Question

20. Помогите решить интеграл $\int\limits {sin^{\frac{1}{3} }x cos^{-\frac{13}{3} } x } \, dx$ . Ответ должен получиться: $\frac{3}{4} tg^{\frac{4}{3} } x + \frac{3}{10} tg^{\frac{10}{3} }x + C$

Answer 1

Ответ:

Пошаговое объяснение:

$\int {\sin^\frac{1}{3}x\cos^{-\frac{13}{3} } } \, dx =\int {\sin^\frac{1}{3}x\cos^{-\frac{7}{3} } *} \, \cos^{-2}dx =\int {\sin^\frac{1}{3}x\cos^{-\frac{7}{3} }} \, d(\rm tg \; x) =\\\int {(\sin x\cos^{-7}x)^\frac{1}{3} } \, d(\rm tg \;x) =\int {\sqrt[3]{\frac{\sin x}{\cos^7x} } } \, d(\rm tg \; x)= \int {\sqrt[3]{\frac{\sin x}{\cos x}*\frac{1}{\cos^6x} } } \, d(\rm tg \; x)=$

$\int {\sqrt[3]{\rm tg \; x(\frac{1}{\cos^2x})^3 } } \, d(\rm tg \; x)=\int {\frac{1}{\cos^2x}\sqrt[3]{\rm tg \; x} } \, dx(\rm tg \; x)=\int {(1+\rm tg^2x)\rm tg^\frac{1}{3}x } \, d(\rm tg \; x)=\\\int {(\rm tg^\frac{1}{3} x+\rm tg^\frac{7}{3}x) } \, d(\rm tg \; x)=\frac{3}{4}\rm tg^\frac{4}{3}x+\frac{3}{10}\rm tg^\frac{10}{3}x+C$

Answer 2

Ответ:

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

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