$\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$

Question

$$\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$$ $$\tan x + \cot x + \csc x + \sec x=\frac{\sin x + 1}{\cos x} +\frac{\cos x + 1}{\sin x} $$ $$= \frac{\sin x +\cos x +1}{\sin x \cos x}$$ $$t= \tan {\frac{x}{2}}$$ On solving , $$\frac{1}{\tan x + \cot x + \csc x + \sec x}=\frac{t(1- t)}{1+ t^2}$$ $$\implies \int \frac{\tan {\frac{x}{2}}(1-\tan {\frac{x}{2}})}{1+\tan^2 {\frac{x}{2}}}{dx}$$

Creo que he hecho las cosas más difíciles. ¿Cómo puedo seguir adelante? ¿Hay alguna sustitución mejor para él?

Answer 1

$\displaystyle\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}=$

$\displaystyle\int\frac{\sin x\cos x}{1+\sin x+\cos x}\,dx=\int\frac{\sin x\cos x}{1+\sin x+\cos x}\cdot\frac{1-(\sin x+\cos x)}{1-(\sin x+\cos x)}\,dx$

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

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