## Problem Statement

Calculate the definite integral:

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

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

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� $\% \mathrm{D} 0 \% 9 \mathrm{~A} \% \mathrm{D} 1 \% 83 \% \mathrm{D0} \% \mathrm{~B} 7 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \mathrm{H} 0 \% 98 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 1 \% 82$ $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+10-8$ » Categories: Kuznetsov Problem Book Integrals Problem $10 \mid$ Integrals

- Last edited: 16:33, 20 May 2009.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals $10-9$

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