Integration of 1/sinx+cosx

{tex}\begin{array}{l}\int\frac1{\sin x\;+\;\cos x}dx\\putting\rightarrow\;\;\sin x\;=\;\frac{2\tan{\displaystyle\frac x2}}{1+\tan^2{\displaystyle\frac x2}}\;\;and\;\cos x\;=\;\frac{1-\tan^2{\displaystyle\frac x2}}{1+t\mathrm{an}^2\frac x2}\;\\we\;get\;\int\frac{1+\tan^2\frac x2}{1+2\tan{\displaystyle\frac x2}-\tan^2{\displaystyle\frac x2}}dx\;=\;\int\frac{\displaystyle sec^2\frac x2}{\displaystyle1+2\tan\frac x2-\tan^2\frac x2}dx\\let\;\tan\frac x2\;=\;t\;\Rightarrow\;sec^2\frac x2dx\;=\;2dt\\\Rightarrow\;\int\frac{2dt}{1\;+\;2t\;\;-\;t^2}\;\;=\;\int\frac{2dt}{{\displaystyle\left(\sqrt2\right)^2}{\displaystyle-}{\displaystyle{\displaystyle\left(t-1\right)}^2}}\\=\;2\;\log\left|\frac{\displaystyle\sqrt2+t-1}{{\displaystyle\sqrt2}{\displaystyle-}t+1}\right|\;+\;c\;=\;2\;\log\;\left|\frac{\sqrt2{\displaystyle-}{\displaystyle1}{\displaystyle+}{\displaystyle\;}{\displaystyle\tan}{\displaystyle\frac x2}}{\sqrt2{\displaystyle+}1-\;\tan\frac x2}\right|\;+\;c\\\\\end{array}{/tex}