## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\pi} 2^{4} \cdot \sin ^{8}\left(\frac{x}{2}\right) d x
$$
## Solution

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

$$
\begin{aligned}
& =4 \pi+\frac{\pi}{4}-0+\left.\frac{1}{2^{3}}\left(x+\frac{1}{4} \sin 4 x\right)\right|_{0} ^{\pi}=4 \pi+\frac{\pi}{4}+\frac{1}{2^{3}}\left(\pi+\frac{1}{4} \cdot 0\right)-\frac{1}{2^{3}}\left(0+\frac{1}{4} \cdot 0\right)= \\
& =4 \pi+\frac{\pi}{4}+\frac{\pi}{2^{3}}-0=\frac{32 \pi+2 \pi+\pi}{8}=\frac{35 \pi}{8}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+10-11$ » Categories: Kuznetsov's Problem Book Integrals Problem $10 \mid$ Integrals

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- Last edited: 11:37, 26 May 2009.
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## Problem Kuznetsov Integrals 10-12

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