Question Number 169250 by mathocean1 last updated on 26/Apr/22
$${Calculate}\:{B}=\int\frac{\mathrm{1}}{\mathrm{1}+{sin}\left({x}\right)}{dx}\:{using}\:{u}={tan}\left(\frac{{x}}{\mathrm{2}}\right) \\ $$
Commented by infinityaction last updated on 27/Apr/22
$$\:\:\:{I}\:=\:\int\frac{{dx}}{\left(\mathrm{sin}\:\frac{{x}}{\mathrm{2}}\:+\:\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\:\:{I}\:=\:\int\frac{\mathrm{sec}\:^{\mathrm{2}} {x}/\mathrm{2}}{\left(\mathrm{1}+\mathrm{tan}\:{x}/\mathrm{2}\right)^{\mathrm{2}} }{dx} \\ $$$$\:\:\:{u}\:=\:\mathrm{tan}\:{x}/\mathrm{2} \\ $$$$\:\:\:{du}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}\:^{\mathrm{2}} {x}/\mathrm{2} \\ $$$$\:\:\:{I}\:\:=\:\int\frac{\mathrm{2}{du}}{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} } \\ $$$$\:\:\:{I}\:\:\:=\:\:−\frac{\mathrm{2}}{\mathrm{1}+{u}}\:+\:{c} \\ $$$$\:\:\:\:{I}\:\:\:=\:−\frac{\mathrm{2}}{\mathrm{1}+\mathrm{tan}\:{x}/\mathrm{2}}\:+\:{c} \\ $$
Answered by MikeH last updated on 27/Apr/22
$$\mathrm{sin}\:{x}\:=\:\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${dx}\:=\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\Rightarrow\:{B}\:=\:\int\frac{\mathrm{1}}{\left(\mathrm{1}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)}.\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\:\int\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{1}} \\ $$$${B}\:=\:\mathrm{2}\int\frac{{dt}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }\:=\:−\frac{\mathrm{2}}{\mathrm{1}+{t}}\:+\:{k} \\ $$$${B}\:=\:−\frac{\mathrm{2}}{\mathrm{1}+\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}\right)}+{k} \\ $$