Question

integration of 1 divided by 1- tanx​

Answer 1

$\bigstar$ Solution :

$\bigstar$ To Find :

   $\int \frac{1}{1-tanx} dx$

$\bigstar$ Explanation  :

$\int \frac{1}{1-tanx}dx\\\\ =>\int \frac{cosx}{cosx-sinx}dx \\\\=>\frac{1}{2}\int (\frac{2.cosx}{cosx-sinx} )dx \\\\=>\frac{1}{2}\int [\frac{(cosx+sinx)+(cosx-sinx)}{cosx-sinx}]dx\\\\=>\frac{1}{2}\int [\frac{(cosx+sinx)}{cosx-sinx}+\frac{(cosx-sinx)}{cosx-sinx}]dx\\\\=>\frac{1}{2}\int [-\frac{\frac{d}{dx}(sinx-cosx) }{sinx-cosx} +1]dx\\\\=> \frac{1}{2}[-ln(sinx-cosx)+x]+c\\\\=>\frac{1}{2}(x-ln(sinx-cosx))+c$

                                                 

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