Question Number 164612 by bounhome last updated on 19/Jan/22
$${solve}: \\ $$$$\:\mathrm{1}.\:\int\frac{\mathrm{1}}{{sinx}}{dx} \\ $$$$\:\mathrm{2}.\int\frac{\mathrm{1}}{{cosx}}{dx} \\ $$
Answered by Ar Brandon last updated on 19/Jan/22
$$\int\frac{\mathrm{1}}{\mathrm{sin}{x}}{dx},\:{x}=\mathrm{2}{t}\Rightarrow{dx}=\mathrm{2}{dt} \\ $$$$=\int\frac{\mathrm{2}{dt}}{\mathrm{sin2}{t}}=\int\frac{\mathrm{1}}{\mathrm{sin}{t}\mathrm{cos}{t}}{dt}=\int\frac{\mathrm{cos}^{\mathrm{2}} {t}+\mathrm{sin}^{\mathrm{2}} {t}}{\mathrm{sin}{t}\mathrm{cos}{t}}{dt} \\ $$$$=\int\left(\frac{\mathrm{cos}{t}}{\mathrm{sin}{t}}+\frac{\mathrm{sin}{t}}{\mathrm{cos}{t}}\right){dt}=\mathrm{ln}\left(\mathrm{sin}{t}\right)−\mathrm{ln}\left(\mathrm{cos}{t}\right)+{C} \\ $$$$=\mathrm{ln}\mid\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}\right)\mid+{C} \\ $$
Answered by Ar Brandon last updated on 19/Jan/22
$$\int\frac{\mathrm{1}}{\mathrm{cos}{x}}{dx}=\:{Q}\mathrm{163829} \\ $$
Answered by Eulerian last updated on 20/Jan/22
$$\: \\ $$$$\:\boldsymbol{\mathrm{Solution}}: \\ $$$$\:\mathrm{1}.\:\int\:\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\:\mathrm{dx}\:=\:\int\:\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\:=\:−\mathrm{tanh}^{−\mathrm{1}} \left(\mathrm{cos}\:\mathrm{x}\right)\:+\:\mathrm{C} \\ $$$$\: \\ $$$$\:\mathrm{2}.\:\int\:\frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\:\mathrm{dx}\:=\:\int\:\frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\:\mathrm{dx}\:=\:\mathrm{tanh}^{−\mathrm{1}} \left(\mathrm{sin}\:\mathrm{x}\right)\:+\:\mathrm{C} \\ $$$$\: \\ $$