## Problem Statement

Calculate the definite integral:

$$
\int_{\pi / 2}^{\pi} 2^{8} \cdot \sin ^{2} x \cos ^{6} x d x
$$
## Solution

$$
\begin{aligned}
& \int_{\pi / 2}^{\pi} 2^{8} \sin ^{2} x \cos ^{6} x d x=2^{8} \int_{\pi / 2}^{\pi} \frac{1-\cos 2 x}{2} \cdot \frac{(1+\cos 2 x)^{3}}{8} d x= \\
& =2^{4} \int_{\pi / 2}^{\pi / 2}(1-\cos 2 x)(1+\cos 2 x)\left(1+2 \cos 2 x+\cos ^{2} 2 x\right) d x= \\
& =2^{4} \int_{\pi / 2}^{\pi}\left(1-\cos ^{2} 2 x\right)\left(1+2 \cos 2 x+\cos ^{2} 2 x\right) d x= \\
& =2^{4} \int_{\pi / 2}^{\pi}\left(1+2 \cos 2 x+\cos ^{2} 2 x-\cos ^{2} 2 x-2 \cos ^{3} 2 x-\cos ^{4} 2 x\right) d x= \\
& =2^{4}\left[\frac{\pi}{2}+\left.\sin 2 x\right|_{\pi / 2} ^{\pi}-2 \int_{\pi / 2}^{\pi} \cos ^{2} 2 x d(\sin 2 x)-\frac{1}{4} \int_{\pi / 2}^{\pi}(1+\cos 4 x)^{2} d x\right]= \\
& =2^{4}\left[\frac{\pi}{2}-\frac{1}{4} \int_{\pi / 2}^{\pi}\left(1+2 \cos 4 x+\cos ^{2} 4 x\right) d x\right]=2^{4}\left[\frac{\pi}{2}-\frac{1}{4} \cdot \frac{\pi}{2}-\frac{1}{8} \int_{\pi / 2}^{\pi}(1+\cos 8 x) d x\right]= \\
& =2^{4}\left[\frac{\pi}{2}-\frac{\pi}{8}-\frac{1}{8} \cdot \frac{\pi}{2}\right]=2^{4}\left[\frac{8 \pi-2 \pi-\pi}{16}\right]=5 \pi
\end{aligned}
$$

Source — "http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\� \%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5 $\% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+10-19$ "

Categories: Kuznetsov's Problem Book Integrals Problem $10 \mid$ Integrals

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## Problem Kuznetsov Integrals 10-20

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