tex]...

July 2022 2 9 Report

Помогите решить интеграл: [tex]\int\limits {\frac{sin^3x}{cosx} } \, dx[/tex], ответ должен быть: [tex]\frac{cos^2x}{2} - ln|cosx|+C[/tex]

LymarIvan

Ответ:

$\displaystyle \int \frac{sin^3x}{cosx}\, dx=\int \frac{sin^2x\cdot sinx}{cosx}\, dx=\int \frac{(1-cos^2x)\cdot sinx}{cosx}\, dx=\\\\\\=\Big[\ t=cosx\ ,\ dt=-sinx\, dx\ \Big]=-\int \frac{1-t^2}{t}\cdot dt=\int \Big(t-\frac{1}{t}\Big)\, dt=\\\\\\=\frac{t^2}{2}-ln|t|+C=\frac{cos^2x}{2}-ln|cosx|+C$

1 votes Thanks 1

NNNLLL54

Verified answer

Ответ:

Пошаговое объяснение:

$\int {\frac{\sin^3x}{\cos x} } \, dx =\int {\frac{\sin^2x*\sin x}{\cos x} } \, dx = -\int {\frac{\sin^2x}{\cos x} } \, d(\cos x) =\\-\int {\frac{1-\cos^2x}{\cos x} } \, d(\cos x) =-\int {(\frac{1}{\cos x} -\frac{\cos^2x}{\cos x} )} \, d(\cos x) =\\\int {(\cos x-\frac{1}{\cos x}) } \, d(\cos x)=\frac{\cos^2x}{2}-\ln|\cos x|+C$

3 votes Thanks 1

Answers & Comments

Verified answer

tex]...

tex]...

tex]...

tex]...

tex]...

tex]...

tex]...

raspishite shagi resheniya

tex]...

tex]...

рекомендуемые вопросы

Helpful Links

Helpful Social