How do you integrate $tanx/(1+cosx)$ ?

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Answer 1 · 2016-07-04T01:59:37

$inttanx/(1+cosx)dx=intsinx/(cosx(1+cosx))dx$

let

$u=cosx$
$du=-sinxdx$

partial fractions

$-int(du)/(u(u+1))=-int(1/u-1/(u+1))du$

$-int(1/u-1/(u+1))du=int(du)/(u+1)-int(du)/u$

$-int(du)/u=-ln|u|$

For $int(du)/(u+1)$

set
$w=u+1$
$dw=du$

$int(dw)/w = ln|w|$

Plug back in for u and x

$ln|w|= ln|u+1|$

So far we have

$ln|u+1|-ln|u|$

Remember $u=cosx$

$ln|cosx+1|-ln|cosx|$

And $+ C$

$inttanx/(1+cosx)dx=ln|cosx+1|-ln|cosx|+C$

How do you integrate tanx/(1+cosx)tanx/(1+cosx)?