dx-tan-x-sin-x-

Question Number 113757 by bemath last updated on 15/Sep/20

$$\:\int\:\frac{{dx}}{\mathrm{tan}\:{x}−\mathrm{sin}\:{x}}\:?\: \\ $$

Answered by bemath last updated on 15/Sep/20

Answered by Dwaipayan Shikari last updated on 15/Sep/20

2∫(1/(((2t)/(1−t^2 ))−((2t)/(1+t^2 )))).(1/(1+t^2 ))dt tan(x/2)=t 2∫(1/(2t(((1+t^2 −1+t^2 )/((1−t^2 )(1+t^2 )))))).(1/(1+t^2 ))dt ∫((1−t^2 )/(2t^3 ))dt=−(1/(4t^2 ))−(1/2)logt+C=−(1/(4tan^2 (x/2)))−(1/2)log(tan(x/2))+C

$$\mathrm{2}\int\frac{\mathrm{1}}{\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }−\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}.\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{tan}\frac{{x}}{\mathrm{2}}={t}\:\:\: \\ $$$$\mathrm{2}\int\frac{\mathrm{1}}{\mathrm{2}{t}\left(\frac{\mathrm{1}+{t}^{\mathrm{2}} −\mathrm{1}+{t}^{\mathrm{2}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\right)}.\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\int\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}{t}^{\mathrm{3}} }{dt}=−\frac{\mathrm{1}}{\mathrm{4}{t}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}}{logt}+{C}=−\frac{\mathrm{1}}{\mathrm{4}{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{2}}{log}\left({tan}\frac{{x}}{\mathrm{2}}\right)+{C} \\ $$

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