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#