## Problem Statement

Calculate the definite integral:

$$
\int_{\pi / 2}^{\pi} 2^{4} \cdot \sin ^{6} x \cos ^{2} x d x
$$
## Solution

$$
\begin{aligned}
& \int_{\pi / 2}^{\pi} 2^{4} \cdot \sin ^{6} x \cos ^{2} x d x=\left|\sin x \cos x=\frac{1}{2} \sin 2 x, \sin ^{2} x=\frac{1}{2}(1-\cos 2 x)\right|= \\
& =\int_{\pi / 2}^{\pi} 2^{4} \cdot \frac{1}{2^{2}} \sin ^{2} 2 x \cdot \frac{1}{2^{2}}(1-\cos 2 x)^{2} d x=\int_{\pi / 2}^{\pi} \sin ^{2} 2 x\left(1-2 \cos 2 x+\cos ^{2} 2 x\right) d x= \\
& =\int_{\pi / 2}^{\pi} \sin ^{2} 2 x d x-2 \int_{\pi / 2}^{\pi} \sin ^{2} 2 x \cos 2 x d x+\int_{\pi / 2}^{\pi} \sin ^{2} 2 x \cos ^{2} 2 x d x= \\
& =\left|\sin ^{2} 2 x=\frac{1}{2}(1-\cos 4 x), \sin 2 x \cos 2 x=\frac{1}{2} \sin 4 x, \frac{1}{2} d(\sin 2 x)=\cos 2 x d x\right|= \\
& =\int_{\pi / 2}^{\pi} \frac{1}{2}(1-\cos 4 x) d x-2 \int_{\pi}^{\pi} \sin ^{2} 2 x \cdot \frac{1}{2} d(\sin 2 x)+\int_{\pi / 2}^{2^{2}} \sin ^{2} 4 x d x= \\
& =\left|\sin ^{2} 4 x=\frac{1}{2}(1-\cos 8 x)\right|= \\
& =\left.\frac{1}{2}\left(x-\frac{1}{4} \sin 4 x\right)\right|_{\pi / 2} ^{\pi}-\left.\frac{1}{3} \sin ^{3} 2 x\right|_{\pi / 2} ^{\pi}+\int_{\pi / 2}^{\pi} \frac{1}{2^{2}} \cdot \frac{1}{2}(1-\cos 8 x) d x= \\
& =\frac{1}{2}\left(\pi-\frac{1}{4} \sin 4 \pi\right)-\frac{1}{2}\left(\frac{\pi}{2}-\frac{1}{4} \sin \left(4 \cdot \frac{\pi}{2}\right)\right)-\frac{1}{3} \sin ^{3} 2 \pi+\frac{1}{3} \sin ^{3}\left(2 \cdot \frac{\pi}{2}\right)+\left.\frac{1}{2^{3}} \cdot\left(x-\frac{1}{8} \sin 8 x\right)\right|_{\pi / 2} ^{\pi}= \\
& =\frac{1}{2}(\pi-0)-\frac{1}{2}\left(\frac{\pi}{2}-0\right)-\frac{1}{3} \cdot 0+\frac{1}{3} \cdot 0+\frac{1}{2^{3}} \cdot\left(\pi-\frac{1}{8} \sin ^{\pi} \pi\right)-\frac{1}{2^{3}} \cdot\left(\frac{\pi}{2}-\frac{1}{8} \sin \left(8 \cdot \frac{\pi}{2}\right)\right)= \\
& =\frac{\pi}{2}-\frac{\pi}{4}+\frac{1}{2^{3}} \cdot(\pi-0)-\frac{1}{2^{3}} \cdot\left(\frac{\pi}{2}-0\right)=\frac{\pi}{4}+\frac{\pi}{8}-\frac{\pi}{16}=\frac{\pi}{4}+\frac{\pi}{16}=\frac{5 \pi}{16}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\� \%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 $\% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}=10-7$ >

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

- Last edited: 16:28, May 20, 2009.
- Content is available under CC-BY-SA 3.0.


## Problem Kuznetsov Integrals 10-8

## Material from PlusPi