## Problem Statement

Calculate the definite integral:

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

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

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 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_10-23$ " Categories: Kuznetsov's Problem Book Integrals Problem $10 \mid$ Integrals

- Last edited: 19:28, June 14, 2009.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 10-24

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