How do you evaluate $\int \frac { 1} { \csc x - 1} d x$ ?

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Answer 1 · 2017-12-02T03:38:43

$sec x + tanx -x +C$

Multiplying by $csc x +1$ in both numerator and denominator we have
$1/(csc x -1) = (cscx +1)/((csc x -1)(csc x +1))$

$= (cscx+1)/(csc^2 x -1)$

But we have the trigonometric identity that $csc^2x - 1 = cot^2x$ . Hence

$=(csc x +1) / (cot^2x)$
$= csc x /cot^2x + tan^2x$
$= sin x / cos^2x +(sec^2x - 1)$
$= secx tanx + sec^2x -1$

Now integrate

$\int (dx)/(csc x -1) = \int secx tanx dx + \int sec^2x dx - \int dx$

$= sec x + tanx -x +C$

How do you evaluate \int \frac { 1} { \csc x - 1} d x\int \frac { 1} { \csc x - 1} d x?