How do you find $int sec^2x/(1-sinx)$ ?

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Answer 1 · 2016-03-22T15:14:33

= $tanx +tan^3x/3+sec^3x/3+C$

$int(sec^2x)/(1-sinx)dx$
$=int(sec^2x)/(1-sinx)*(1+sinx)/(1+sinx)dx$

$=int(sec^2x)/(1-sin^2x)*(1+sinx)dx$

$=int(sec^2x/cos^2x)*(1+sinx)dx$

$=intsec^2x*(1/cos^2x+sinx/cos^2x)dx$
$=intsec^2x*sec^2xdx+intsec^2xsecxtanxdx$

$=intsec^2x*(1+tan^2x)dx+intsec^2xsecxtanxdx$
For first part let $u=tanx =>du=sec^2xdx$
For second part let $v = secx =>dv=secxtanxdx$

The whole integral becomes
= $int(1+u^2)du+intv^2dv$
= $u +u^3/3+v^3/3+C$

= $tanx +tan^3x/3+sec^3x/3+C$

How do you find int sec^2x/(1-sinx) int sec^2x/(1-sinx) ?