Proof ¶
$\displaystyle \int\tan x dx$
$\displaystyle =\int\frac{\sin x}{\cos x}dx$
$\displaystyle =-\int\frac1udu$
$\displaystyle =-\ln|u|+C$
$\displaystyle =-\ln|\cos x|+C$
$\displaystyle =\ln|\sec x|+C$
$\displaystyle =\int\frac{\sin x}{\cos x}dx$
치환:
$\displaystyle u=\cos x$
$\displaystyle du=-\sin x dx$
$\displaystyle \sin x dx=-du$
$\displaystyle =\int\frac1u(-du)$$\displaystyle u=\cos x$
$\displaystyle du=-\sin x dx$
$\displaystyle \sin x dx=-du$
$\displaystyle =-\int\frac1udu$
$\displaystyle =-\ln|u|+C$
$\displaystyle =-\ln|\cos x|+C$
$\displaystyle =\ln|\sec x|+C$
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