## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\pi} 2^{4} \cdot \sin ^{6} x \cos ^{2} x d x
$$
## Solution

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

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 $\% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+10-2$ »

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

- Last modified: 08:06, 15 May 2009.
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## Problem Kuznetsov Integrals 10-3

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