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Question Number 59998 by Mr X pcx last updated on 16/May/19

find ∫ (dx/(acosx +bsinx)) with a and b reals

$${find}\:\int\:\frac{{dx}}{{acosx}\:+{bsinx}}\:\:{with}\:{a}\:{and}\:{b}\:{reals} \\ $$

Commented by mr W last updated on 16/May/19

a cos x+b sin x=(√(a^2 +b^2 ))((a/(√(a^2 +b^2 ))) cos x+(b/(√(a^2 +b^2 ))) sin x) =(√(a^2 +b^2 ))(sin θ cos x+cos θ sin x) =(√(a^2 +b^2 )) sin (x+θ) with θ=tan^(−1) (a/b) ∫ (dx/(acosx +bsinx)) =(1/(√(a^2 +b^2 )))∫ (dx/(sin(x+θ))) =(1/(2(√(a^2 +b^2 ))))×ln ((1−cos (x+θ))/(1+cos (x+θ)))+C

$${a}\:\mathrm{cos}\:{x}+{b}\:\mathrm{sin}\:{x}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\left(\frac{{a}}{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\:\mathrm{cos}\:{x}+\frac{{b}}{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\:\mathrm{sin}\:{x}\right) \\ $$$$=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\left(\mathrm{sin}\:\theta\:\mathrm{cos}\:{x}+\mathrm{cos}\:\theta\:\mathrm{sin}\:{x}\right) \\ $$$$=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\mathrm{sin}\:\left({x}+\theta\right) \\ $$$${with}\:\theta=\mathrm{tan}^{−\mathrm{1}} \frac{{a}}{{b}} \\ $$$$\int\:\frac{{dx}}{{acosx}\:+{bsinx}} \\ $$$$=\frac{\mathrm{1}}{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\int\:\frac{{dx}}{{sin}\left({x}+\theta\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}×\mathrm{ln}\:\frac{\mathrm{1}−\mathrm{cos}\:\left({x}+\theta\right)}{\mathrm{1}+\mathrm{cos}\:\left({x}+\theta\right)}+{C} \\ $$

Commented by maxmathsup by imad last updated on 17/May/19

thanks you sir .

$${thanks}\:{you}\:{sir}\:. \\ $$

Commented by maxmathsup by imad last updated on 17/May/19

let I =∫ (dx/(acosx +bsinx)) changement tan((x/2))=t give I =∫ (1/(a ((1−t^2 )/(1+t^2 )) +b((2t)/(1+t^2 )))) ((2dt)/(1+t^2 )) = ∫ ((2dt)/(a−at^2 +2bt)) =∫ ((−2dt)/(at^2 −2bt−a)) =∫ F(t)dt roots of at^2 −2bt −a =0 →Δ^′ =b^2 +a^2 for a≠0 ⇒Δ^′ >0 ⇒ t_1 =((b+(√(a^2 +b^2 )))/a) and t_2 =((b−(√(a^2 +b^2 )))/a) ⇒F(t) =((−2)/(a(t−t_1 )(t−t_2 ))) =((−2)/(a(t_1 −t_2 ))){ (1/(t−t_1 )) −(1/(t−t_2 ))} =((−2)/(2(√(a^2 +b^2 )))){(1/(t−t_1 )) −(1/(t−t_2 ))} ⇒ I =(1/(√(a^2 +b^2 )))ln∣((t−t_1 )/(t−t_2 ))∣ +C =(1/(√(a^2 +b^2 )))ln∣((tan((x/2))−((b+(√(a^2 +b^2 )))/a))/(tan((x/2))−((b−(√(a^2 +b^2 )))/a)))∣ +C =(1/(√(a^2 +b^2 )))ln∣((atan((x/2))−b−(√(a^2 +b^2 )))/(atan((x/2))−b +(√(a^2 +b^2 ))))∣ +C .

$${let}\:{I}\:=\int\:\:\frac{{dx}}{{acosx}\:+{bsinx}}\:\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:\:{give} \\ $$$${I}\:=\int\:\:\frac{\mathrm{1}}{{a}\:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:+{b}\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\:\int\:\:\frac{\mathrm{2}{dt}}{{a}−{at}^{\mathrm{2}} \:+\mathrm{2}{bt}}\:=\int\:\frac{−\mathrm{2}{dt}}{{at}^{\mathrm{2}} −\mathrm{2}{bt}−{a}}\:=\int\:{F}\left({t}\right){dt} \\ $$$${roots}\:{of}\:{at}^{\mathrm{2}} −\mathrm{2}{bt}\:−{a}\:=\mathrm{0}\:\rightarrow\Delta^{'} ={b}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \:\:{for}\:{a}\neq\mathrm{0}\:\:\Rightarrow\Delta^{'} >\mathrm{0}\:\Rightarrow \\ $$$${t}_{\mathrm{1}} =\frac{{b}+\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}{{a}}\:\:{and}\:{t}_{\mathrm{2}} =\frac{{b}−\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}{{a}}\:\Rightarrow{F}\left({t}\right)\:=\frac{−\mathrm{2}}{{a}\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)} \\ $$$$=\frac{−\mathrm{2}}{{a}\left({t}_{\mathrm{1}} −{t}_{\mathrm{2}} \right)}\left\{\:\frac{\mathrm{1}}{{t}−{t}_{\mathrm{1}} }\:−\frac{\mathrm{1}}{{t}−{t}_{\mathrm{2}} }\right\}\:=\frac{−\mathrm{2}}{\mathrm{2}\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}\left\{\frac{\mathrm{1}}{{t}−{t}_{\mathrm{1}} }\:−\frac{\mathrm{1}}{{t}−{t}_{\mathrm{2}} }\right\}\:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{1}}{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{ln}\mid\frac{{t}−{t}_{\mathrm{1}} }{{t}−{t}_{\mathrm{2}} }\mid\:+{C}\:=\frac{\mathrm{1}}{\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}{ln}\mid\frac{{tan}\left(\frac{{x}}{\mathrm{2}}\right)−\frac{{b}+\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}{{a}}}{{tan}\left(\frac{{x}}{\mathrm{2}}\right)−\frac{{b}−\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}{{a}}}\mid\:+{C} \\ $$$$=\frac{\mathrm{1}}{\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}{ln}\mid\frac{{atan}\left(\frac{{x}}{\mathrm{2}}\right)−{b}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{atan}\left(\frac{{x}}{\mathrm{2}}\right)−{b}\:+\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}\mid\:+{C}\:. \\ $$

Answered by kaivan.ahmadi last updated on 16/May/19

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