## Problem Statement

Calculate the definite integral:

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

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+10-14$ » Categories: Kuznetsov's Problem Book Integrals Problem $10 \mid$ Integrals

- Last edited: 18:35, 16 June 2009.
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## Problem Kuznetsov Integrals 10-15

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