## Problem Statement

Calculate the definite integral:

$$
\int_{\pi / 2}^{2 \pi} 2^{8} \cdot \sin ^{4} x \cos ^{4} x d x
$$
## Solution

$$
\begin{aligned}
& \int_{\pi / 2}^{2 \pi} 2^{8} \cdot \sin ^{4} x \cos ^{4} x d x=\int_{\pi / 2}^{2 \pi} 2^{4} \cdot(2 \sin x \cdot \cos x)^{4} d x=\int_{\pi / 2}^{2 \pi} 2^{4} \cdot \sin ^{4} 2 x d x= \\
& =16 \int_{\pi / 2}^{2 \pi} \sin ^{4} 2 x d x=16 \int_{\pi / 2}^{2 \pi} \frac{1}{4} \cdot(1-\cos 4 x)^{2} d x=4 \int_{\pi / 2}^{2 \pi}\left(1-2 \cos 4 x+\cos ^{2} 4 x\right) d x= \\
& =4 \int_{\pi / 2}^{2 \pi}\left(1-2 \cos 4 x+\frac{1+\cos 8 x}{2}\right) d x=2 \int_{\pi / 2}^{2 \pi}(2-4 \cos 4 x+1+\cos 8 x) d x= \\
& =2 \int_{\pi / 2}^{2 \pi}(3-4 \cos 4 x+\cos 8 x) d x=\left.2\left(3 x-\sin 4 x+\frac{\sin 8 x}{8}\right)\right|_{\pi / 2} ^{2 \pi}= \\
& =2\left(6 \pi-\sin 8 \pi+\frac{\sin 16 \pi}{8}\right)-2\left(\frac{3 \pi}{2}-\sin 2 \pi+\frac{\sin 4 \pi}{8}\right)=12 \pi-3 \pi=9 \pi
\end{aligned}
$$

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%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%B0_10-13"

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

- Last edited: 07:53, 25 May 2009.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 10-14

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