## Problem Statement

Calculate the definite integral:

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

$$
\begin{aligned}
& \int_{0}^{2 \pi} \sin ^{6} x \cos ^{2} x d x=\left|\sin x \cos x=\frac{1}{2} \sin 2 x, \sin ^{2} x=\frac{1}{2}(1-\cos 2 x)\right|= \\
& =\int_{0}^{2 \pi} \frac{1}{2^{2}} \sin ^{2} 2 x \cdot \frac{1}{2^{2}}(1-\cos 2 x)^{2} d x=\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi} \sin ^{2} 2 x\left(1-2 \cos 2 x+\cos ^{2} 2 x\right) d x= \\
& =\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi} \sin ^{2} 2 x d x-\frac{1}{2^{3}} \cdot \int_{0}^{2 \pi} \sin ^{2} 2 x \cos 2 x d x+\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi} \sin ^{2} 2 x \cos ^{2} 2 x d x= \\
& =\left|\sin ^{2} 2 x=\frac{1}{2}(1-\cos 4 x), \sin 2 x \cos 2 x=\frac{1}{2} \sin 4 x, \frac{1}{2} d(\sin 2 x)=\cos 2 x d x\right|= \\
& =\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi} \frac{1}{2}(1-\cos 4 x) d x-\frac{1}{2^{3}} \cdot \int_{0}^{2 \pi} \sin ^{2} 2 x \cdot \frac{1}{2} d(\sin 2 x)+\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi} \frac{1}{2^{2}} \sin ^{2} 4 x d x= \\
& =\left|\sin ^{2} 4 x=\frac{1}{2}(1-\cos 8 x)\right|= \\
& =\left.\frac{1}{2^{4}} \cdot \frac{1}{2}\left(x-\frac{1}{4} \sin 4 x\right)\right|_{0} ^{2 \pi}-\left.\frac{1}{2^{3}} \cdot \frac{1}{3} \sin ^{3} 2 x\right|_{0} ^{2 \pi}+\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi} \frac{1}{2^{2}} \cdot \frac{1}{2}(1-\cos 8 x) d x= \\
& =\frac{1}{2^{2}} \cdot\left(2 \pi-\frac{1}{4} \sin 8 \pi\right)-\frac{1}{2^{5}} \cdot\left(0-\frac{1}{4} \sin (4 \cdot 0)\right)-\frac{1}{2^{4}} \cdot \frac{1}{3} \sin ^{3} 4 \pi+\frac{1}{2^{4}} \cdot \frac{1}{3} \sin ^{3}(2 \cdot 0)+\left.\frac{1}{2^{7}} \cdot\left(x-\frac{1}{8} \sin 8 x\right)\right|_{0} ^{2 \pi}= \\
& =\frac{1}{2^{5}}(\pi-0)-\frac{1}{2^{5}}(0-0)-\frac{1}{2^{4}} \cdot \frac{1}{3} \cdot 0+\frac{1}{2^{4}} \cdot \frac{1}{3} \cdot 0+\frac{1}{2^{7}} \cdot\left(2 \pi-\frac{1}{8} \sin 16 \pi\right)-\frac{1}{2^{7}} \cdot\left(0-\frac{1}{8} \sin (8 \cdot 0)\right)= \\
& =\frac{\pi}{2^{4}}+\frac{1}{2^{7}} \cdot(2 \pi-0)-\frac{1}{2^{7}} \cdot(0-0)=\frac{4 \pi}{2^{6}}+\frac{\pi}{2^{6}}=\frac{5 \pi}{64}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\� \%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 $\% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+10-27$ »

Categories: Kuznetsov's Problem Book Integrals Problem 10 | Integrals

Ukrainian Banner Network

- Last edited: 19:41, 14 June 2009.
- Content is available under CC-BY-SA 3.0.