How do you evaluate the integral $(secx tanx) / (sec^2(x) - secx)$ dx?

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Answer 1 · 2018-05-10T11:52:38

$I=ln|1-cosx|+c$

We know that,
$color(red)((1)tanx=sinx/cosx and secx= 1/cosx$
Here,

$I=int(secxtanx)/(sec^2x-secx)dx$

$=int(cancelsecxtanx)/(cancelsecx(secx-1))dx$

$=inttanx/(secx-1)dx$

$=int(sinx/cosx)/(1/cosx-1)dx...tocolor(red)(Apply(1)$

$I=intsinx/(1-cosx)dx$

Let,

$1-cosx=u=>sinxdx=du$

$I=int1/udu$

$=ln|u|+c, where, u=1-cosx$

$I=ln|1-cosx|+c$

How do you evaluate the integral (secx tanx) / (sec^2(x) - secx)(secx tanx) / (sec^2(x) - secx) dx?