## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\pi} 2^{4} \cdot \sin ^{8} x d x
$$
## Solution

$$
\begin{aligned}
& \int_{0}^{\pi} 2^{4} \sin ^{8} x d x=\int_{0}^{\pi} 2^{4}\left(\sin ^{2} x\right)^{4} d x=\int_{0}^{\pi} 2^{4}\left(\frac{1-\cos 2 x}{2}\right)^{4} d x=\int_{0}^{\pi}(1-\cos 2 x)^{4} d x= \\
& =\int_{0}^{\pi}\left(1-4 \cos 2 x+6 \cos ^{2} 2 x-4 \cos ^{3} 2 x+\cos ^{4} 2 x\right) d x= \\
& =\int_{0}^{\pi}\left(1-4 \cos 2 x+6 \frac{1+\cos 4 x}{2}-4 \cos ^{3} 2 x+\cos ^{4} 2 x\right) d x= \\
& =\int_{0}^{\pi}\left(4-4 \cos 2 x+3 \cos 4 x-4 \cos ^{3} 2 x+\cos ^{4} 2 x\right) d x= \\
& =\left.\left(4 x-2 \sin 2 x+\frac{3 \sin 4 x}{4}\right)\right|_{0} ^{\pi}-2 \int_{0}^{\pi}\left(1-\sin ^{2} 2 x\right) d(\sin 2 x)+\int_{0}^{\pi} \cos ^{4} 2 x d x= \\
& =4 \pi-\left.2\left(\sin 2 x-\frac{\sin ^{3} 2 x}{3}\right)\right|_{0} ^{\pi}+\int_{0}^{\pi}\left(\frac{1+\cos 4 x}{2}\right)^{2} d x= \\
& =4 \pi+\frac{1}{4} \cdot \int_{0}^{\pi}\left(1+2 \cos 4 x+\cos ^{2} 4 x\right) d x=4 \pi+\frac{1}{4} \cdot \int_{0}^{\pi}\left(1+2 \cos 4 x+\frac{1+\cos 8 x}{2}\right) d x= \\
& =4 \pi+\frac{1}{8} \cdot \int_{0}^{\pi}(3+4 \cos 4 x+\cos 8 x) d x=4 \pi+\left.\frac{1}{8}\left(3 x+\sin 4 x+\frac{\sin 8 x}{8}\right)\right|_{0} ^{\pi}= \\
& =4 \pi+\frac{3 \pi}{8}=\frac{35 \pi}{8}
\end{aligned}
$$

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%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%B0_10-26" Categories: Kuznetsov's Problem Book Integrals Problem 10 | Integrals

## Problem Kuznetsov Integrals 10-27

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