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

$$\mathcal{J}=\int \frac{dx}{1+\tan x+\csc x+\cot x+\sec x}$$

Answered by john_santu last updated on 27/Feb/21

$$\mathcal{J}=\int \frac{dx}{1+\frac{\sin x}{\cos x}+\frac{1}{\sin x}+\frac{\cos x}{\sin x}+\frac{1}{\cos x}}$$$$=\int \frac{\cos x\sin x}{\sin x\cos x+\sin {}^{2}x+\cos {}^{2}x+\sin x}$$$$=\int \frac{\sin x\cos x(1-\sin x)(1-\cos x)}{(\sin x\cos x+1+\sin x)(1-\sin x)(1-\cos x)}dx$$$$=\int \frac{\sin x\cos x(1-\sin x)(1-\cos x)dx}{(1+\sin x+\sin x\cos x)(1-\sin x)(1-\cos x)}$$$$=\int \frac{\sin x\cos x(1-\sin x)(1-\cos x)dx}{(1+\sin x+\sin x\cos x)(1-\sin x-\cos x+\sin x\cos x)}$$$$=\int \frac{\sin x\cos x(1-\sin x)(1-\cos x)}{\sin {}^{2}x\cos {}^{2}x}dx$$$$=\int \frac{(1-\sin x)(1-\cos x)}{\sin x\cos x}dx$$$$=\int (1-\sec x-\csc x+\sec x\csc x)dx$$$$=2\int \csc 2xdx-\ln \mid \sec x+\tan x\mid +x+\ln \mid \cot x+\mathrm{cs}cx\mid$$$$=-\ln \mid \cot 2x+\csc 2x\mid -\ln \mid \sec x+\tan x\mid +$$$$\ln \mid \cot x+\csc x\mid +x+c$$

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