20. Помогите решить интеграл [tex]\int\limits {sin^{\frac{1}{3} }x cos^{-\frac{13}{3} } x } \, dx[/tex] . Ответ должен получиться: [tex]\frac{3}{4} tg^{\frac{4}{3} } x + \frac{3}{10} tg^{\frac{10}{3} }x + C[/tex]. Created by milakamilat1m. matematika-ru

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Ответ:

$\displaystyle \int (sinx)^{\frac{1}{3}}\cdot (cosx)^{-\frac{13}{3}}\, dx=\int \sqrt[3]{\frac{sinx}{cosx}\cdot \frac{1}{cos^{12}x}}=\int \sqrt[3]{tgx}\cdot \cdot \frac{dx}{cos^4x}=\\\\\\=\int \sqrt[3]{tgx}\cdot \underbrace {\frac{1}{cos^2x}}_{1+tg^2x}\cdot \underbrace {\frac{dx}{cos^2x}}_{d(tgx)}=\int \sqrt[3]{tgx}\cdot (1+tg^2x)\cdot d(tgx)=\Big[\ t=tgx\ \Big]=$

$\displaystyle =\int t^{\frac{1}{3}}\cdot (1+t^2)\, dt=\int \Big(t^{\frac{1}{3}}+t^{\frac{7}{3}}\Big)\, dt=\frac{3\, t^{\frac{4}{3}}}{4}+\frac{3\, t^{\frac{10}{3}}}{10}+C=\\\\\\=\frac{3}{4}\, \Big(tgx\Big)^{\frac{4}{3}}+\frac{3}{10}\, \Big(tgx\Big)^{\frac{10}{3}}+C=\frac{3\, \sqrt[3]{tg^4x}}{4}+\frac{3\sqrt[3]{tg^{10}x}}{10}+C$

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