## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\pi} 2^{4} \cdot \sin ^{6}\left(\frac{x}{2}\right) \cos ^{2}\left(\frac{x}{2}\right) d x
$$
## Solution

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

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 $\% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+10-17$ >

Categories: Kuznetsov's Integral Problems 10 | Integrals

- Last edited: 10:38, 29 May 2009.
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## Problem Kuznetsov Integrals 10-18

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