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July 2022 2 9 Report

Помогите решить интеграл: [tex]\int\limits {\frac{1-tgx}{1+tgx} } \, dx[/tex] Ответ: [tex]ln|sinx+cosx|+ C[/tex]

LymarIvan

Ответ:

$\displaystyle \int \frac{1-tgx}{1+tgx}\, dx=\int \frac{1-\frac{sinx}{cosx}}{1+\frac{sinx}{cosx}}=\int \frac{cosx-sinx}{cosx+sinx}\, dx=\Big[\ t=cosx+sinx\ ,\\\\\\dt=(-sinx+cosx)\, dx\ \Big]=\int \frac{dt}{t}=ln|t|+C=ln|cosx+sinx|+C\ ;$

1 votes Thanks 1

NNNLLL54

Verified answer

Ответ:

Пошаговое объяснение:

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

1 votes Thanks 0

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