∫ d x tan x + cot x + sec x + cosec x =

Let,

$I=\int \frac{dx}{\tan x+\cot x+\sec x+cosecx}$

  $=\int \frac{dx}{\left({\displaystyle \frac{\sin x}{\cos x}}+{\displaystyle \frac{\cos x}{\sin x}}+{\displaystyle \frac{1}{\cos x}}+{\displaystyle \frac{1}{\sin x}}\right)}$

   $=\int \left\{\frac{\sin x\cos x}{{\displaystyle {\sin }^{2}x+{\cos }^{2}x+\sin x+\cos x}}\right\}dx$

  $=\int \left\{\frac{\sin x\cos x}{{\displaystyle 1+\sin x+\cos x}}\right\}dx$

  $=\int \left\{\frac{\sin x\cos x}{{\displaystyle 1+\sin x+\cos x}}\times \left(\frac{\sin x+\cos x-1}{\sin x+\cos x-1}\right)\right\}dx$

$=\int \left\{\frac{\sin x\cos x\left(\sin x+\cos x-1\right)}{{\displaystyle {\left(\sin x+\cos x\right)}^2-1}}\right\}dx$

 $=\int \left\{\frac{\sin x\cos x\left(\sin x+\cos x-1\right)}{{\displaystyle {\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x-1}}\right\}dx$

 $=\int \left\{\frac{\sin x\cos x\left(\sin x+\cos x-1\right)}{{\displaystyle 2\sin x\cos x}}\right\}dx$

  $=\frac{1}{2}\int \left(\sin x+\cos x-1\right)dx$

 $=\frac{1}{2}\left(-\cos x+\sin x-x\right)+c$