## Problem Statement

Calculate the definite integral:

$$
\int_{-\pi / 2}^{0} 2^{8} \cdot \cos ^{8} x d x
$$
## Solution

$\int_{-\pi / 2}^{0} 2^{8} \cdot \cos ^{8} x d x=\left|\cos ^{2} x=\frac{1}{2}(1+\cos 2 x)\right|=2^{8} \cdot \int_{-\pi / 2}^{0} \frac{1}{2^{4}}(1+\cos 2 x)^{4} d x=$

$=2^{4} \cdot \int_{-\pi / 2}^{0}\left(1+4 \cos 2 x+6 \cos ^{2} 2 x+4 \cos ^{3} 2 x+\cos ^{4} 2 x\right) d x=$

$=2^{4} \cdot \int_{-\pi / 2}^{0}(1+4 \cos 2 x) d x+2^{4} \cdot \int_{-\pi / 2}^{0} 6 \cos ^{2} 2 x d x+2^{4} \cdot \int_{-\pi / 2}^{0} 4 \cos ^{3} 2 x d x+2^{4} \cdot \int_{-\pi / 2}^{0} \cos ^{4} 2 x$

$=\left|\cos ^{2} 2 x=\frac{1}{2}(1+\cos 4 x), \cos ^{2} 2 x=1-\sin ^{2} 2 x, \frac{1}{2} d(\sin 2 x)=\cos 2 x d x\right|=$

$=\left.2^{4} \cdot(x+2 \sin 2 x)\right|_{-\pi / 2} ^{0}+2^{4} \cdot \int_{-\pi / 2}^{0} 6 \cdot \frac{1}{2}(1+\cos 4 x) d x+2^{4} \cdot \int_{-\pi / 2}^{0} 4\left(1-\sin ^{2} 2 x\right) \cdot \frac{1}{2} d(\sin 2 x)$

$+2^{4} \cdot \int_{-\pi / 2}^{0} \frac{1}{2^{2}}(1+\cos 4 x)^{2} d x=$

$=2^{4} \cdot(0+2 \cdot 0)-2^{4} \cdot\left(-\frac{\pi}{2}+2 \cdot 0\right)+\left.2^{4} \cdot 3\left(x+\frac{1}{4} \sin 4 x\right)\right|_{-\pi / 2} ^{0}+2^{4} \cdot 2\left(\sin 2 x-\frac{1}{3} \sin ^{3} 2 x\right)$

$+2^{5} \cdot\left(0-\frac{1}{3} \cdot 0^{3}\right)-2^{5} \cdot\left(0-\frac{1}{3} \cdot 0^{3}\right)+$

$+2^{2} \cdot \int_{-\pi / 2}^{0}(1+2 \cos 4 x) d x+2^{2} \cdot \int_{-\pi / 2}^{0} \cos ^{2} 4 x d x=$

$=0+8 \pi+48 \cdot\left(0+\frac{1}{4} \cdot 0\right)-48 \cdot\left(-\frac{\pi}{2}+\frac{1}{4} \cdot 0\right)+2^{5} \cdot\left(0-\frac{1}{3} \cdot 0^{3}\right)-2^{5} \cdot\left(0-\frac{1}{3} \cdot 0^{3}\right)+$

$+2^{2} \cdot \int_{-\pi / 2}^{0}(1+2 \cos 4 x) d x+2^{2} \cdot \int_{-\pi / 2}^{0} \cos ^{2} 4 x d x=$

$=8 \pi+24 \pi+\left.2^{2} \cdot\left(x+\frac{1}{2} \sin 4 x\right)\right|_{-\pi / 2} ^{0}+2^{2} \cdot \int_{-\pi / 2}^{0} \cos ^{2} 4 x d x=\left|\cos ^{2} 4 x=\frac{1}{2}(1+\cos 8 x)\right|$

$$
\begin{aligned}
& =32 \pi+2^{2} \cdot\left(0+\frac{1}{2} \cdot 0\right)-2^{2} \cdot\left(-\frac{\pi}{2}+\frac{1}{2} \cdot 0\right)+2^{2} \cdot \int_{-\pi / 2}^{0} \frac{1}{2}(1+\cos 8 x) d x= \\
& =32 \pi+0+2 \pi+\left.2\left(x+\frac{1}{8} \sin 8 x\right)\right|_{-\pi / 2} ^{0}=32 \pi+2 \pi+2\left(0+\frac{1}{8} \cdot 0\right)-2\left(-\frac{\pi}{2}+\frac{1}{8} \cdot 0\right)= \\
& =34 \pi+0+\pi=35 \pi
\end{aligned}
$$

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 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+10-30$ » Categories: Kuznetsov's Problem Book Integrals Problem $10 \mid$ Integrals

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## Problem Kuznetsov Integrals 10-31

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