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Question Number 68868 by mathmax by abdo last updated on 16/Sep/19

find ∫ (dx/(a+cosx)) with a>0

$${find}\:\int\:\:\:\:\frac{{dx}}{{a}+{cosx}}\:\:{with}\:{a}>\mathrm{0} \\ $$

Commented bymathmax by abdo last updated on 17/Sep/19

let I = ∫ (dx/(a+cosx)) changement tan((x/2))=t give I =∫ (1/(a+((1−t^2 )/(1+t^2 ))))((2dt)/(1+t^2 )) =∫ ((2dt)/(a+at^2 +1−t^2 )) =∫ ((2dt)/((a−1)t^2 +a+1)) =(2/(a−1)) ∫ (dt/(t^2 +((a+1)/(a−1)))) case 1 0<a<1 ⇒I =(2/(a−1))∫ (dt/(t^2 −((1+a)/(1−a)))) =(2/(a−1)) ∫ (dt/(t^2 −((√((1+a)/(1−a))))^2 )) =_(t=(√((1+a)/(1−a)))u) (2/(a−1))((1−a)/(1+a)) ∫ (1/(u^2 −1))(√((1+a)/(1−a)))du =(2/(1+a))×((√(1+a))/(√(1−a))) ∫((1/(u−1))−(1/(u+1)))du =(1/(√(1−a^2 ))) ln∣((u−1)/(u+1))∣ +c =(1/(√(1−a^2 ))) ln∣(((√((1−a)/(1+a )))tan((x/2))−1)/((√((1−a)/(1+a)))tan((x/2)) +1))∣ +c case 2 a>1 ⇒ I =_(t=(√((a+1)/(a−1)))u) (2/(a−1))((a−1)/(a+1)) ∫ (1/(u^2 +1))(√((a+1)/(a−1)))du =(2/(√(a^2 −1))) arctan(u)+c =(2/(√(a^2 −1))) arctan((√((a−1)/(a+1)))t) +c .

$${let}\:{I}\:=\:\int\:\:\frac{{dx}}{{a}+{cosx}}\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$ $${I}\:=\int\:\:\:\frac{\mathrm{1}}{{a}+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\int\:\:\:\:\:\frac{\mathrm{2}{dt}}{{a}+{at}^{\mathrm{2}} +\mathrm{1}−{t}^{\mathrm{2}} }\:=\int\:\:\frac{\mathrm{2}{dt}}{\left({a}−\mathrm{1}\right){t}^{\mathrm{2}} \:+{a}+\mathrm{1}} \\ $$ $$=\frac{\mathrm{2}}{{a}−\mathrm{1}}\:\int\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}}\:\:\:\:{case}\:\mathrm{1}\:\:\:\:\mathrm{0}<{a}<\mathrm{1}\:\Rightarrow{I}\:=\frac{\mathrm{2}}{{a}−\mathrm{1}}\int\:\:\frac{{dt}}{{t}^{\mathrm{2}} −\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}} \\ $$ $$=\frac{\mathrm{2}}{{a}−\mathrm{1}}\:\int\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} −\left(\sqrt{\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}}\right)^{\mathrm{2}} }\:=_{{t}=\sqrt{\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}}{u}} \:\:\:\:\frac{\mathrm{2}}{{a}−\mathrm{1}}\frac{\mathrm{1}−{a}}{\mathrm{1}+{a}}\:\int\:\:\:\frac{\mathrm{1}}{{u}^{\mathrm{2}} −\mathrm{1}}\sqrt{\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}}{du} \\ $$ $$=\frac{\mathrm{2}}{\mathrm{1}+{a}}×\frac{\sqrt{\mathrm{1}+{a}}}{\sqrt{\mathrm{1}−{a}}}\:\int\left(\frac{\mathrm{1}}{{u}−\mathrm{1}}−\frac{\mathrm{1}}{{u}+\mathrm{1}}\right){du}\:=\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\:{ln}\mid\frac{{u}−\mathrm{1}}{{u}+\mathrm{1}}\mid\:\:+{c} \\ $$ $$=\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\:{ln}\mid\frac{\sqrt{\frac{\mathrm{1}−{a}}{\mathrm{1}+{a}\:}}{tan}\left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{1}}{\sqrt{\frac{\mathrm{1}−{a}}{\mathrm{1}+{a}}}{tan}\left(\frac{{x}}{\mathrm{2}}\right)\:+\mathrm{1}}\mid\:+{c} \\ $$ $${case}\:\mathrm{2}\:\:\:{a}>\mathrm{1}\:\Rightarrow\:{I}\:=_{{t}=\sqrt{\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}}{u}} \:\:\:\frac{\mathrm{2}}{{a}−\mathrm{1}}\frac{{a}−\mathrm{1}}{{a}+\mathrm{1}}\:\int\:\:\:\frac{\mathrm{1}}{{u}^{\mathrm{2}} +\mathrm{1}}\sqrt{\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}}{du} \\ $$ $$=\frac{\mathrm{2}}{\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}\:{arctan}\left({u}\right)+{c}\:=\frac{\mathrm{2}}{\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}\:{arctan}\left(\sqrt{\frac{{a}−\mathrm{1}}{{a}+\mathrm{1}}}{t}\right)\:+{c}\:. \\ $$

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