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Question Number 93098 by Rio Michael last updated on 10/May/20
find the general solution to ∫ (1/(a sin x + b cos x)) dx and ∫ (1/(a cos x − bsin x)) dx where a , b are constants.
$$\mathrm{find}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{to} \\ $$$$\:\int\:\frac{\mathrm{1}}{{a}\:\mathrm{sin}\:{x}\:+\:{b}\:\mathrm{cos}\:{x}}\:{dx}\:\:\mathrm{and}\:\int\:\frac{\mathrm{1}}{{a}\:\mathrm{cos}\:{x}\:−\:{b}\mathrm{sin}\:{x}}\:{dx} \\ $$$$\mathrm{where}\:{a}\:,\:{b}\:\mathrm{are}\:\mathrm{constants}. \\ $$$$ \\ $$
Commented by prakash jain last updated on 10/May/20
asin x+bcos x =(√(a^2 +b^2 ))((a/( (√(a^2 +b^2 ))))sin x+(b/( (√(a^2 +b^2 ))))cos x) (a/( (√(a^2 +b^2 ))))=cos α (b/( (√(a^2 +b^2 ))))=sin α (√(a^2 +b^2 ))=(√(a^2 +b^2 ))sin (α+x) (1/(asin x+bcos x))=(√(a^2 +b^2 ))cosec (x+α) now use cosec formula for integration. −−−−−− Same procedure can be use (1/(asin x−bcos x))
$${a}\mathrm{sin}\:{x}+{b}\mathrm{cos}\:{x} \\ $$$$=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\left(\frac{{a}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\mathrm{sin}\:{x}+\frac{{b}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\mathrm{cos}\:{x}\right) \\ $$$$\frac{{a}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}=\mathrm{cos}\:\alpha \\ $$$$\frac{{b}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}=\mathrm{sin}\:\alpha \\ $$$$\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\mathrm{sin}\:\left(\alpha+{x}\right) \\ $$$$\frac{\mathrm{1}}{{a}\mathrm{sin}\:{x}+{b}\mathrm{cos}\:{x}}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\mathrm{cosec}\:\left({x}+\alpha\right) \\ $$$$\mathrm{now}\:\mathrm{use}\:\mathrm{cosec}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{integration}. \\ $$$$−−−−−− \\ $$$$\mathrm{Same}\:\mathrm{procedure}\:\mathrm{can}\:\mathrm{be}\:\mathrm{use} \\ $$$$\frac{\mathrm{1}}{{a}\mathrm{sin}\:{x}−{b}\mathrm{cos}\:{x}} \\ $$
Commented by Rio Michael last updated on 10/May/20
thank you sir.
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by abdomathmax last updated on 11/May/20
I =∫ (dx/(acosx +bsinx)) we suppose a≠0 ⇒ I =(1/a)∫ (dx/(cosx +λ sinx)) =A_λ (λ=(b/a)) changement tan((x/2))=t give A_λ =(1/a)∫ ((2dt)/((1+t^2 )(((1−t^2 )/(1+t^2 )) +λ((2t)/(1+t^2 ))))) aA_λ =∫ ((2dt)/(1−t^2 +2λt )) =−∫ ((2dt)/(t^2 −2λt −1)) let solve t^2 −2λt −1 =0 Δ^′ =λ^2 +1 ⇒t_1 =λ+(√(1+λ^2 )) t_2 =λ−(√(1+λ^2 )) aA_λ =−∫ ((2dt)/((t−t_1 )(t−t_2 ))) =−(2/(2(√(1+λ^2 )))) ∫ ((1/(t−t_1 ))−(1/(t−t_2 )))dt =(1/( (√(1+λ^2 ))))ln∣((t−t_2 )/(t−t_1 ))∣ +C =(1/( (√(1+λ^2 ))))ln∣((t−λ +(√(1+λ^2 )))/(t−λ−(√(1+λ^2 ))))∣ +C =(1/( (√(1+(b^2 /a^2 )))))ln∣((t−(b/a)+(√(1+(b^2 /a^2 ))))/(t−(b/a)−(√(1+(b^2 /a^2 )))))∣ +C =((∣a∣)/( (√(a^2 +b^2 ))))ln∣((((at−b)/a)+((√(a^2 +b^2 ))/(∣a∣)))/(((at−b)/a)−((√(a^2 +b^2 ))/(∣a∣))))∣ +C if we take a>0 we get I =(1/( (√(a^2 +b^2 ))))ln∣((atan((x/2))−b+(√(a^2 +b^2 )))/(atan((x/2))−b−(√(a^2 +b^2 ))))∣ +C = =
$${I}\:=\int\:\:\frac{{dx}}{{acosx}\:+{bsinx}}\:{we}\:{suppose}\:{a}\neq\mathrm{0}\:\:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{1}}{{a}}\int\:\:\:\:\:\frac{{dx}}{{cosx}\:+\lambda\:{sinx}}\:\:={A}_{\lambda} \:\:\left(\lambda=\frac{{b}}{{a}}\right) \\ $$$${changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${A}_{\lambda} =\frac{\mathrm{1}}{{a}}\int\:\:\:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:+\lambda\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)} \\ $$$${aA}_{\lambda} =\int\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}−{t}^{\mathrm{2}} \:+\mathrm{2}\lambda{t}\:}\:=−\int\:\:\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} −\mathrm{2}\lambda{t}\:−\mathrm{1}} \\ $$$${let}\:{solve}\:{t}^{\mathrm{2}} −\mathrm{2}\lambda{t}\:−\mathrm{1}\:=\mathrm{0} \\ $$$$\Delta^{'} \:=\lambda^{\mathrm{2}} \:+\mathrm{1}\:\:\Rightarrow{t}_{\mathrm{1}} =\lambda+\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} } \\ $$$${t}_{\mathrm{2}} =\lambda−\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} } \\ $$$${aA}_{\lambda} =−\int\:\:\:\frac{\mathrm{2}{dt}}{\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)} \\ $$$$=−\frac{\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }}\:\int\:\:\left(\frac{\mathrm{1}}{{t}−{t}_{\mathrm{1}} }−\frac{\mathrm{1}}{{t}−{t}_{\mathrm{2}} }\right){dt} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }}{ln}\mid\frac{{t}−{t}_{\mathrm{2}} }{{t}−{t}_{\mathrm{1}} }\mid\:+{C} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }}{ln}\mid\frac{{t}−\lambda\:+\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }}{{t}−\lambda−\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }}\mid\:+{C} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}}{ln}\mid\frac{{t}−\frac{{b}}{{a}}+\sqrt{\mathrm{1}+\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}}{{t}−\frac{{b}}{{a}}−\sqrt{\mathrm{1}+\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}}\mid\:+{C} \\ $$$$=\frac{\mid{a}\mid}{\:\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}{ln}\mid\frac{\frac{{at}−{b}}{{a}}+\frac{\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}{\mid{a}\mid}}{\frac{{at}−{b}}{{a}}−\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{\mid{a}\mid}}\mid\:+{C} \\ $$$${if}\:{we}\:{take}\:{a}>\mathrm{0}\:{we}\:{get} \\ $$$${I}\:=\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{ln}\mid\frac{{atan}\left(\frac{{x}}{\mathrm{2}}\right)−{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{atan}\left(\frac{{x}}{\mathrm{2}}\right)−{b}−\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}\mid\:+{C} \\ $$$$= \\ $$$$= \\ $$
Commented by Rio Michael last updated on 11/May/20
wow sir spendid
$$\mathrm{wow}\:\mathrm{sir}\:\mathrm{spendid} \\ $$
Commented by mathmax by abdo last updated on 11/May/20
thankx
$${thankx} \\ $$

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