# find-dx-a-cosx-with-a-gt-0-

Question Number 68868 by mathmax by abdo last updated on 16/Sep/19
$${find}\:\int\:\:\:\:\frac{{dx}}{{a}+{cosx}}\:\:{with}\:{a}>\mathrm{0} \\$$
Commented by mathmax by abdo last updated on 17/Sep/19
$${let}\:{I}\:=\:\int\:\:\frac{{dx}}{{a}+{cosx}}\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\$$$${I}\:=\int\:\:\:\frac{\mathrm{1}}{{a}+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\int\:\:\:\:\:\frac{\mathrm{2}{dt}}{{a}+{at}^{\mathrm{2}} +\mathrm{1}−{t}^{\mathrm{2}} }\:=\int\:\:\frac{\mathrm{2}{dt}}{\left({a}−\mathrm{1}\right){t}^{\mathrm{2}} \:+{a}+\mathrm{1}} \\$$$$=\frac{\mathrm{2}}{{a}−\mathrm{1}}\:\int\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}}\:\:\:\:{case}\:\mathrm{1}\:\:\:\:\mathrm{0}<{a}<\mathrm{1}\:\Rightarrow{I}\:=\frac{\mathrm{2}}{{a}−\mathrm{1}}\int\:\:\frac{{dt}}{{t}^{\mathrm{2}} −\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}} \\$$$$=\frac{\mathrm{2}}{{a}−\mathrm{1}}\:\int\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} −\left(\sqrt{\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}}\right)^{\mathrm{2}} }\:=_{{t}=\sqrt{\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}}{u}} \:\:\:\:\frac{\mathrm{2}}{{a}−\mathrm{1}}\frac{\mathrm{1}−{a}}{\mathrm{1}+{a}}\:\int\:\:\:\frac{\mathrm{1}}{{u}^{\mathrm{2}} −\mathrm{1}}\sqrt{\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}}{du} \\$$$$=\frac{\mathrm{2}}{\mathrm{1}+{a}}×\frac{\sqrt{\mathrm{1}+{a}}}{\:\sqrt{\mathrm{1}−{a}}}\:\int\left(\frac{\mathrm{1}}{{u}−\mathrm{1}}−\frac{\mathrm{1}}{{u}+\mathrm{1}}\right){du}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\:{ln}\mid\frac{{u}−\mathrm{1}}{{u}+\mathrm{1}}\mid\:\:+{c} \\$$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\:{ln}\mid\frac{\sqrt{\frac{\mathrm{1}−{a}}{\mathrm{1}+{a}\:}}{tan}\left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}−{a}}{\mathrm{1}+{a}}}{tan}\left(\frac{{x}}{\mathrm{2}}\right)\:+\mathrm{1}}\mid\:+{c} \\$$$${case}\:\mathrm{2}\:\:\:{a}>\mathrm{1}\:\Rightarrow\:{I}\:=_{{t}=\sqrt{\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}}{u}} \:\:\:\frac{\mathrm{2}}{{a}−\mathrm{1}}\frac{{a}−\mathrm{1}}{{a}+\mathrm{1}}\:\int\:\:\:\frac{\mathrm{1}}{{u}^{\mathrm{2}} +\mathrm{1}}\sqrt{\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}}{du} \\$$$$=\frac{\mathrm{2}}{\:\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}\:{arctan}\left({u}\right)+{c}\:=\frac{\mathrm{2}}{\:\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}\:{arctan}\left(\sqrt{\frac{{a}−\mathrm{1}}{{a}+\mathrm{1}}}{t}\right)\:+{c}\:. \\$$