{2 \pi}(2-4 \cos 12 x+1+\cos 24 x) d x= \\
& =\frac{1}{2^{7}} \cdot \int_{0}^{2 \pi}(3-4 \cos 12 x+\cos 24 x) d x=\left.\frac{1}{2^{7}} \cdot\left(3 x-\frac{1}{3} \cdot \sin 12 x+\frac{\sin 24 x}{24}\right)\right|_{0} ^{2 \pi}= \\
& =\frac{1}{2^{7}} \cdot\left(6 \pi-\frac{1}{3} \cdot \sin 24 \pi+\frac{\sin 48 \pi}{24}\right)-\frac{1}{2^{7}} \cdot\left(0-\frac{1}{3} \cdot 0+\frac{0}{24}\right)=\frac{3 \pi}{64}
\end{aligned}
$$

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 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+10-31$ » Categories: Kuznetsov's Problem Book Integrals Problem $10 \mid$ Integrals

- Last edited: 20:28, June 16, 2009.
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