Question #f9641

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Answer 1 · 2018-02-14T06:11:01

$int\ cos(x)/(sin^2(x)+sin(x))\ "d"x=ln|sin(x)/(sin(x)+1)|+C$

$\ \ \ \ \ \ int\ cos(x)/(sin^2(x)+sin(x))\ "d"x$

Substitute $u=sin(x)$ and $"d"u=cos(x)\ "d"x$ . This gives
$=int\ ("d"u)/(u^2+u)$
$=int\ ("d"u)/(u(u+1))$

Separate to partial fractions since $1/(u(u+1))=1/u-1/(u+1)$ :
$=int\ (1/u-1/(u+1))\ "d"u$

$=ln|u|-ln|u+1|+C$

$=ln|u/(u+1)|+C$

Substitute back $u=sin(x)$ :

$=ln|sin(x)/(sin(x)+1)|+C$