Question

Please, check. I don't know what's wrong.
\[\int tanx dx \]

Answer 1

if I go this way 1/ let u = cos x so du = -sinx dx , int = - ln sec x +C
but the formula give out ln sec x + c only, why?

Answer 2

You mean you get -ln(cosx) + c right? this is equivalent to ln(secx) + c

Answer 3

oh ya, yes for ln cos x but no for = ln sec x

Answer 4

\[\large u=\cos x\qquad du=-\sin x\,dx \]
\[\large \int\tan x\,dx=\int\frac{\sin x}{\cos x}\,dx=\int\frac{-du}{u}=-\ln u=
-\ln\cos x \]
\[\large =\ln(\cos x)^{-1}=\ln\sec x \]

Answer 5

oh yea, I got it. I know the reason, hehehe.. sorry, my bad.

Answer 6

thanks everybody.

Answer 7

No problem!