Zadacha Kuznecov Integraly 10-5

Условие задачи

Вычислить определенный интеграл:

$\int \limits _{0}^{\pi }2^{4}\cdot \cos ^{8}{\left({\frac {x}{2}}\right)}dx$

Решение

$\int \limits _{0}^{\pi }2^{4}\cdot \cos ^{8}{\left({\frac {x}{2}}\right)}dx=\left|\cos ^{2}{\left({\frac {x}{2}}\right)}={\frac {1}{2}}(1+\cos {x})\right|=\int \limits _{0}^{\pi }2^{4}\cdot {\frac {1}{2^{4}}}(1+\cos {x})^{4}dx=$

$=\int \limits _{0}^{\pi }\left(1+4\cos {x}+6\cos ^{2}{x}+4\cos ^{3}{x}+\cos ^{4}{x}\right)dx=$

$=\int \limits _{0}^{\pi }(1+4\cos {x})dx+\int \limits _{0}^{\pi }6\cos ^{2}{x}\;dx+\int \limits _{0}^{\pi }4\cos ^{3}{x}\;dx+\int \limits _{0}^{\pi }\cos ^{4}{x}\;dx=$

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

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

$+\int \limits _{0}^{\pi }{\frac {1}{2^{2}}}(1+\cos {2x})^{2}dx=$

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

$+\int \limits _{0}^{\pi }{\frac {1}{2^{2}}}\left(1+2\cos {2x}+\cos ^{2}{2x}\right)dx=$

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

$+{\frac {1}{2^{2}}}\cdot \int \limits _{0}^{\pi }\left(1+2\cos {2x}\right)dx+\int \limits _{0}^{\pi }\cos ^{2}{2x}dx=$

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

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

$=4\pi +{\frac {1}{2^{2}}}\cdot \left(\pi +2\cdot {\frac {1}{2}}\sin {2\pi }\right)-{\frac {1}{2^{2}}}\cdot \left(0+2\cdot {\frac {1}{2}}\sin {(2\cdot 0)}\right)+{\frac {1}{2^{2}}}\cdot \int \limits _{0}^{\pi }{\frac {1}{2}}(1+\cos {4x})dx=$

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

$={\frac {17\pi }{4}}+{\frac {\pi }{8}}-0={\frac {35\pi }{8}}$