Question Number 187526 by sciencestudentW last updated on 18/Feb/23
$$\int\frac{{dx}}{{cscx}−\mathrm{1}}=? \\ $$
Answered by ARUNG_Brandon_MBU last updated on 18/Feb/23
$${I}=\int\frac{{dx}}{\mathrm{cosec}{x}−\mathrm{1}} \\ $$$$\:\:=\int\frac{\mathrm{sin}{x}}{\mathrm{1}−\mathrm{sin}{x}}{dx}=\int\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{sin}{x}}−\mathrm{1}\right){dx} \\ $$$$\:\:=\int\frac{\mathrm{1}+\mathrm{sin}{x}}{\mathrm{cos}^{\mathrm{2}} {x}}{dx}−{x}=\mathrm{tan}{x}+\frac{\mathrm{1}}{\mathrm{cos}{x}}−{x}+{C} \\ $$
Answered by floor(10²Eta[1]) last updated on 18/Feb/23
$$ \\ $$$$\int\frac{\mathrm{senx}}{\mathrm{1}−\mathrm{senx}}\mathrm{dx} \\ $$$$\mathrm{u}=\mathrm{tg}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\Rightarrow\frac{\mathrm{2}}{\mathrm{u}^{\mathrm{2}} +\mathrm{1}}\mathrm{du}=\mathrm{dx} \\ $$$$\int\frac{\mathrm{4u}}{\left(\mathrm{u}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{du}=\int\frac{\mathrm{2}}{\left(\mathrm{u}−\mathrm{1}\right)^{\mathrm{2}} }\mathrm{du}−\int\frac{\mathrm{2}}{\mathrm{u}^{\mathrm{2}} +\mathrm{1}}\mathrm{du} \\ $$$$=−\frac{\mathrm{2}}{\mathrm{u}−\mathrm{1}}−\mathrm{2arctg}\left(\mathrm{u}\right)+\mathrm{C} \\ $$$$=\frac{\mathrm{2}}{\mathrm{1}−\mathrm{tg}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)}−\mathrm{x}+\mathrm{C} \\ $$$$ \\ $$