## Problem Statement

Calculate the definite integral:

$$
\int_{-\pi / 2}^{0} 2^{8} \cdot \sin ^{4} x \cos ^{4} x d x
$$
## Solution

$$
\begin{aligned}
& \int_{-\pi / 2}^{0} 2^{8} \cdot \sin ^{4} x \cos ^{4} x d x=\int_{-\pi / 2}^{0} 2^{4} \cdot(2 \sin x \cdot \cos x)^{4} d x=\int_{-\pi / 2}^{0} 2^{4} \cdot \sin ^{4} 2 x d x= \\
& =16 \int_{-\pi / 2}^{0} \sin ^{4} 2 x d x=16 \int_{-\pi / 2}^{0} \frac{1}{4} \cdot(1-\cos 4 x)^{2} d x=4 \int_{-\pi / 2}^{0}\left(1-2 \cos 4 x+\cos ^{2} 4 x\right) d x= \\
& =4 \int_{-\pi / 2}^{0}\left(1-2 \cos 4 x+\frac{1+\cos 8 x}{2}\right) d x=2 \int_{-\pi / 2}^{0}(2-4 \cos 4 x+1+\cos 8 x) d x= \\
& =2 \int_{-\pi / 2}^{0}(3-4 \cos 4 x+\cos 8 x) d x=\left.2\left(3 x-\sin 4 x+\frac{\sin 8 x}{8}\right)\right|_{-\pi / 2} ^{0}= \\
& =2\left(0-\sin 0+\frac{\sin 0}{8}\right)-2\left(-\frac{3 \pi}{2}-\sin (-2 \pi)+\frac{\sin (-4 \pi)}{8}\right)=0+3 \pi=3 \pi
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� $\% \mathrm{D} 0 \% 9 \mathrm{~A} \% \mathrm{D} 1 \% 83 \% \mathrm{D0} \% \mathrm{~B} 7 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \mathrm{O} 0 \% 98 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 1 \% 82$ $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+10-18 »$ Categories: Kuznetsov Problem Book Integrals Problem $10 \mid$ Integrals

- Last edited: 10:43, 29 May 2009.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals $10-19$

## Material from PlusPi