integration of dx / (tanx + cotx+ sec x+ cosec x) - Maths - Inverse Trigonometric Functions

Question

integration of dx / (tanx + cotx+ sec x+ cosec x)

Poonam Mittal · Accepted Answer

$\int \frac{1}{\tan (x) + \cot (x) + \sec (x) + \csc (x)} ⅆ x = \int \frac{1}{\frac{\sin (x)}{\cos (x)} + \frac{\cos (x)}{\sin (x)} + \frac{1}{\cos (x)} + \frac{1}{\sin (x)}} ⅆ x = \int \frac{\sin (x) \cos (x)}{\sin^{2} (x) + \cos^{2} (x) + \sin (x) + \cos (x)} ⅆ x = \int \frac{\sin (x) \cos (x)}{1 + \sin (x) + \cos (x)} ⅆ x = \int \frac{2 \sin (\frac{x}{2}) \cos (\frac{x}{2}) \cos (x)}{2 \cos^{2} (\frac{x}{2}) + 2 \sin (\frac{x}{2}) \cos (\frac{x}{2})} ⅆ x = \int \frac{\sin (\frac{x}{2}) \cos (x)}{\cos (\frac{x}{2}) + \sin (\frac{x}{2})} ⅆ x = \int \frac{\sin (\frac{x}{2}) (\cos^{2} (\frac{x}{2}) - \sin^{2} (\frac{x}{2}))}{\cos (\frac{x}{2}) + \sin (\frac{x}{2})} ⅆ x = \int \sin (\frac{x}{2}) (\cos (\frac{x}{2}) - \sin (\frac{x}{2})) ⅆ x$

$= \int \sin (\frac{x}{2}) \cos (\frac{x}{2}) ⅆ x - \int \sin^{2} (\frac{x}{2}) ⅆ x = \frac{1}{2} \int 2 \sin (\frac{x}{2}) \cos (\frac{x}{2}) ⅆ x - \int \frac{1 - \cos (x)}{2} ⅆ x = \frac{1}{2} \int \sin (x) ⅆ x - \int \frac{1}{2} ⅆ x + \int \frac{\cos (x)}{2} ⅆ x = \frac{- \cos (x)}{2} - \frac{x}{2} + \frac{\sin (x)}{2} = \frac{\sin (x) - \cos (x) - x}{2}$