J-dx-1-tan-x-csc-x-cot-x-sec-x-

Question Number 134058 by john_santu last updated on 27/Feb/21

J = \int \frac{d x}{1 + \tan x + \csc x + \cot x + \sec x}

Answered by john_santu last updated on 27/Feb/21

J = \int \frac{d x}{1 + \frac{\sin x}{\cos x} + \frac{1}{\sin x} + \frac{\cos x}{\sin x} + \frac{1}{\cos x}}

= \int \frac{\cos x \sin x}{\sin x \cos x + \sin^{2} x + \cos^{2} x + \sin x}

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

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

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

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

= \int \frac{(1 - \sin x) (1 - \cos x)}{\sin x \cos x} d x

= \int (1 - \sec x - \csc x + \sec x \csc x) d x

= 2 \int \csc 2 x d x - \ln ∣ \sec x + \tan x ∣ + x + \ln ∣ \cot x + cs c x ∣

= - \ln ∣ \cot 2 x + \csc 2 x ∣ - \ln ∣ \sec x + \tan x ∣ +

\ln ∣ \cot x + \csc x ∣ + x + c