## Problem Statement

Calculate the definite integral:

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

$$
\begin{aligned}
& \int_{\pi / 2}^{2 \pi} 2^{8} \cdot \cos ^{8} x d x=\left|\cos ^{2} x=\frac{1}{2}(1+\cos 2 x)\right|=2^{8} \cdot \int_{\pi / 2}^{2 \pi} \frac{1}{2^{4}}(1+\cos 2 x)^{4} d x= \\
& =2^{4} \cdot \int_{\pi / 2}^{2 \pi}\left(1+4 \cos 2 x+6 \cos ^{2} 2 x+4 \cos ^{3} 2 x+\cos ^{4} 2 x\right) d x= \\
& =2^{4} \cdot \int_{\pi / 2}^{2 \pi}(1+4 \cos 2 x) d x+2^{4} \cdot \int_{\pi / 2}^{2 \pi} 6 \cos ^{2} 2 x d x+2^{4} \cdot \int_{\pi / 2}^{2 \pi} 4 \cos ^{3} 2 x d x+2^{4} \cdot \int_{\pi / 2}^{2 \pi} \cos ^{4} 2 x d x \\
& =\left|\cos ^{2} 2 x=\frac{1}{2}(1+\cos 4 x), \cos ^{2} 2 x=1-\sin ^{2} 2 x, \frac{1}{2} d(\sin 2 x)=\cos 2 x d x\right|= \\
& =\left.2^{4} \cdot(x+2 \sin 2 x)\right|_{\pi / 2} ^{2 \pi}+2^{4} \cdot \int_{\pi / 2}^{2 \pi} 6 \cdot \frac{1}{2}(1+\cos 4 x) d x+2^{4} \cdot \int_{\pi / 2}^{2 \pi} 4\left(1-\sin ^{2} 2 x\right) \cdot \frac{1}{2} d(\sin 2 x) \\
& +2^{4} \cdot \int_{\pi / 2}^{2 \pi} \frac{1}{2^{2}}(1+\cos 4 x)^{2} d x= \\
& =2^{4} \cdot(2 \pi+2 \cdot 0)-2^{4} \cdot\left(\frac{\pi}{2}+2 \cdot 0\right)+\left.2^{4} \cdot 3\left(x+\frac{1}{4} \sin 4 x\right)\right|_{\pi / 2} ^{2 \pi}+2^{4} \cdot 2\left(\sin 2 x-\frac{1}{3} \sin ^{3} 2 x\right) \\
& +2^{4} \cdot \int_{\pi / 2}^{2 \pi} \frac{1}{2^{2}}\left(1+2 \cos 4 x+\cos ^{2} 4 x\right) d x= \\
& =32 \pi-8 \pi+48 \cdot\left(2 \pi+\frac{1}{4} \cdot 0\right)-48 \cdot\left(\frac{\pi}{2}+\frac{1}{4} \cdot 0\right)+2^{5} \cdot\left(2 \pi-\frac{1}{3} \cdot 0^{3}\right)-2^{5} \cdot\left(0-\frac{1}{3} \cdot 0^{3}\right) \\
& +2^{2} \cdot \int_{\pi / 2}^{2 \pi}(1+2 \cos 4 x) d x+2^{2} \cdot \int_{\pi / 2}^{2 \pi} \cos ^{2} 4 x d x= \\
& =24 \pi+96 \pi-24 \pi+\left.2^{2} \cdot\left(x+\frac{1}{2} \sin ^{2} 4 x\right)\right|_{\pi / 2} ^{2 \pi}+2^{2} \cdot \int_{\pi / 2}^{2 \pi} \cos ^{2} 4 x d x=\left|\cos ^{2} 4 x=\frac{1}{2}(1+\cos 8 x)\right. \\
\end{aligned}
$$

$$
\begin{aligned}
& =96 \pi+2^{2} \cdot\left(2 \pi+\frac{1}{2} \cdot 0\right)-2^{2} \cdot\left(\frac{\pi}{2}+\frac{1}{2} \cdot 0\right)+2^{2} \cdot \int_{\pi / 2}^{2 \pi} \frac{1}{2}(1+\cos 8 x) d x= \\
& =96 \pi+8 \pi-2 \pi+\left.2\left(x+\frac{1}{8} \sin 8 x\right)\right|_{\pi / 2} ^{2 \pi}=102 \pi+2\left(2 \pi+\frac{1}{8} \cdot 0\right)-2\left(\frac{\pi}{2}+\frac{1}{8} \cdot 0\right)= \\
& =102 \pi+4 \pi-\pi=105 \pi
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+10-25$ » Categories: Kuznetsov's Problem Book Integrals Problem $10 \mid$ Integrals

- Last edited: 08:18, June 13, 2010.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals $10-26$

## Material from PlusPi