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Question Number 86428 by jagoll last updated on 28/Mar/20

∫ (dx/(a cos x + b sin x))?

$$\int\:\:\frac{\mathrm{dx}}{\mathrm{a}\:\mathrm{cos}\:\mathrm{x}\:+\:\mathrm{b}\:\mathrm{sin}\:\mathrm{x}}? \\ $$

Commented by jagoll last updated on 28/Mar/20

standard solving a cos x + b sin x = k cos (x−θ) ∫ (dx/(k cos (x−θ))) = (1/k)∫ sec (x−θ) dx = (1/k) ln ∣ sec (x−θ)+tan (x−θ)∣ + c = (1/(√(a^2 +b^2 ))) ln ∣ sec (x−tan^(−1) ((b/a))+ tan (x−tan^(−1) ((b/a)))∣ +c

$$\mathrm{standard}\:\mathrm{solving} \\ $$$$\mathrm{a}\:\mathrm{cos}\:\mathrm{x}\:+\:\mathrm{b}\:\mathrm{sin}\:\mathrm{x}\:=\:\mathrm{k}\:\mathrm{cos}\:\left(\mathrm{x}−\theta\right) \\ $$$$\int\:\frac{\mathrm{dx}}{\mathrm{k}\:\mathrm{cos}\:\left(\mathrm{x}−\theta\right)}\:=\:\frac{\mathrm{1}}{\mathrm{k}}\int\:\mathrm{sec}\:\left(\mathrm{x}−\theta\right)\:\mathrm{dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{k}}\:\mathrm{ln}\:\mid\:\mathrm{sec}\:\left(\mathrm{x}−\theta\right)+\mathrm{tan}\:\left(\mathrm{x}−\theta\right)\mid\:+\:\mathrm{c} \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }}\:\mathrm{ln}\:\mid\:\mathrm{sec}\:\left(\mathrm{x}−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{b}}{\mathrm{a}}\right)+\:\mathrm{tan}\:\left(\mathrm{x}−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{b}}{\mathrm{a}}\right)\right)\mid\:+\mathrm{c}\:\right. \\ $$

Commented by abdomathmax last updated on 28/Mar/20

I =∫ (dx/(acosx +bsinx)) ⇒ I=_(tan((x/2))=t) ∫ ((2dt)/((1+t^2 )(a((1−t^2 )/(1+t^2 )) +((2bt)/(1+t^2 ))))) =∫ ((2dt)/(a−at^2 +2bt)) =∫ ((−2dt)/(at^2 −2bt −a)) Δ^′ = b^2 +a^2 >0 ⇒t_1 =((b+(√(a^2 +b^2 )))/a) (we suppose a≠0) t_2 =((b−(√(a^2 +b^2 )))/a) ⇒I =−2∫ (dt/((t−t_1 )(t−t_2 ))) =((−2)/((2(√(a^2 +b^2 )))/a)) ∫ ((1/(t−t_1 ))−(1/(t−t_2 )))dt =−(a/(√(a^2 +b^2 )))ln∣((t−t_1 )/(t−t_2 ))∣ +C =−(a/(√(a^2 +b^2 )))ln∣((t−((b+(√(a^2 +b^2 )))/a))/(t−((b−(√(a^2 +b^2 )))/a)))∣ +C I=−(a/(√(a^2 +b^2 )))ln∣((at−b−(√(a^2 +b^2 )))/(at−b+(√(a^2 +b^2 ))))∣ +C

$${I}\:=\int\:\:\frac{{dx}}{{acosx}\:+{bsinx}}\:\Rightarrow \\ $$$${I}=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\:\:\:\:\int\:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left({a}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:+\frac{\mathrm{2}{bt}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)} \\ $$$$=\int\:\:\:\:\frac{\mathrm{2}{dt}}{{a}−{at}^{\mathrm{2}} +\mathrm{2}{bt}}\:=\int\:\:\frac{−\mathrm{2}{dt}}{{at}^{\mathrm{2}} −\mathrm{2}{bt}\:−{a}} \\ $$$$\Delta^{'} =\:{b}^{\mathrm{2}} +{a}^{\mathrm{2}} \:>\mathrm{0}\:\Rightarrow{t}_{\mathrm{1}} =\frac{{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{a}}\:\:\left({we}\:{suppose}\:{a}\neq\mathrm{0}\right) \\ $$$${t}_{\mathrm{2}} =\frac{{b}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{a}}\:\Rightarrow{I}\:=−\mathrm{2}\int\:\:\frac{{dt}}{\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)} \\ $$$$=\frac{−\mathrm{2}}{\frac{\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{a}}}\:\int\:\:\left(\frac{\mathrm{1}}{{t}−{t}_{\mathrm{1}} }−\frac{\mathrm{1}}{{t}−{t}_{\mathrm{2}} }\right){dt} \\ $$$$=−\frac{{a}}{\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}{ln}\mid\frac{{t}−{t}_{\mathrm{1}} }{{t}−{t}_{\mathrm{2}} }\mid\:+{C} \\ $$$$=−\frac{{a}}{\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}{ln}\mid\frac{{t}−\frac{{b}+\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}{{a}}}{{t}−\frac{{b}−\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}{{a}}}\mid\:+{C} \\ $$$${I}=−\frac{{a}}{\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}{ln}\mid\frac{{at}−{b}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{at}−{b}+\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}\mid\:+{C} \\ $$

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