Question Number 33009 by kyle_TW last updated on 09/Apr/18
$${help}\:!\:!\:! \\ $$$$\int\:\frac{{dx}}{{csc}\left({x}\right)−\mathrm{1}}\:=\:? \\ $$$$ \\ $$$$\left[\:{my}\:{way}\:\right] \\ $$$$\int\left(\:\frac{{dx}}{\frac{\mathrm{1}}{{sinx}}\:−\:\mathrm{1}}\:\right) \\ $$$$=\int\frac{{sinx}}{\mathrm{1}−{sinx}}\:{dx} \\ $$$$=−\int\:\frac{{sinx}−\mathrm{1}+\mathrm{1}}{{sinx}−\mathrm{1}}\:{dx} \\ $$$$=−\int\mathrm{1}+\frac{\mathrm{1}}{{sinx}−\mathrm{1}}\:{dx} \\ $$$$=−\left(\int\mathrm{1}{dx}+\int\frac{{sinx}+\mathrm{1}}{\left({sinx}−\mathrm{1}\right)\left({sinx}+\mathrm{1}\right)}\:{dx}\right) \\ $$$$=−\left({x}+{C}−\int\frac{{sinx}+\mathrm{1}}{\mathrm{1}−{sin}^{\mathrm{2}} {x}}\:{dx}\right) \\ $$$$=−\left({x}+{C}−\int\:\frac{{sinx}}{{cos}^{\mathrm{2}} {x}}\:{dx}−\int\:\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}\:{dx}\right) \\ $$$$=−\left({x}+{C}+\int\left({cosx}\right)^{−\mathrm{2}} {dcosx}−\int\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}{dx}\right) \\ $$$$=−\left({x}−\left({cosx}\right)^{−\mathrm{1}} +{C}−\int\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}{dx}\right) \\ $$$$…{and}\:{I}\:{can}'{t}\:{solve}\:{the}\:\int\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$ \\ $$$${oh}\:{i}\:{just}\:{found}\:{that}\:{is}\:{tanx}+{C} \\ $$
Commented by prof Abdo imad last updated on 09/Apr/18
$${let}\:{put}\:\:{I}\:=\:\int\:\:\:\frac{{dx}}{\frac{\mathrm{1}}{{sinx}}\:−\mathrm{1}} \\ $$$${I}\:=\:\int\:\:\:\:\:\:\frac{{sinx}}{\mathrm{1}−{sinx}}{dx}\: \\ $$$$=−\int\:\:\frac{\mathrm{1}−{sinx}\:−\mathrm{1}}{\mathrm{1}−{sinx}}{dx}\:\:=\:−{x}+\int\:\:\frac{{dx}}{\mathrm{1}−{sinx}} \\ $$$${the}\:{ch}.{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$$\int\:\:\:\:\frac{{dx}}{\mathrm{1}−{sinx}}\:=\:\int\:\:\frac{\:\mathrm{1}}{\mathrm{1}\:−\:\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\int\:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} −\mathrm{2}{t}}\:=\:\mathrm{2}\:\int\:\:\frac{{dt}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }\:=\:\frac{−\mathrm{2}}{{t}−\mathrm{1}}\:\:=\:\frac{\mathrm{2}}{\mathrm{1}−{t}} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{1}−{tan}\left(\frac{{x}}{\mathrm{2}}\right)}\:\Rightarrow \\ $$$${I}\:=\:\:−{x}\:\:\:+\:\:\:\frac{\mathrm{2}}{\mathrm{1}\:−{tan}\left(\frac{{x}}{\mathrm{2}}\right)}\:+\lambda\:. \\ $$
Answered by Joel578 last updated on 09/Apr/18
$${I}\:=\:\int\:\frac{{dx}}{\left(\frac{\mathrm{1}}{\mathrm{sin}\:{x}}\:−\:\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:{x}}\right)}\:\:=\:\int\:\frac{\mathrm{sin}\:{x}}{\mathrm{1}\:−\:\mathrm{sin}\:{x}}\:{dx} \\ $$$$\:\:\:=\:\int\:\left(\frac{\mathrm{sin}\:{x}}{\mathrm{1}\:−\:\mathrm{sin}\:{x}}\right)\left(\frac{\mathrm{1}\:+\:\mathrm{sin}\:{x}}{\mathrm{1}\:+\:\mathrm{sin}\:{x}}\right)\:{dx} \\ $$$$\:\:\:=\:\int\:\frac{\mathrm{sin}\:{x}\:+\:\mathrm{sin}^{\mathrm{2}} \:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}}\:{dx} \\ $$$$\:\:\:=\:\int\:\mathrm{tan}\:{x}\:\mathrm{sec}\:{x}\:{dx}\:+\:\int\:\mathrm{tan}^{\mathrm{2}} \:{x}\:{dx} \\ $$$$\:\:\:=\:\int\:\mathrm{tan}\:{x}\:\mathrm{sec}\:{x}\:{dx}\:+\:\int\:\mathrm{sec}^{\mathrm{2}} \:{x}\:−\:\mathrm{1}\:{dx} \\ $$$$\:\:\:=\:\mathrm{sec}\:{x}\:+\:\mathrm{tan}\:{x}\:−\:{x}\:+\:{C} \\ $$
Commented by kyle_TW last updated on 09/Apr/18
$${wow}\:{thank}\:{you}\:{very}\:{much} \\ $$
Commented by kyle_TW last updated on 09/Apr/18
$${I}\:{forgot}\:{secx}\:=\:\frac{\mathrm{1}}{{cosx}}… \\ $$
Answered by math1967 last updated on 09/Apr/18
$$\int\frac{\mathrm{1}}{{cosex}−\mathrm{1}}{dx} \\ $$$$\int\frac{{cosex}+\mathrm{1}}{{cot}^{\mathrm{2}} {x}}{dx} \\ $$$$\int\frac{{cosecx}}{{cot}^{\mathrm{2}} {x}}{dx}\:+\int{tan}^{\mathrm{2}} {xdx} \\ $$$$\int\frac{{sinx}}{{cos}^{\mathrm{2}} {x}}{dx}\:+\int{sec}^{\mathrm{2}} {x}\:−\int{dx} \\ $$$$−\int\frac{{d}\left({cosx}\right)}{{cos}^{\mathrm{2}} {x}}\:+{tanx}\:−{x} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{1}}{{cos}^{\mathrm{3}} {x}}\:+{tanx}\:−{x}\:+{C} \\ $$